ECO607 — Midterm Summary (Lectures 1–22)

📘 Lecture 01 — INTRODUCTION TO DYNAMIC ANALYSIS

📖 Overview: This lecture introduces the concept of dynamic analysis in economics, contrasting it with static and comparative-static approaches. It establishes why incorporating time as a dimension is essential for realistic economic modeling and demonstrates the foundational mathematical concepts through a population example.

🗂️ Topics Covered

The lecture begins with a comparison of static, comparative-static, and dynamic analyses across six grounds: time dimension, equilibrium, transition, nature of model, real-life situation, and complexity. It then provides the rationale for dynamic analysis and real-world examples. The second topic demonstrates dynamic analysis through a population example, introducing time paths, differential equations, arbitrary constants, and the importance of initial conditions for obtaining a definite time path.

📝 Lecture Summary

TOPIC 001: NEED AND IMPORTANCE OF DYNAMIC ANALYSIS IN ECONOMICS

Pre-requisites: Basic Mathematical Economics

Types of Economic Analyses: Static, Comparative-Static & Dynamic

• Static (ساکن): Not moving (over time)
• Comparative-Static (نکاس ناتلبقام): Moving (due to exogenous variables)
• Dynamic (متحرک): Moving (over time)

COMPARISON

Rationale for Dynamic Analysis: Time is ‘unstoppable’ and affect variables of all disciplines including those of economics. Better to conduct analysis over time rather than one point in time only.

Examples of Dynamic Analysis in Economics:

1. Decline in global GDP due to Covid-19 during first half of 2020.
2. Rise in unemployment in Pakistan due to Covid-19 during first half of 2020.
3. Decline of fuel (Petrol) prices due to Covid-19 during first half of 2020.
4. Change of share value over time.
5. Daily activity in Pakistan Stock Exchange (PSX).

💡 Why this matters: Understanding the distinction between static and dynamic analysis is crucial because most real-world economic phenomena unfold over time. Dynamic analysis captures the process of change, not just the endpoints.

TOPIC 002: DYNAMIC ANALYSIS: EXAMPLE FROM POPULATION

Say, population size (H) changes over time at the rate: $$\frac{dH}{dt} = t^{-\frac{1}{2}}$$

What time path of population (H = H(t)) can yield the rate of change (\frac{dH}{dt} = t^{-\frac{1}{2}})? Here (H(t)) can be called time path as ‘population’ depends on ‘time’.

If (H(t) = 2t^{\frac{1}{2}}), then its derivative is: $$\frac{dH(t)}{dt} = 2\left(\frac{1}{2}\right)t^{\frac{1}{2} - 1} \Rightarrow \frac{dH}{dt} = t^{-\frac{1}{2}}$$

Interpretation: If rate of change of population over time is (\frac{dH}{dt} = t^{-\frac{1}{2}}), then the relationship (time path) between population and time is (H(t) = 2t^{\frac{1}{2}}).

However, the same (\frac{dH}{dt}) can be had with a large number of similar (H(t)):

• If (H(t) = 2t^{\frac{1}{2}} \pm 15) then (\frac{dH}{dt} = t^{-\frac{1}{2}})
• If (H(t) = 2t^{\frac{1}{2}} \pm 99) again (\frac{dH}{dt} = t^{-\frac{1}{2}})
• ... ... ... ... ... ...
• (H(t) = 2t^{\frac{1}{2}} \pm c) or (H(t) = 2t^{\frac{1}{2}} + c^*) again (\frac{dH}{dt} = t^{-\frac{1}{2}})

(H(t) = 2t^{\frac{1}{2}} \pm c) is time path of population, however, without determination of (c) the time path will remain ‘general’ in nature and not ‘definite’.

Therefore, we try to determine value of (c), which is possible. We need the starting point of the function for it – Initial condition.

🔑 Definition — Initial condition: Value of the dependent variable when independent variable is zero. Here dependent variable is (H) and independent variable is (t). Therefore, initial condition for function (H(t)) is (H(0)) or (H_o).

Now for the initial condition, general time path (H(t)) becomes: $$H(t) = 2t^{\frac{1}{2}} + c \Rightarrow H(0) = 2(0)^{\frac{1}{2}} + c \Rightarrow H(0) = c$$

Value of arbitrary constant (c) is (H(0)).

Plugging it back in ‘general’ time path: (H(t) = 2t^{\frac{1}{2}} + c)

We get ‘definite’ time path: $$\underbrace{H(t)}{\text{Definite Time Path}} = \underbrace{2t^{\frac{1}{2}}}{\text{Term involving time}} + \underbrace{H(0)}_{\text{Initial Population}}$$

Arbitrary constant determined is not arbitrary any more: (c = H(0)).

📌 Example: $$H(t) = 2t^{\frac{1}{2}} + H(0) \quad \text{where } c = H(0)$$

Note: Instead of mentioning (-c), (-ve) sign is merged into (c). (c) is used as general notation for all possible constant values. It’s better to call (c) as the ‘arbitrary constant’. As arbitrary is something which is based on ‘random choice’, rather than any reason. (c) is arbitrary as it can assume any value at random.

Multiple arbitrary constants can exist for which initial conditions should be available. Complete Map of the function – Solution of the dynamic model.

We can experiment some value of (H(0)). e.g. (H(0) = 100). $$H(t) = 2t^{\frac{1}{2}} + 100$$

⭐ Key Takeaways

The most critical distinction in this lecture is between static, comparative-static, and dynamic analysis: static ignores time completely, comparative-static examines different equilibrium points, while dynamic analysis studies the actual time path of adjustment. A dynamic model requires a differential equation (showing the rate of change) and an initial condition to obtain a definite time path. The arbitrary constant (c) represents the family of possible solutions, but it becomes determined once an initial condition like (H(0)) is specified. The population example demonstrates the fundamental process: starting from a rate of change ((dH/dt)), integrating to find a general time path, and using an initial condition to find the unique definite time path.

🧠 Quick Revision Questions

1. What are the three types of economic analyses discussed in the lecture? How does each treat the time dimension?
2. On what six grounds are static/comparative-static analysis compared with dynamic analysis? Give one key difference for each.
3. In the population example, if (\frac{dH}{dt} = t^{-\frac{1}{2}}), what is the general time path for (H(t))? Why does it contain an arbitrary constant?
4. What is an initial condition and how does it convert a general time path into a definite one? Use the population example to explain.
5. If the initial population (H(0) = 100) in the example, what is the definite time path equation?

📘 Lecture 2 — Integration and Rules

📖 Overview: This lecture introduces integral calculus as the inverse of differentiation, focusing on reconstructing a primitive function from its derivative. It covers the rationale for integration and establishes fundamental rules—constant function rule, constant multiple rule, power rule, and derivative over function rule—essential for solving basic integration problems in economics.

🗂️ Topics Covered

The lecture begins with the historical perspective and mathematical rationale for integration, positioning it as the anti-derivative process. It then systematically introduces four core rules of integration: the constant function rule (integrating a constant), the constant multiple rule (factoring out constants), the power rule (integrating x^n for n ≠ -1), ending with the derivative over function rule (integrating a ratio involving a derivative and its primitive).

📝 Lecture Summary

TOPIC 003: INTRODUCTION TO INTEGRAL CALCULUS: INDEFINITE INTEGRALS

This section establishes the fundamental concept of integration as the inverse operation of differentiation. It explains why integration is necessary in economics—specifically, to reconstruct a function from its known derivative, such as finding a total cost function from marginal cost. The section also provides historical context, noting that ancient Babylonians grappled with area calculations for curved surfaces, linking early geometry to the integration problem.

The indefinite integral is defined as the set of all anti-derivatives of a function. The process of integration moves from a derived function back to the original (primitive) function: 𝒇(𝒙) ⇒ 𝑭(𝒙). This is why integration is also called anti-derivative. The elongated 'S' symbol (∫) represents the integration sign, and 𝒅𝒙 indicates the variable of integration. The function being integrated is called the integrand.

The formal relationship between differentiation and integration is stated as: 𝒅/𝒅𝒙 𝑭(𝒙) = 𝒇(𝒙) ⇒ ∫ 𝒇(𝒙) 𝒅𝒙 = 𝑭(𝒙) + 𝒄

Here, 𝒄 is the arbitrary constant of integration, which accounts for the fact that many different primitives can have the same derivative (multiple parentage of the integrand).

TOPIC 004: BASIC RULES OF INTEGRATION: CONSTANT FUNCTION RULE & CONSTANT MULTIPLE RULE

This section introduces the first two fundamental integration rules, which are direct counterparts to their differentiation counterparts. These rules simplify the process by allowing us to handle constants systematically during integration.

1. Constant Function Rule: The integral of a constant function is the constant multiplied by the variable of integration, plus the constant of integration.

🔑 Definition — Constant Function Rule: ∫ 𝒂 𝒅𝒙 = 𝒂𝒙 + 𝒄, where 𝒂 is any real number.

📐 Formula: ∫ 𝒂 𝒅𝒙 = 𝒂𝒙 + 𝒄 → Meaning: Integrating a constant 𝒂 with respect to 𝒙 yields a linear function, 𝒂𝒙, plus the integration constant.

📌 Example: Let 𝒇(𝒙) = 𝟏, so 𝒂 = 𝟏. Step 1: Set up the integral: ∫ 𝟏 𝒅𝒙 Step 2: Apply the constant function rule: = 𝟏∙𝒙 + 𝒄 Step 3: Simplify: = 𝒙 + 𝒄 Result: 𝑭(𝒙) = 𝒙 + 𝒄

2. Constant Multiple Rule: If a constant is multiplied by a function within the integral, the constant can be factored out.

🔑 Definition — Constant Multiple Rule: ∫ 𝒂 ∙ 𝒇(𝒙) 𝒅𝒙 = 𝒂 ∙ ∫ 𝒇(𝒙) 𝒅𝒙

📐 Formula: ∫ 𝒂 ∙ 𝒇(𝒙) 𝒅𝒙 = 𝒂 ∙ ∫ 𝒇(𝒙) 𝒅𝒙 → Meaning: A constant multiplier outside the integral does not affect the integration process itself.

📌 Example: Let 𝒂 = 𝟓 and 𝒇(𝒙) be any function. Step 1: Set up the integral: ∫ 𝟓 ∙ 𝒇(𝒙) 𝒅𝒙 Step 2: Apply the constant multiple rule: = 𝟓 ∫ 𝒇(𝒙) 𝒅𝒙 Step 3: Assume ∫ 𝒇(𝒙) 𝒅𝒙 = 𝑭(𝒙) + 𝒄 (where 𝒄 accounts for the combined constant of integration) Result: ∫ 𝟓 ∙ 𝒇(𝒙) 𝒅𝒙 = 𝟓𝑭(𝒙) + 𝒄

💡 Why this matters: These rules mirror their differentiation counterparts, making the inverse relationship clear. A constant function (degree 0) integrates to a linear function (degree 1), demonstrating how integration increases the degree of a polynomial by one.

TOPIC 005: POWER RULE OF INTEGRATION

This section covers the power rule for integrating polynomial functions, which is the most commonly used integration technique. It also explicitly explains the restriction on the power (𝒏 ≠ −1) and shows how the constant function rule can be derived from the power rule.

🔑 Definition — Power Rule of Integration: For any power 𝒏, except 𝒏 = −1, the integral of 𝒙^𝒏 is 𝒙^(𝒏+𝟏) divided by (𝒏+𝟏), plus the constant of integration.

📐 Formula: ∫ 𝒙^𝒏 𝒅𝒙 = 𝒙^(𝒏+𝟏) / (𝒏+𝟏) + 𝒄, where 𝒏 ≠ −1 → Meaning: To integrate a power function, add 1 to the exponent and divide by the new exponent.

📌 Parametric Stipulation & Example for 𝒏 = −1: If 𝒏 = −1, the power rule would give: ∫ 𝒙^(−1) 𝒅𝒙 = 𝒙^(0) / 0 + 𝒄 = 1/0 + 𝒄 = ∞ + 𝒄. This is undefined (∞), hence the rule only applies for 𝒏 ≠ −1.

📌 Example: Let 𝒇(𝒙) = 𝒙^𝟑 (power 𝒏 = 3). Step 1: Set up the integral: ∫ 𝒚 𝒅𝒚 = ∫ 𝒙^𝟑 𝒅𝒙 Step 2: Apply the power rule: = 𝒙^(𝟑+𝟏) / (𝟑+𝟏) + 𝒄 Step 3: Simplify the numerator and denominator: = 𝒙^𝟒 / 𝟒 + 𝒄 Result: 𝑭(𝒙) = 𝒙^𝟒/𝟒 + 𝒄

📌 Deriving Constant Function Rule from Power Rule: Step 1: Start with the constant function 𝒚 = 𝒂, where a constant = 1. Step 2: Rewrite as: 𝒚 = (𝟏)𝒙^𝟎 Step 3: Set up the integral: ∫ 𝒚 𝒅𝒚 = ∫ (𝒙^𝟎) 𝒅𝒙 Step 4: Apply the power rule: = 𝒙^(𝟎+𝟏) / (𝟎+𝟏) + 𝒄 = 𝒙^𝟏 / 𝟏 + 𝒄 Result: ∫ 𝒚 𝒅𝒚 = 𝒙 + 𝒄 (matching the constant function rule)

TOPIC 006: DERIVATIVE OVER FUNCTION RULE OF INTEGRATION

This section introduces the derivative over function rule, used when the integrand is a ratio where the numerator is the derivative of the denominator. The rule is powerful for integrating expressions that resemble logarithmic derivatives.

🔑 Definition — Derivative over Function Rule: If the integrand is a fraction where the numerator is exactly the derivative of the denominator, the integral is the natural logarithm of the absolute value of the denominator.

📐 Formula: ∫ [𝒇′(𝒙) / 𝒇(𝒙)] 𝒅𝒙 = 𝒍𝒏|𝒇(𝒙)| + 𝒄 → Meaning: The integral of a derivative-function ratio simplifies to the natural log of the absolute value of the function in the denominator.

📌 Example: 𝒚 = 𝟒𝒙 / (𝒙^𝟐+𝟏) Step 1: Set up the integral: ∫ 𝒚 𝒅𝒚 = ∫ [𝟒𝒙 / (𝒙^𝟐+𝟏)] 𝒅𝒙 Step 2: Identify the primitive (denominator): 𝒇(𝒙) = 𝒙^𝟐+𝟏; its derivative is: 𝒇′(𝒙) = 𝟐𝒙. Step 3: Adjust the numerator to match the derivative: The original numerator is 𝟒𝒙. Since 𝟒𝒙 = 𝟐(𝟐𝒙), rewrite 𝟒𝒙 as 𝟐∙𝒇′(𝒙). Step 4: Set up the integral: ∫ 𝒚 𝒅𝒚 = ∫ 𝟐(𝟐𝒙) / (𝒙^𝟐+𝟏) 𝒅𝒙 = 𝟐 ∫ [𝟐𝒙/(𝒙^𝟐+𝟏)] 𝒅𝒙 Step 5: Apply the derivative over function rule: = 𝟐 [𝒍𝒏|𝒙^𝟐+𝟏|] + 𝒄 Result: 𝑭(𝒙) = 𝟐(𝒍𝒏|𝒙^𝟐+𝟏|) + 𝒄

⭐ Key Takeaways

1. Integration is the inverse of differentiation; the constant of integration (c) accounts for the indeterminate shift of the primitive function, meaning multiple functions share the same derivative.
2. The constant function rule (∫a dx = ax + c) and constant multiple rule (∫a·f(x) dx = a∫f(x) dx) are foundational and directly reverse their differentiation counterparts.
3. The power rule (∫x^n dx = x^(n+1)/(n+1) + c) is valid for all n ≠ -1; the case n = -1 is undefined by this rule and requires the derivative-over-function rule (yielding a natural log).
4. The derivative over function rule (∫f′(x)/f(x) dx = ln|f(x)| + c) is a critical technique for integrating rational expressions where the numerator is a multiple of the denominator's derivative.
5. A constant function (degree 0) integrates to a linear function (degree 1), confirming that integration increases the polynomial degree by one (for power rule cases).

🧠 Quick Revision Questions

1. What does the arbitrary constant of integration (c) represent, and why is it necessary for indefinite integrals?
2. State and apply the constant function rule and constant multiple rule to the expression: ∫ 𝟑(𝒙 + 𝟐) 𝒅𝒙.
3. Why does the power rule fail when 𝒏 = −1? What rule is used instead to integrate ∫ 𝟏/𝒙 𝒅𝒙?
4. Apply the derivative over function rule to evaluate: ∫ [𝟔𝒙 / (𝒙^𝟐 + 𝟒)] 𝒅𝒙 (hint: adjust the numerator to match 2x).
5. If 𝑭′(𝒙) = 𝒙^𝟐 + 𝟑𝒙 − 𝟓 and 𝑭(𝟎) = 𝟏𝟎, find the explicit function 𝑭(𝒙) using the power rule and the initial condition.

📘 Lecture 3 — Integration and Rules (Continued)

📖 Overview: This lecture continues the study of integration rules, covering exponential and natural exponential function rules, sum-difference rule, and more complex techniques including integration by substitution and integration by parts. These rules are essential for solving a wider range of integral problems encountered in mathematical economics.

🗂️ Topics Covered

This lecture covers five main topics: the Exponential Function Rule of Integration, the Natural Exponential Function Rule of Integration, the Sum-Difference Rule of Integration, Integration by Substitution (for products/quotients where a function and its derivative are present), and Integration by Parts (for products/quotients where substitution fails). Each topic includes numerical examples and graphical visualization of integrands and their integrals.

📝 Lecture Summary

TOPIC 007: EXPONENTIAL FUNCTION RULE OF INTEGRATION

The Exponential Function Rule states that if the integrand is an exponential function of the form (a^{f(x)}), then its integral is given by: (\int a^{f(x)} dx = \frac{a^{f(x)}}{f'(x) \cdot \ln|a|} + c). This rule applies when the base 'a' is a constant.

🔑 Definition — Exponential Function Rule: (\int a^{f(x)} dx = \frac{a^{f(x)}}{f'(x) \ln|a|} + c) 📐 Formula: (\int a^{f(x)} dx = \frac{a^{f(x)}}{f'(x) \ln|a|}) → The integral of an exponential function equals the original function divided by the derivative of the exponent times the natural log of the base. 📌 Example: (\int 2^{3x} dx). Here, (a=2) and (f(x)=3x), so (f'(x)=3). Then (\ln|a| = \ln|2| \approx 0.693). Applying the formula: (\int 2^{3x} dx = \frac{2^{3x}}{3 \cdot \ln|2|} + c). The lecture also writes this as (\frac{2^{3x}}{3 \cdot 0.693} + c). The pre-graph is (f(x)=2^{3x}) and the post-graph (integral) is (F(x)= \frac{2^{3x}}{3 \cdot \ln|2|} + c) with (c=0).

TOPIC 008: NATURAL EXPONENTIAL FUNCTION RULE OF INTEGRATION

The Natural Exponential Function Rule is a special case of the exponential rule where the base is Euler's number 'e'. Since (\ln|e| = 1), the rule simplifies. It states: (\int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x)} + c).

🔑 Definition — Natural Exponential Function Rule: (\int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x) \cdot \ln|e|} + c = \frac{e^{f(x)}}{f'(x)} + c) 📐 Formula: (\int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x)} + c) → The integral of a natural exponential function equals the original function divided by the derivative of its exponent. 📌 Example: Integrate (y = 13e^{4x}). Apply the rule: (\int 13e^{4x} dx = 13 \cdot \frac{e^{4x}}{\frac{d}{dx}(4x)} + c = 13 \cdot \frac{e^{4x}}{4} + c = \frac{13}{4} e^{4x} + c). The pre-graph is (f(x)=13e^{4x}) and the post-graph is (F(x)=\frac{13}{4}e^{4x}+c) with (c=1).

TOPIC 009: SUM-DIFFERENCE RULE OF INTEGRATION

The Sum-Difference Rule states that the integral of a sum or difference of two or more functions is equal to the sum or difference of the integrals of those individual functions.

🔑 Definition — Sum-Difference Rule: (\int {g(x) \pm h(x)} dx = \int g(x) dx \pm \int h(x) dx + c) 📐 Formula: (\int {g(x) \pm h(x)} dx = \int g(x) dx \pm \int h(x) dx + c) → The integral of a sum/difference is the sum/difference of the integrals. 📌 Example: Integrate (f(x) = 3x^3 - x + 1). This involves three functions and both operations. (\int (3x^3 - x + 1) dx = \int 3x^3 dx - \int x dx + \int 1 dx = 3 \int x^3 dx - \int x dx + \int 1 dx). Applying the power rule: (3 \cdot \frac{x^{3+1}}{3+1} - \frac{x^{1+1}}{1+1} + x + c = 3 \cdot \frac{x^4}{4} - \frac{x^2}{2} + x + c = \frac{3x^4}{4} - \frac{x^2}{2} + x + c). The pre-graph is (f(x)=3x^3 - x + 1) and the post-graph is (F(x)=\frac{3x^4}{4} - \frac{x^2}{2} + x + c) with (c=1).

TOPIC 010: MORE COMPLEX RULES: INTEGRATION BY SUBSTITUTION

Integration by Substitution is used for the integral of a product or quotient of two differentiable functions where one function is a constant multiple of the derivative of the other. The method involves setting the inner function (u = h(x)), finding its derivative (du = h'(x) dx), and rewriting the integral in terms of (u).

🔑 Definition — Integration by Substitution: If an integral can be expressed as (\int n \cdot {h(x) \cdot h'(x)} dx), let (u = h(x)) and (du = h'(x) dx). Then (\int n \cdot u \cdot du = n \int u du). 💡 Why this matters: This technique simplifies integrals by reducing them to a basic form. 📌 Example: (\int 12x^2 (x^3 + 2) dx). Here, (u = x^3 + 2) and (\frac{du}{dx} = 3x^2). The integrand has (12x^2) which is (4 \times 3x^2). So, (\int 12x^2 (x^3 + 2) dx = \int 4 \cdot 3x^2 \cdot (x^3 + 2) dx = \int 4 \cdot u \cdot \frac{du}{dx} dx = 4 \int u du = 4 \cdot \frac{u^2}{2} + c = 2u^2 + c). Restoring in terms of x: (2(x^3 + 2)^2 + c). Verification by differentiation: (\frac{d}{dx}[2(x^3+2)^2 + c] = 2 \cdot 2(x^3+2) \cdot 3x^2 = 12x^2(x^3+2)), which is the original integrand.

TOPIC 011: MORE COMPLEX RULES: INTEGRATION BY PARTS

Integration by Parts is used when the integrand is a product or quotient of differentiable functions, but Integration by Substitution fails (e.g., the derivative of one function is not a constant multiple of the other). The formula is derived by reversing the product rule for differentiation.

🔑 Definition — Integration by Parts: (\int {h(x) \cdot g'(x)} dx = g(x) \cdot h(x) - \int {g(x) \cdot h'(x)} dx) Hack: Let (h(x)) be the simpler function and (g'(x)) be the more complicated one to make the new integral (\int g(x) h'(x) dx) simpler. 📐 Formula: (\int {h(x) g'(x)} dx = g(x) h(x) - \int {g(x) h'(x)} dx) 📌 Example: (\int 4x (x+1)^3 dx). Here, (h(x) = 4x) (simpler) and (g'(x) = (x+1)^3) (more complicated). First, find (g(x) = \int g'(x) dx = \int (x+1)^3 dx = \frac{(x+1)^4}{4} + c_1) and (h'(x) = \frac{d}{dx}(4x) = 4). Plug into the formula: (\int 4x (x+1)^3 dx = g(x) \cdot h(x) - \int g(x) \cdot h'(x) dx = \frac{(x+1)^4}{4} \cdot 4x - \int \frac{(x+1)^4}{4} \cdot 4 dx + \text{constants} = x(x+1)^4 - \int (x+1)^4 dx). Then integrate: (x(x+1)^4 - \frac{(x+1)^5}{5} + c). Verification by differentiation (using product and power rules) yields (\frac{d}{dx}[x(x+1)^4 - \frac{(x+1)^5}{5} + c] = 4x(x+1)^3 + (x+1)^4 - (x+1)^4 = 4x(x+1)^3), which is the original integrand.

⭐ Key Takeaways

The most critical concepts from this lecture are the four distinct integration rules. First, the Exponential and Natural Exponential rules handle functions where the variable is in the exponent, with the natural exponential being a simpler special case. Second, the Sum-Difference rule allows breaking complex polynomials into simpler, individual integrals. Third, Integration by Substitution is the go-to method when the integrand contains a function and its derivative (or a constant multiple of it), simplifying it to a basic power rule. Fourth, when substitution fails, Integration by Parts is the superior alternative, derived from reversing the product rule and requiring careful selection of (h(x)) and (g'(x)). Every rule must be verified by differentiating the result to ensure it yields the original integrand.

🧠 Quick Revision Questions

1. What is the formula for integrating a general exponential function (a^{f(x)})?
2. Why does the integration rule for (e^{f(x)}) not contain the term (\ln|e|) in the denominator?
3. How does the Sum-Difference rule simplify the integration of a function like (3x^2 - 2x + 5)?
4. Under what specific condition is Integration by Substitution the appropriate method to use?
5. What is the formula for Integration by Parts, and how is it derived from the product rule?

📘 Lecture 4 — Demand and Supply Analysis Using Integration

📖 Overview: This lecture covers the application of integration by substitution to derive inverse demand, supply, and revenue functions from their marginal rates of change. It demonstrates how economic functions can be recovered from derivatives using definite integration and initial conditions, which is essential for practical pricing and production decisions.

🗂️ Topics Covered

This lecture covers four major applications of integration by substitution in economics: demand analysis using the inverse demand function derived from a price change rate, supply analysis recovering the supply function from marginal price, revenue analysis obtaining total revenue from marginal revenue, and investment portfolio analysis finding the portfolio value function from its rate of change. Each topic includes a step-by-step demonstration of the substitution method, handling of constants of integration, and use of boundary conditions to obtain particular solutions.

📝 Lecture Summary

TOPIC 012: DEMAND ANALYSIS USING INTEGRATION BY SUBSTITUTION

The price p (dollars) of each unit of a commodity changes at the rate: dp/dx = -135x/√(9 + x²), where x is consumer demand in hundred units. When x = 4, price p = PKR 30. We need to find the demand function p(x), prices at specific demands, and quantity demanded at a given price.

To find the inverse demand function p(x), integrate the rate of change: p(x) = ∫ (-135x/√(9 + x²)) dx

Integration by substitution is needed since basic rules don't apply. Let the relatively higher degree polynomial be substituted: u = 9 + x², which implies du/dx = 2x, so du/2 = x dx.

Substituting into the integrand: p(x) = ∫ (-135/√u)(du/2) = -135/2 ∫ u^(-1/2) du

Using the power rule: p(x) = -135/2 [u^(1/2)/(1/2)] + c = -135√u + c

Restoring in terms of x: p(x) = -135√(9 + x²) + c (general solution)

Using the initial condition p(4) = 30: 30 = -135√(9 + 16) + c → 30 = -135(5) + c → c = 705

The definite solution is: p(x) = -135√(9 + x²) + 705

🔑 Definition — Inverse Demand Function: The function p(x) that gives the price per unit as a function of quantity demanded.

📌 Example: For finding the price when 300 units are demanded (x = 3 hundred): p(3) = -135√(9 + 9) + 705 = -135√18 + 705 ≈ PKR 132.27

📌 Example: For finding the price when no units are demanded (x = 0): p(0) = -135√9 + 705 = -135(3) + 705 = PKR 300

📌 Example: For finding units demanded at price PKR 20: 20 = -135√(9 + x²) + 705 → -685 = -135√(9 + x²) → √(9 + x²) ≈ 5.074 → x² ≈ 16.75 → x ≈ 4.09 hundred units (409 units)

💡 Why this matters: This inverse demand function allows businesses to set prices based on expected demand and predict how many units will be sold at any given price.

TOPIC 013: SUPPLY ANALYSIS USING INTEGRATION BY SUBSTITUTION

The owner of a fast-food chain finds that if x thousand units of a new meal item are supplied, the marginal price is given by p'(x) = x/(x + 3)². Currently, 5,000 units (x = 5) are supplied at PKR 2.20 per unit.

To find the supply function p(x), integrate: p(x) = ∫ [x/(x + 3)²] dx

Let u = x + 3, so du = dx and x = u - 3. p(x) = ∫ [(u - 3)/u²] du = ∫ (1/u - 3/u²) du = ln|u| + 3/u + c

Restoring: p(x) = ln|x + 3| + 3/(x + 3) + c (general solution)

Using the initial condition p(5) = 2.20: 2.20 = ln|8| + 3/8 + c → 2.20 = 2.0794 + 0.375 + c → c ≈ -0.2544

The definite solution is approximately: p(x) = ln|x + 3| + 3/(x + 3) - 0.2544

📌 Example: If 10,000 meal units (x = 10) are supplied: p(10) = ln|13| + 3/13 - 0.2544 ≈ 2.5649 + 0.2308 - 0.2544 ≈ PKR 2.54 per unit

💡 Why this matters: The supply function helps determine the minimum price needed to supply a given quantity, which is essential for pricing strategies and production planning.

TOPIC 014: REVENUE ANALYSIS USING INTEGRATION BY SUBSTITUTION

The marginal revenue from the sale of x units is: R'(x) = 50 + 3.5xe^(-0.01x²) dollars per unit, with R(0) = 0.

To find the revenue function R(x): R(x) = ∫ [50 + 3.5xe^(-0.01x²)] dx = 50x + c₁ + 3.5∫ xe^(-0.01x²) dx

For the second integral, let u = -0.01x², so du = -0.02x dx, and x dx = du/(-0.02)

Substituting: 3.5∫ xe^(-0.01x²) dx = 3.5∫ e^u [du/(-0.02)] = -175∫ e^u du = -175e^u + c₂

R(x) = 50x - 175e^(-0.01x²) + c (general solution)

Using R(0) = 0: 0 = 50(0) - 175e^0 + c → c = 175

The definite solution is: R(x) = 50x - 175e^(-0.01x²) + 175

📌 Example: Revenue from selling 1,000 units: R(1000) = 50(1000) - 175e^(-0.01(1000)²) + 175 = 50,000 - 175e^(-10,000) + 175 ≈ PKR 50,175

Note: e^(-10,000) is essentially zero.

💡 Why this matters: The revenue function allows firms to forecast total revenue at different production levels, which is crucial for profit planning and break-even analysis.

TOPIC 015: INVESTMENT PORTFOLIO ANALYSIS USING INTEGRATION BY SUBSTITUTION

An investment portfolio changes value at rate: V'(t) = 12e^(-0.05t)(e^(0.3t) - 3) dollars per unit time.

To find V(t): V(t) = ∫ 12e^(-0.05t)(e^(0.3t) - 3) dt = 12∫ e^(-0.05t)(e^(0.3t) - 3) dt

Let u = -t/20, so du = -dt/20 and dt = -20du. Also, t = -20u.

Substituting: V(u) = 12∫ e^u(e^(-6u) - 3)(-20du) = -240∫ (e^(-5u) - 3e^u) du = 240∫ (3e^u - e^(-5u)) du

V(u) = 240[3e^u + e^(-5u)/5] + c = 720e^u + 48e^(-5u) + c

Restoring in terms of t: V(t) = 720e^(-t/20) + 48e^(t/4) + c (general solution)

📌 Example: If the constant c = 0, the portfolio value function is V(t) = 720e^(-t/20) + 48e^(t/4)

💡 Why this matters: This allows investors to model how their portfolio value changes over time, combining both decaying and growing exponential components to capture complex investment dynamics.

⭐ Key Takeaways

The lecture demonstrates that integration by substitution is essential when marginal functions contain composite expressions that cannot be integrated using basic rules. The key steps are always: identify the substitution (usually setting u equal to the inner function), solve for the differential, rewrite the entire integral in terms of u, integrate using appropriate rules, restore the original variable, and use initial conditions to find the constant of integration. The four applications—demand, supply, revenue, and investment portfolio analysis—follow the same methodical approach but differ in their economic interpretation and boundary conditions. For the exam, remember that the substitution method transforms complex integrands into forms solvable by power rule or natural exponential function rule, and the constant of integration is always determined by plugging in a known point (initial condition).

🧠 Quick Revision Questions

1. In demand analysis using integration by substitution, what substitution was made for the integrand -135x/√(9+x²)?

2. Given the marginal revenue function R'(x) = 50 + 3.5xe^(-0.01x²), what is the value of the constant of integration c if R(0) = 0?

3. In the supply analysis example, what substitution converts ∫ x/(x+3)² dx into a simpler form?

4. For the investment portfolio problem, what two exponential terms appear in the final general solution V(t)?

5. If the demand function p(x) = -135√(9+x²) + 705, what is the price when 400 units (x = 4) are demanded?

📘 Lecture 05 — DEFINITE INTEGRALS

📖 Overview: This lecture introduces the concept of definite integrals as signed areas under curves between specified limits. It covers the fundamental definition, numerical evaluation, approximation techniques using Riemann sums, essential properties, and the handling of improper integrals with infinite limits or integrands, which have direct applications in economics.

🗂️ Topics Covered

This lecture begins by defining definite integrals as the signed area between a curve, the x-axis, and vertical lines at the limits of integration, showing that the constant of integration cancels out. It then explains the Riemann sum method for approximating the area under a curve by dividing it into vertical rectangles. The lecture details five key properties of definite integrals, including the effects of interchanging limits and splitting integrals. Finally, it covers improper integrals, which arise from infinite limits of integration or infinite integrands, and provides numerical examples.

📝 Lecture Summary

TOPIC 016: DEFINITE INTEGRALS

Unlike indefinite integrals, which give a family of functions, a definite integral calculates the signed area between a curve (f(x)), the x-axis, and two vertical lines (x=a) and (x=b). This signed area means the sign is positive above the x-axis and negative below it. The lower limit is (a) and the upper limit is (b), where (a < b).

🔑 Definition — Definite Integral: The signed area between the curve (f(x)), the x-axis, and the lines (x=a) and (x=b). 📐 Formula: (\int_a^b f(x) dx = F(b) - F(a)), where (F(x)) is the antiderivative of (f(x)). The arbitrary constant of integration (c) cancels out. 💡 Why this matters: The result of a definite integral is a specific numerical value, not a function plus a constant.

📌 Example 1: (\int_1^5 3x^2 dx)

1. Find the antiderivative: (x^3 + c).
2. Evaluate at the upper limit: ( (5)^3 + c = 125 + c).
3. Evaluate at the lower limit: ( (1)^3 + c = 1 + c).
4. Subtract: ((125 + c) - (1 + c) = \mathbf{124}).

📌 Example 2: (\int_a^b k e^x dx)

1. The antiderivative of (k e^x) is (k e^x).
2. Apply limits: (k e^x ]_a^b = k(e^b - e^a)).

📌 Example 3: (\int_0^4 (\frac{1}{1+x} + 2x) dx)

1. Find the Derivative over Function Rule for (\frac{1}{1+x}), which yields (\ln|1+x|). For (2x), use the Power Rule to get (x^2).
2. The integral is (\ln|1+x| + x^2 ]_0^4).
3. Evaluate: ((\ln|1+4| + 4^2) - (\ln|1+0| + 0^2) = (\ln|5| + 16) - (\ln|1| + 0) = \ln 5 - 16).

TOPIC 017: DEFINITE INTEGRAL: AREA UNDER CURVE

The area under a curve between points (a) and (b) can be approximated using the Riemann Sum. This involves dividing the area into (n) vertical rectangles of equal width ((\Delta x_n)) but varying heights (f(x_n)). The total approximated area (A) is the sum of the areas of all rectangles.

🔑 Definition — Riemann Sum: A method to approximate the total area under a curve by summing the areas of a series of adjacent rectangles. 📐 Formula: (A = \sum_{i=o}^n { f(x_n) \times (\Delta x_n) }) 💡 Why this matters: Increasing the number of rectangles (n) makes them narrower and the approximation more precise, approaching the exact calculus-based value.

📌 Example: Approximate the area under (f(x) = 1/x) from (x=1) to (x=6) using 5 rectangles.

1. The endpoints are (x_1=1, x_2=2, x_3=3, x_4=4, x_5=5, x_6=6).
2. Heights are (f(1)=1, f(2)=1/2, f(3)=1/3, f(4)=1/4, f(5)=1/5).
3. Each width is 1.
4. Area (A = 1(1) + \frac{1}{2}(1) + \frac{1}{3}(1) + \frac{1}{4}(1) + \frac{1}{5}(1) = 1 + 0.5 + 0.333 + 0.25 + 0.2 = \mathbf{2.283}).
5. This is an over-approximation compared to the exact integral (\int_1^6 \frac{1}{x} dx = \ln 6 - \ln 1 = \mathbf{1.792}).

TOPIC 018: PROPERTIES OF DEFINITE INTEGRALS

Property 1: Interchanging Limits Interchanging the upper and lower limits changes the sign of the definite integral. 📐 Formula: (\int_a^b f(x) dx = -\int_b^a f(x) dx) Proof: (\int_a^b f(x) dx = F(b) - F(a) = -[F(a) - F(b)] = -\int_b^a f(x) dx).

Property 2: Identical Limits The definite integral is zero when the upper and lower limits are equal. 📐 Formula: (\int_a^a f(x) dx = 0) Proof: (\int_a^a f(x) dx = F(a) - F(a) = 0).

Property 3: Additivity A definite integral over an interval can be split into a sum of integrals over sub-intervals. 📐 Formula: (\int_a^d f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx + \int_c^d f(x) dx), where (a < b < c < d).

Property 4: Constant Multiple A constant coefficient can be factored out of the integral. 📐 Formula: (\int_a^b k f(x) dx = k \int_a^b f(x) dx)

Property 5: Sum/Difference Rule The integral of a sum or difference is the sum or difference of the integrals. 📐 Formula: (\int_a^b { g(x) \pm h(x) } dx = \int_a^b g(x) dx \pm \int_a^b h(x) dx)

📌 Example: (\int_{-1}^2 (9 - x^2) dx)

1. Integrate: (9x - \frac{x^3}{3} ]_{-1}^2).
2. Evaluate upper limit: (9(2) - \frac{2^3}{3} = 18 - \frac{8}{3} = \frac{46}{3}).
3. Evaluate lower limit: (9(-1) - \frac{(-1)^3}{3} = -9 - (-\frac{1}{3}) = -9 + \frac{1}{3} = -\frac{26}{3}).
4. Subtract: (\frac{46}{3} - (-\frac{26}{3}) = \frac{46}{3} + \frac{26}{3} = \frac{72}{3} = \mathbf{24}).

TOPIC 019: IMPROPER INTEGRALS

An integral is considered improper if it is "unfitting" or not standard, mathematically involving an undefined element like infinity ((\infty)). This occurs for two reasons: infinite limits of integration or an infinite integrand.

1. INFINITE LIMITS OF INTEGRATION This occurs when one or both of the limits of integration are infinite. 📐 General Forms: (\int_a^\infty f(x) dx) or (\int_{-\infty}^b f(x) dx)

📌 Example 1: (\int_1^\infty \frac{dx}{x^2})

1. Integrate: (-\frac{1}{x} ]_1^\infty).
2. Evaluate upper limit: (-\frac{1}{\infty} = 0).
3. Evaluate lower limit: (-(\frac{1}{1}) = -1).
4. Subtract: (0 - (-1) = \mathbf{1}).

📌 Example 2: (\int_1^\infty \frac{dx}{x})

1. Integrate: (\ln|x| ]_1^\infty).
2. Evaluate: (\ln|\infty| - \ln|1| = \infty - 0 = \mathbf{\infty}). This integral diverges.

2. INFINITE INTEGRAND This occurs when the function being integrated (the integrand) becomes infinite within the interval of integration, even if the limits are finite.

📌 Example 1: (\int_0^1 \frac{dx}{x})

1. Integrate: (\ln|x| ]_0^1).
2. Evaluate: (\ln|1| - \ln|0| = 0 - (-\infty) = \mathbf{\infty}). This integral diverges.

📌 Example 2: (\int_0^9 \frac{dx}{\sqrt{x}})

1. Rewrite: (\int_0^9 x^{-1/2} dx).
2. Integrate: (2x^{1/2} ]_0^9 = 2\sqrt{x} ]_0^9).
3. Evaluate: (2(\sqrt{9} - \sqrt{0}) = 2(3 - 0) = \mathbf{6}).

⭐ Key Takeaways

Definite integrals calculate the exact signed area between a curve and the x-axis, yielding a single numerical value. The area can be approximated by the Riemann sum, which becomes more accurate with an increasing number of rectangles. Five key properties—interchanging limits, identical limits, additivity, constant multiple, and sum/difference—are essential for manipulating and solving definite integrals efficiently. Improper integrals handle cases with infinite limits or integrands that become infinite, requiring careful evaluation as they may converge to a finite number or diverge to infinity.

🧠 Quick Revision Questions

1. State the Fundamental Theorem of Calculus as it applies to evaluating a definite integral (\int_a^b f(x) dx).
2. Why does the constant of integration (c) not appear in the final answer of a definite integral?
3. Using the Riemann sum with 4 rectangles of equal width, approximate the area under (f(x) = x^2) from (x=0) to (x=2).
4. Simplify the integral (\int_2^5 3x^2 dx) using the constant multiple property of definite integrals.
5. Explain the difference between a proper definite integral and an improper integral. Give one example of an integral that is improper due to an infinite limit and one that is improper due to an infinite integrand.

📘 Lecture 6 — Marginal to Total Analysis Using Integrals

📖 Overview: This lecture demonstrates how to recover total functions (Cost, Revenue, Savings, and Production) from their marginal counterparts using integration. It is essential because economists often have data on marginal changes and need to reconstruct the underlying total functions for policy analysis and optimization.

🗂️ Topics Covered

This lecture covers four key applications of integration for moving from marginal to total functions: deriving the total cost function from marginal cost using a fixed cost initial condition, deriving the total revenue function from marginal revenue given zero revenue at zero output, deriving the savings function from marginal propensity to save with a specific income-level condition, and deriving the short-run production function from the marginal product of labor. Each topic emphasizes the role of the constant of integration and the use of initial conditions to obtain a unique (definite) solution.

📝 Lecture Summary

MARGINAL TO TOTAL COST USING INTEGRALS

Given the marginal cost function ( C'(Q) = 2e^{0.2Q} ) and fixed cost ( C_F = 90 ), we find the total cost function ( C(Q) ). The integral of ( C'(Q) ) is ( \int 2e^{0.2Q} dQ ). Using the rule ( \int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x)} ), we get ( 2 \left( \frac{e^{0.2Q}}{0.2} \right) + c = 10e^{0.2Q} + c ). Since derivative and integral are inverse operators, this is the general solution, but the constant ( c ) is arbitrary. The initial condition uses fixed cost: at ( Q = 0 ), total cost equals fixed cost. Substituting: ( C(0) = 10e^0 + c = 10 + c = C_F = 90 ). Solving gives ( c = 90 - 10 = 80 ).

🔑 Definition — Initial Condition: A known value of the dependent variable at a specific value of the independent variable, used to determine the constant of integration.

📐 Formula: ( C(Q) = \int C'(Q) dQ = 10e^{0.2Q} + c ) → The total cost is the integral of marginal cost plus a constant.

📌 Example: Given ( C'(Q) = 2e^{0.2Q} ) and ( C_F = 90 ). Integrating: ( C(Q) = 10e^{0.2Q} + c ). At ( Q=0 ), ( C(0) = 10 + c = 90 ), so ( c = 80 ). The definite solution is ( C(Q) = 10e^{0.2Q} + 80 ).

💡 Why this matters: In economics, marginal cost is easier to observe, but total cost decisions require the fixed cost component, which the constant of integration captures.

MARGINAL TO TOTAL REVENUE USING INTEGRALS

Given the marginal revenue function ( R'(Q) = 28Q - e^{0.3Q} ), we find total revenue ( R(Q) ). Integrate: ( \int (28Q - e^{0.3Q}) dQ = 28 \int Q dQ - \int e^{0.3Q} dQ ). For the first term: ( 28 \cdot \frac{Q^2}{2} = 14Q^2 ). For the second, using the same exponential rule: ( \int e^{0.3Q} dQ = \frac{1}{0.3} e^{0.3Q} = \frac{10}{3} e^{0.3Q} ). However, the lecture applies a substitution: ( \int e^{0.3Q} dQ = \int \frac{1}{0.3} \cdot 0.3 e^{0.3Q} dQ = \frac{1}{0.3} e^{0.3Q} = \frac{10}{3} e^{0.3Q} ). Thus, ( R(Q) = 14Q^2 - \frac{10}{3} e^{0.3Q} + c ). The initial condition is ( R(0) = 0 ) (no revenue at zero output). Substituting ( Q=0 ): ( R(0) = 14(0)^2 - \frac{10}{3} e^0 + c = 0 - \frac{10}{3} + c = 0 ), so ( c = \frac{10}{3} ).

🔑 Definition — Marginal Revenue: The additional revenue generated from selling one more unit; its integral yields total revenue.

📐 Formula: ( R(Q) = \int R'(Q) dQ = 14Q^2 - \frac{10}{3} e^{0.3Q} + c ) → Total revenue is the integral of marginal revenue.

📌 Example: Given ( R'(Q) = 28Q - e^{0.3Q} ). Integrating gives ( R(Q) = 14Q^2 - \frac{10}{3} e^{0.3Q} + c ). Using ( R(0) = 0 ): ( -\frac{10}{3} + c = 0 ), so ( c = \frac{10}{3} ). The definite solution is ( R(Q) = 14Q^2 - \frac{10}{3} e^{0.3Q} + \frac{10}{3} ).

MARGINAL TO TOTAL SAVINGS USING INTEGRALS

Given the marginal propensity to save ( S'(Y) = 0.3 - 0.1Y^{-\frac{1}{2}} ) and a condition ( (S)_{Y=81} = 0 ) (savings are zero when income is 81), find the savings function ( S(Y) ). A caveat: the condition is ( S(81) = 0 ), not ( Y(81) = 0 ), because income cannot be zero for positive savings. Integrate: ( \int (0.3 - 0.1Y^{-\frac{1}{2}}) dY = 0.3Y - 0.1 \int Y^{-\frac{1}{2}} dY ). The integral of ( Y^{-\frac{1}{2}} ) is ( \frac{Y^{\frac{1}{2}}}{1/2} = 2Y^{\frac{1}{2}} ). So, ( S(Y) = 0.3Y - 0.1(2Y^{\frac{1}{2}}) + c = 0.3Y - 0.2Y^{\frac{1}{2}} + c ). Apply the condition ( S(81) = 0 ): ( 0.3(81) - 0.2(81^{\frac{1}{2}}) + c = 24.3 - 0.2(9) + c = 24.3 - 1.8 + c = 22.5 + c = 0 ). Thus, ( c = -22.5 ).

🔑 Definition — Marginal Propensity to Save (MPS): The change in savings resulting from a one-unit change in income; its integral yields the total savings function.

📐 Formula: ( S(Y) = \int S'(Y) dY = 0.3Y - 0.2Y^{\frac{1}{2}} + c ) → Total savings is the integral of MPS.

📌 Example: Given ( S'(Y) = 0.3 - 0.1Y^{-\frac{1}{3}} ) (note: the text has ( -\frac{1}{2} ) exponent). Integrating: ( S(Y) = 0.3Y - 0.2Y^{\frac{1}{2}} + c ). Using ( S(81) = 0 ): ( 24.3 - 1.8 + c = 0 ), so ( c = -22.5 ). The definite solution is ( S(Y) = 0.3Y - 0.2Y^{\frac{1}{2}} - 22.5 ).

SHORT RUN PRODUCTION FUNCTION FROM MARGINAL PRODUCTS USING INTEGRALS

In the short run, capital (( K )) is fixed, so the production function depends only on labor (( L )). Given the marginal product of labor ( MP_L = \frac{d}{dL} { Q(K, L) } = 1000 - 3L^2 ), integrate to find ( Q(K, L) ). Integrate w.r.t. ( L ): ( \int (1000 - 3L^2) dL = 1000 \int L^0 dL - 3 \int L^2 dL = 1000L - 3 \left( \frac{L^3}{3} \right) + c = 1000L - L^3 + c ). The constant ( c ) represents the combined constants from each integral.

🔑 Definition — Marginal Product of Labor (MPL): The additional output produced by one more unit of labor; its integral, holding capital constant, gives the short-run production function.

📐 Formula: ( Q(K, L) = \int MP_L dL = 1000L - L^3 + c ) → Short-run total product is the integral of marginal product of labor.

📌 Example: Given ( MP_L = 1000 - 3L^2 ). Integrating: ( Q(K, L) = 1000L - L^3 + c ). The constant ( c ) is arbitrary unless additional information (e.g., output when labor is zero) is given. The graph shows the function with ( c = 1 ).

⭐ Key Takeaways

The most critical concept is that integrating marginal functions yields total functions plus an arbitrary constant, which is determined by an initial condition specific to the economic context, such as fixed cost for cost functions or zero revenue at zero output for revenue functions. The integration of exponential and power functions is applied repeatedly: ( \int e^{kx} dx = \frac{1}{k} e^{kx} ) and ( \int x^n dx = \frac{x^{n+1}}{n+1} + c ). For savings, the condition is placed on the dependent variable (savings) at a specific independent variable value (income), not the reverse. In short-run production, capital is held constant, so the integral of MPL with respect to labor gives the production function. Finally, the definite solution is the unique function that satisfies both the marginal relationship and the initial condition.

🧠 Quick Revision Questions

1. If marginal cost is ( C'(Q) = 4e^{0.5Q} ) and fixed cost is 50, what is the total cost function ( C(Q) )?
2. Given marginal revenue ( R'(Q) = 15Q^2 - 2e^{0.4Q} ), find the total revenue function assuming ( R(0) = 0 ).
3. For a savings function with MPS ( S'(Y) = 0.4 - 0.2Y^{-1/2} ) and ( S(100) = 0 ), what is the constant of integration?
4. If the marginal product of labor is ( MP_L = 2000 - 5L^2 ) in the short run, what is the general form of the production function?
5. Why is the constant of integration necessary when moving from marginal to total functions in economics?

📘 Lecture 7 — Marginal to Total Analysis Using Integrals (Continued)

📖 Overview: This lecture continues the application of integration to derive total functions from marginal functions in economics. It specifically covers how to find the consumption function from the marginal propensity to consume (MPC), the imports function from its marginal function, and the profit function from marginal profit, using initial conditions to determine the constant of integration.

🗂️ Topics Covered

The lecture is divided into three main topics: Topic 024 covers deriving the consumption function from the MPC using integration and a given consumption value at a specific income level, resulting in a determinate consumption function. Topic 025 demonstrates how to derive the total imports function from the marginal propensity to import using an initial condition (imports at zero income), leading to a linear imports function. Topic 026 shows the derivation of the total profit function from marginal profit, with a given profit at a specific quantity, resulting in a quadratic profit function that can be evaluated at any quantity.

📝 Lecture Summary

Topic 024: Consumption Function from MPC Using Integrals

Given the marginal propensity to consume (MPC) as a function of income (Y), the total consumption function (C(Y)) can be found by integration. The lecture uses the example: ( \textbf{MPC} = 20 + 10Y^{-\frac{1}{4}} ), with the condition that when (Y = 16), (C = 420). The integral of the MPC gives the indeterminate consumption function, which includes an arbitrary constant c. The integral is solved as: (C(Y) = \int (20 + 10Y^{\frac{3}{4}}) dY = 20Y + 40Y^{\frac{1}{4}} + c).

🔑 Definition — Indeterminate Consumption Function: The general solution obtained by integrating the MPC, which contains an arbitrary constant of integration (c), making the function not fully defined.

📐 Formula: The general form is (C(Y) = \int MPC , dY + c), where c is the constant of integration.

📌 Example: Given (MPC = 20 + 10Y^{\frac{3}{4}}), and (C(16) = 420). First, integrate: (C(Y) = 20Y + 40Y^{\frac{1}{4}} + c). Then, substitute (Y=16) and (C=420): (420 = 20(16) + 40(16^{\frac{1}{4}}) + c \Rightarrow 420 = 320 + 40(2) + c \Rightarrow 420 = 320 + 80 + c). Solving gives (c = 20). The determinate consumption function is (C(Y) = 20Y + 40Y^{\frac{1}{4}} + 20), which can also be written as (C(Y) = 20(Y + 2Y^{\frac{1}{4}} + 1)). 💡 Why this matters: The constant c represents autonomous consumption, i.e., consumption when income is zero ((Y=0)). The lecture clarifies a caveat: savings can be positive with some income, but with zero income, savings cannot be positive (it must be zero or negative, meaning dissaving).

Topic 025: Marginal to Total Imports Using Integrals

The marginal propensity to import ((M'(Y))) can be integrated to find the total imports function (M(Y)). In the example, the marginal propensity to import is given as (M'(Y) = 0.1), and the initial condition is that when income (Y=0), total imports (M = 20). The integral of the marginal function is: (\int M'(Y) dY = \int 0.1 dY = 0.1Y + c), which is the indeterminate imports function.

🔑 Definition — Indeterminate Imports Function: The general solution obtained by integrating the marginal propensity to import, containing an arbitrary constant.

📐 Formula: (M(Y) = \int M'(Y) dY + c)

📌 Example: (M'(Y) = 0.1), (M(0) = 20). Integrate to get (M(Y) = 0.1Y + c). Using the initial condition: (M(0) = 0.1(0) + c = 20), so (c = 20). The determinate imports function is (M(Y) = 0.1Y + 20). 💡 Why this matters: The constant c represents autonomous imports (imports that occur even with zero income).

Topic 026: Marginal to Total Profit Using Integrals

The marginal profit function (\pi'(q)) can be integrated to find the total profit function (\pi(q)). In this example, marginal profit is (\pi'(q) = 100 - 2q), and the condition is that when (q=10) units are produced, the total profit is (\pi(10) = 700). Integrating gives the indeterminate profit function: (\pi(q) = \int (100 - 2q) dq = 100q - q^2 + c).

🔑 Definition — Indeterminate Profit Function: The general solution obtained by integrating the marginal profit function, containing an arbitrary constant.

📐 Formula: (\pi(q) = \int \pi'(q) dq + c)

📌 Example: (\pi'(q) = 100 - 2q), (\pi(10) = 700). Integrate: (\pi(q) = 100q - q^2 + c). Substitute the given condition: (100(10) - (10)^2 + c = 700 \Rightarrow 1000 - 100 + c = 700 \Rightarrow 900 + c = 700). Solving, (c = -200). The determinate profit function is (\pi(q) = 100q - q^2 - 200). The lecture then evaluates this function for (q=50): (\pi(50) = 100(50) - (50)^2 - 200 = 5000 - 2500 - 200 = 2300). So, profit at (q=50) is $2300.

⭐ Key Takeaways

The core concept of this lecture is that through integration, we can reverse the process of differentiation in economics, moving from a given marginal function (like MPC, marginal propensity to import, or marginal profit) to the total function (consumption, imports, or profit). This process always yields an indeterminate function with an arbitrary constant of integration, which must be determined using a specific initial condition, such as the value of the total function at a particular point. The determined constant is economically significant, representing, for example, autonomous consumption or autonomous imports. These techniques are essential for constructing complete economic models from rate-of-change information.

🧠 Quick Revision Questions

1. What is the general process for finding a total function from a given marginal function?
2. In the context of deriving the consumption function from the MPC, what does the constant of integration represent?
3. If the marginal propensity to import is a constant (e.g., 0.1), what is the general shape of the total imports function?
4. For the marginal profit function (\pi'(q) = 100 - 2q) and a given point (\pi(10) = 700), what is the value of the constant of integration (c) in the total profit function?
5. What is the first step in finding a determinate total function after integrating the marginal function?

📘 Lecture 8 — Various Economic Applications of Integrals

📖 Overview: This lecture demonstrates the practical application of integral calculus to solve real-world economic problems. It covers determining the total number of shops over time from a growth rate, calculating total oil extracted from a given extraction rate, understanding capital formation from an investment flow, and computing total storage costs for a retailer. These examples illustrate how integration transforms rates of change into total accumulated quantities.

🗂️ Topics Covered

The lecture explores four distinct economic applications of integrals: the rate of growth of a chain of shops, the total extraction of oil from a field over time, the formation of capital stock from a flow of investment, and the calculation of a retailer's total storage costs for a shipment of rice. Each topic involves setting up a differential equation, integrating to find an indefinite solution, and then using an initial condition to determine the specific, definite solution.

📝 Lecture Summary

TOPIC 027: RATE OF GROWTH OF SHOPS USING INTEGRALS

The rate of growth in the number of shops, n, in a chain of local grocery stores after t months is given by dn/dt = 3/√t. To find the total number of shops at any time t, we integrate this rate. This yields an indeterminate solution with an arbitrary constant, c. To find a unique solution, we introduce an initial condition, specifying the number of shops at time t=0, n(0). The definite solution is n(t) = 6t^(1/2) + n(0).

🔑 Definition — Indeterminate solution: A general solution to an integral that includes an arbitrary constant (c), representing a family of possible functions. 📐 Formula: n(t) = ∫ (dn/dt) dt = ∫ 3t^(-1/2) dt = 6t^(1/2) + c → The total number of shops is the integral of the growth rate. 📌 Example: If initially there are n(0) = 2 shops, the definite solution is n(t) = 6t^(1/2) + 2. To find the number of shops after t = 9 months, substitute: n(9) = 6(9)^(1/2) + 2 = 6(3) + 2 = 20. This means that after 9 months, there will be 20 shops in total. The 18 shops were established during the 9 months, while 2 shops were pre-existing.

TOPIC 028: OIL EXTRACTION ANALYSIS USING INTEGRALS

A company begins extracting oil from a new field at t=0. The rate of extraction, measured in thousands of barrels per year, is given by dN/dt = 30t² - 4t³. To find the total oil extracted, N(t), we integrate the rate function. The resulting indeterminate solution is N(t) = 10t³ - t⁴ + c. Using the initial condition that at t=0, no oil has been extracted (N(0) = 0), we find c = 0.

🔑 Definition — Rate of extraction: The instantaneous rate at which a resource (like oil) is being removed from a field, measured in volume per unit of time. 📐 Formula: N(t) = ∫ (dN/dt) dt = ∫ (30t² - 4t³) dt = 10t³ - t⁴ + c → Total oil extracted is the integral of the extraction rate. 📌 Example: The definite solution is N(t) = 10t³ - t⁴. To find the oil extracted after t = 4 years, substitute: N(4) = 10(4)³ - (4)⁴ = 10(64) - 256 = 640 - 256 = 384 thousand barrels. This equals 384,000 barrels of oil extracted over four years.

TOPIC 029: CAPITAL FORMATION USING INTEGRALS

The capital formation model assumes that at any time t, the net investment flow, I(t), is the rate of change of the capital stock, K(t), i.e., dK/dt = I(t). Given an investment function, I(t) = 3t^(1/2), and an initial capital stock K(0) (here, K(0) = 0), the total capital accumulated at time t is found by integration.

🔑 Definition — Capital formation (K(t)): The process of building up the stock of capital assets (e.g., machinery, factories) over time. The net investment is the rate of change of the capital stock. 📐 Formula: K(t) = ∫ I(t) dt = ∫ 3t^(1/2) dt = 2t^(3/2) + c → The capital stock is the integral of the net investment flow. 💡 Why this matters: This transforms a flow variable (investment per year) into a stock variable (accumulated capital). The initial capital stock is K(0), which acts as the constant of integration. 📌 Example: If K(0) = 0, the definite solution is K(t) = 2t^(3/2). This shows how capital accumulates from zero over time.

TOPIC 030: RETAILER'S STORAGE COSTS USING INTEGRALS

A retailer receives a shipment of 10,000 kg of rice, used at a constant rate of 2,000 kg per month over 5 months. Storage costs are 1 cent per kg per month. The storage cost function at any time t is S(t) = 0.01(10,000 - 2,000t). This represents the cost of storing the remaining inventory at that instant. The total storage cost over 5 months is the integral of this function from t=0 to t=5.

🔑 Definition — Storage cost function (S(t)): The instantaneous cost of holding the current stock of inventory at a specific point in time. 📐 Formula: Total Cost = ∫ from 0 to 5 of [0.01(10,000 - 2,000t)] dt = ∫ [100 - 20t] dt → The total cost is the integral (accumulation) of the instantaneous storage cost over the relevant time period. 📌 Example: The indefinite integral is S(t) = 100t - 10t² + c. At t=0, the ship has just arrived and no storage has occurred, so S(0) = 0, which implies c = 0. The definite solution is S(t) = 100t - 10t². To find the total cost for the 5-month period (t=5) , substitute: S(5) = 100(5) - 10(5)² = 500 - 250 = $250. The retailer will pay $250 in storage costs.

⭐ Key Takeaways

The core lesson is that integration is the essential tool for converting a rate of change (a derivative) into the total accumulated quantity (the original function). For economic problems, this means we can find total quantities like the number of shops, extracted oil, capital stock, or storage costs from their respective rates of growth, extraction, investment, or cost. The general solution from integration always includes an arbitrary constant, which must be determined using an initial condition to find the specific, definite solution for the scenario. Mastering the integration of power functions and understanding the role of the constant of integration and initial conditions are critical skills for solving these applied problems.

🧠 Quick Revision Questions

1. In the shops growth problem, what is the mathematical meaning of the constant n(0) that appears in the final solution n(t) = 6t^(1/2) + n(0)?
2. For the oil extraction example, why is the initial condition N(0) = 0 a logical assumption for this specific problem?
3. In the capital formation example, what is the relationship between the investment function I(t) and the capital stock function K(t)?
4. In the storage cost problem, what does the expression 0.01(10,000 - 2,000t) represent at a given time 't'?
5. For all four examples, explain the fundamental role of the constant of integration (c) and the initial condition in finding the final, definite solution.

📘 Lecture 9 — CONSUMER AND PRODUCER SURPLUS ANALYSIS USING INTEGRALS

📖 Overview: This lecture develops the concept of consumer and producer surplus as areas under demand and supply curves, calculated using definite integrals. It demonstrates how to compute willingness to pay, actual expenditure, and the surplus differentials for consumers and producers, including the impact of price discrimination and taxation. These concepts are fundamental to welfare economics and policy analysis.

🗂️ Topics Covered

The lecture covers five topics: consumer willingness to pay using integrals, consumer surplus using integrals, differential of consumer surplus using integrals, producer and social surplus using integrals, and producer surplus with tax using integrals. Each topic applies integral calculus to compute surplus areas from given demand and supply functions, with numerical examples and graphical interpretations.

📝 Lecture Summary

TOPIC 031: CONSUMER WILLINGNESS TO PAY USING INTEGRALS

The consumers' demand function for a commodity is given by p(q) = 4(25 − q²) dollars per unit. We find the consumer surplus when the consumer is willing to spend 3 units. Given q = 3, the price is p(3) = 4(25 − 9) = 64.

Graphically, the area under the demand curve from 0 to q* forms a trapezoid OqEP, representing consumers' willingness to pay for 3 units. The rectangle **OqEP*** (price × quantity) represents actual expenditure. The triangle P*EP is the consumer surplus.

The area under the curve is the sum of the rectangular area and the triangular area: ∫₀³ p(q) dq = ▯OqEP + ⊿P*EP. Therefore, consumer surplus is the difference between total willingness to pay and actual expenditure.

🔑 Definition — Consumer Surplus (CS): The difference between what consumers are willing to pay for a good (the area under the demand curve) and what they actually pay (market price × quantity). 📐 Formula: CS = ∫₀^(q*) p(q) dq − (p* × q*) 📌 Example: Given p(q) = 4(25 − q²), q* = 3, p* = 64. ∫₀³ 4(25 − q²) dq = 4[25q − q³/3]₀³ = 4[75 − 9] = 264 Actual expenditure = 3 × 64 = 192 CS = 264 − 192 = 72 💡 Why this matters: Consumer surplus measures the net benefit consumers receive from market exchange.

TOPIC 032: CONSUMER SURPLUS USING INTEGRALS

The demand function is P = 80 / ³√Q dollars per unit. Find consumer's surplus at Q = 64. Given Q = 64, P(64) = 80 / ⁴ = 20.

The area under the curve (consumer's willingness to pay) is ∫₀⁶⁴ (80 / Q^(1/3)) dQ. The actual expenditure is the rectangle: 64 × 20 = 1280. The triangular area P*EP is the consumer surplus.

🔑 Definition — Consumer Surplus: The net gain to consumers, calculated as the definite integral of the demand function from 0 to equilibrium quantity minus the expenditure rectangle. 📐 Formula: CS = ∫₀^(Q*) P(Q) dQ − (P* × Q*) 📌 Example: P = 80Q^(-1/3), Q* = 64, P* = 20. ∫₀⁶⁴ 80Q^(-1/3) dQ = 80 × (3/2) Q^(2/3) |₀⁶⁴ = 120 × (64^(2/3)) = 120 × 16 = 1920 Actual expenditure = 64 × 20 = 1280 CS = 1920 − 1280 = 640

TOPIC 033: DIFFERENTIAL OF CONSUMER SURPLUS USING INTEGRALS

A price-discriminating monopolist faces demand P = 60 − 2Q². The monopolist charges two customers their reservation prices: $42 for the first and $10 for the second. Find how much surplus the first customer loses from not being charged the lower price.

For the first customer (P₁ = 42): 42 = 60 − 2Q₁² → Q₁² = 9 → Q₁ = 3. Actual expenditure = 3 × 42 = 126. Willingness to pay = ∫₀³ (60 − 2Q²) dQ = [60Q − (2/3)Q³]₀³ = 180 − 18 = 162. CS₁ = 162 − 126 = 36.

For the second customer (P₂ = 10): 10 = 60 − 2Q₂² → Q₂² = 25 → Q₂ = 5. Actual expenditure = 5 × 10 = 50. Willingness to pay = ∫₀⁵ (60 − 2Q²) dQ = [60Q − (2/3)Q³]₀⁵ = 300 − (2/3)(125) = 300 − 83.33 = 216.667. CS₂ = 216.667 − 50 = 166.667.

Differential of consumer surpluses: ΔCS = CS₂ − CS₁ = 166.667 − 36 = 130.667.

🔑 Definition — Differential of Consumer Surplus: The difference in consumer surplus between two price-quantity situations, measuring the loss (or gain) from price discrimination. 📌 Example: The first customer loses 130.667 in potential surplus compared to the second customer.

TOPIC 034: PRODUCER & SOCIAL SURPLUS USING INTEGRALS

Demand and supply functions for shirts are P = 60 − 0.6Q and P = 20 + 0.2Q. Find total (social) surplus.

Equilibrium: 60 − 0.6Q = 20 + 0.2Q → 40 = 0.8Q → Q* = 50. Equilibrium price: P* = 20 + 0.2(50) = 30.

Both consumer surplus and producer surplus are right triangles. Consumer Surplus: Area of ⊿A P* E* = (1/2) × 50 × 30 = 750. Producer Surplus: Area of ⊿B P* E* = (1/2) × 50 × 10 = 250.

🔑 Definition — Social (Total) Surplus: The sum of consumer surplus and producer surplus, representing the total net benefit to society from a market transaction. 📐 Formula: Social Surplus = Consumer Surplus + Producer Surplus 📌 Example: Social Surplus = 750 + 250 = 1000 units.

TOPIC 035: PRODUCER SURPLUS WITH TAX USING INTEGRALS

A firm's cost function is C(Q) = Q³/3 − 9Q² + 160Q + 20. The market price is 270. Find producer surplus at Q = 12 and the loss in producer surplus when production is reduced to 9 due to taxation.

Marginal cost MC(Q) = dC/dQ = Q² − 18Q + 160. This is treated as the supply curve.

Producer surplus at Q = 12: Revenue = 12 × 270 = 3240. Area under MC = ∫₀¹² (Q² − 18Q + 160) dQ = [Q³/3 − 9Q² + 160Q]₀¹² = (576 − 1296 + 1920) = 1200. Producer surplus = 3240 − 1200 = 2040.

Loss due to tax: Output falls to Q₁ = 9. Revenue decline = 270 × (12 − 9) = 810. Area under MC from 9 to 12 = ∫₉¹² (Q² − 18Q + 160) dQ = [Q³/3 − 9Q² + 160Q]₉¹² = (576 − 1296 + 1920) − (243 − 729 + 1440) = 1200 − 954 = 246. Loss in producer surplus = 810 − 246 = 564.

🔑 Definition — Producer Surplus (PS): The difference between actual revenue (price × quantity) and the area under the marginal cost (supply) curve, representing the net benefit to producers. 📐 Formula: PS = (P* × Q*) − ∫₀^(Q*) MC(Q) dQ 📌 Example: At Q=12, PS = 2040. After tax (Q=9), loss in PS = 564. 💡 Why this matters: Taxation creates a deadweight loss by reducing producer surplus.

⭐ Key Takeaways

Consumer surplus is the area between the demand curve and the market price line, calculated as the definite integral of the demand function from zero to equilibrium quantity minus the expenditure rectangle. Producer surplus is the area between the market price line and the supply (marginal cost) curve, calculated as revenue minus the integral of marginal cost. Social surplus is the sum of consumer and producer surplus, measuring total welfare. Price discrimination increases producer surplus but reduces consumer surplus for those charged higher prices. Taxation reduces producer surplus and output, creating a loss measured by the difference between the decline in revenue and the area under the marginal cost curve between old and new output levels.

🧠 Quick Revision Questions

1. What is the formula for consumer surplus using integrals, and what does each term represent?
2. How do you find the consumer surplus when the demand function is P = 80/³√Q and equilibrium quantity is 64?
3. In the price discrimination example, why does the first customer lose consumer surplus compared to the second?
4. How is social surplus calculated, and what does it measure in welfare economics?
5. If a tax reduces output from 12 to 9 units, how do you compute the loss in producer surplus?

📘 Lecture 10 — FIRM’S ANALYSIS USING INTEGRALS

📖 Overview: This lecture applies integral calculus to analyze various economic aspects of a firm, including the present value of cash flows, the valuation of perpetual income streams, and the wine storage problem with storage costs. It also covers breakeven analysis and production scale analysis using double integrals, demonstrating how integration provides powerful tools for economic decision-making over time.

🗂️ Topics Covered

The lecture covers five main topics: calculating the present value of cash flows using integrals, determining the present value of perpetual income streams, solving the wine storage problem with storage costs using net present value analysis, performing breakeven analysis of a firm by integrating income and expenditure functions, and analyzing scale of production using double integrals with a Cobb-Douglas production function.

📝 Lecture Summary

TOPIC 036: PRESENT VALUE OF CASH FLOW USING INTEGRALS

The present value of a firm's cash flow is represented by π = ∫₀ᵗ R(t)e⁻ʳᵗ dt, where R(t) is the function defining cash flows over time, t is time (usually in years), r is the discount rate, and e⁻ʳᵗ is the exponential-decay expression depicting discounting. To find the present value of a continuous revenue flow lasting y years at a constant rate of D dollars per year discounted at rate r, we substitute t = y and R(t) = D into the formula.

🔑 Definition — Present Value of Cash Flow: The current worth of a stream of future cash flows discounted at a specific rate, calculated as π = ∫₀ᵗ R(t)e⁻ʳᵗ dt. 📐 Formula: π = (D/r)(1 − e⁻ʳʸ) → The present value of a constant cash flow D over y years discounted at rate r. 📌 Example: D = 3000, r = 0.06, y = 2. Substituting: π = (3000/0.06)(1 − e⁻⁰·⁰⁶ײ) = 50000(1 − e⁻⁰·¹²) = 50000(1 − 0.8869) = 50000(0.1131) = 5655. The present value of revenue in constant dollar flows of $3000 over 2 years with a 6% discount rate is approximately $5655.

TOPIC 037: PRESENT VALUE OF PERPETUAL FLOW USING INTEGRALS

If a cash flow persists forever (perpetual), the present value becomes an improper integral: π = ∫₀^∞ R(t)e⁻ʳᵗ dt. For a constant rate of D dollars per year, we evaluate the limit as y approaches infinity.

🔑 Definition — Present Value of Perpetual Flow: The present value of an income stream that continues indefinitely, calculated as the limit of the finite integral as time approaches infinity. 📐 Formula: π_perpetual = D/r → The present value of a perpetual constant cash flow D discounted at rate r. 📌 Example: D = PKR 1450, r = 5% = 0.05. Substituting: π_perpetual = 1450/0.05 = 29000. The present value of a perpetual income stream of PKR 1450 per year discounted at 5% is PKR 29,000.

TOPIC 038: WINE STORAGE PROBLEM WITH STORAGE COSTS USING INTEGRALS

The wine storage problem now includes storage costs. Let C be the purchase cost, V(t) the future sale value, and s the constant storage cost rate per year. The Net Present Value (NPV) of wine is: NPV_wine = Present value of sale − Present value of storage cost − Purchase cost.

🔑 Definition — Net Present Value (NPV): The difference between the present value of future revenues and the present value of all costs, including purchase and storage costs. 📐 Formula: NPV(t) = {V(t) + s/r}e⁻ʳᵗ − s/r − C → The net present value of wine as a function of storage time t. 📌 First-order condition for maximization: V′(t) = rV(t) + s. This condition equates marginal revenue (MR) = V′(t) with marginal cost (MC) = rV(t) + s, where rV(t) represents the interest cost of holding and s is the storage cost per year. The critical value t* maximizes NPV.

💡 Why this matters: The condition V′(t) = rV(t) + s is a modified profit maximization condition that accounts for both opportunity cost (interest forgone) and explicit storage costs, providing the optimal aging time for wine.

TOPIC 039: BREAKEVEN ANALYSIS OF A FIRM USING INTEGRALS

A firm's annual income Y(t) = 250 + 0.5t and expenditure E(t) = t² − 17t + 287.5, where t is time in years. Net savings equals the area under the revenue curve minus the area under the expenditure curve from t = 2.5 to t = 15.

🔑 Definition — Net Savings: The difference between total income and total expenditure over a given time period, calculated by integrating the difference of the functions. 📐 Formula: Net Savings = ∫₂·₅¹⁵ (250 + 0.5t)dt − ∫₂·₅¹⁵ (t² − 17t + 287.5)dt 📌 Example: Evaluating the integrals: = [250t + 0.25t²]₂·₅¹⁵ − [t³/3 − 8.5t² + 287.5t]₂·₅¹⁵ = 325.5 units. The annual net savings of the firm from 2.5 years to 15 years is 325.5 units.

TOPIC 040: SCALE OF PRODUCTION ANALYSIS USING INTEGRALS

Given the Cobb-Douglas production function f(K,L) = K⁰·⁷L⁰·³, the total increase in output when capital and labor increase from 5 to 10 units each is calculated using a double-definite integral.

🔑 Definition — Double Definite Integral: Integration of a function with respect to two variables sequentially, each over specified limits, used to calculate volume under a 3D surface. 📐 Formula: Q(K,L) = ∫₅¹⁰ [∫₅¹⁰ (K⁰·⁷L⁰·³)dK]dL 📌 Example: Step 1: Integrate with respect to K: ∫₅¹⁰ K⁰·⁷L⁰·³ dK = (L⁰·³/1.7)[K¹·⁷]₅¹⁰ = (L⁰·³/1.7)(34.69) = 20.41(L⁰·³). Step 2: Integrate with respect to L: ∫₅¹⁰ 20.41(L⁰·³)dL = 20.41[L¹·³/1.3]₅¹⁰ = 15.7[(10¹·³) − (5¹·³)] = 186.03 units. Thus, the increase in output when capital and labor increase from 5 to 10 units each is 186.03 units.

💡 Why this matters: Double integration allows economists to measure the total change in output when both inputs change simultaneously, capturing the volume under the production surface rather than just area under a curve.

⭐ Key Takeaways

The lecture demonstrates four critical applications of integrals in firm analysis. First, the present value of finite cash flows uses the formula π = (D/r)(1 − e⁻ʳʸ), while perpetual flows simplify to π = D/r. Second, when analyzing wine storage with costs, the net present value must account for purchase cost C, sale value V(t), and storage cost stream s, with the optimal storage time satisfying V′(t) = rV(t) + s. Third, breakeven analysis is performed by integrating the difference between income and expenditure functions over a specified time interval. Fourth, for multi-input production functions like Cobb-Douglas, double definite integrals calculate the total increase in output when both inputs change simultaneously, integrating first with respect to one input then the other.

🧠 Quick Revision Questions

1. What is the formula for the present value of a constant cash flow D over y years discounted at rate r?
2. How does the formula for perpetual cash flows differ from finite cash flows, and what is the simplified result?
3. In the wine storage problem with storage costs, what is the first-order condition for maximizing net present value, and what does each term represent?
4. How is net savings calculated using integrals in breakeven analysis?
5. Describe the process of using a double definite integral to calculate the increase in output when both capital and labor increase.

📘 Lecture 11 — FIRM’S ANALYSIS USING INTEGRALS (CONTINUED 1)

📖 Overview: This lecture continues the application of definite integrals to firm-level economic analysis. It demonstrates how integration is used to calculate total pollution from a cattle feedlot, the increase in crop value over time, total mobile phone production during a specific month, and the firm's average profit and average cost over specified output ranges.

🗂️ Topics Covered

The lecture covers five topics: pollution in cattle fattening analysis using integrals to find pollutant volume; crop farming analysis to find the increase in crop value over time; mobile production analysis to find units produced during a specific month; firm's average profit analysis using integrals to find average profit over an output range; and firm's average cost analysis using integrals to find average cost over an output range.

📝 Lecture Summary

TOPIC 041: POLLUTION IN CATTLE FATTENING ANALYSIS USING INTEGRALS

Cattle fattening involves environmental costs. An intensive cattle-fattening unit adds pollutant (in hundreds of gallons) to a river at the rate of ( r(t) = \sqrt{t} ), where ( t ) is the number of months the unit has been in operation.

(a) The rate of pollution is graphed for t = 0 to 12 months.

(b) To calculate the volume of pollutant in the river during the first six months (t=6), we find the area of triangle OABC minus the area under the curve.

🔑 Definition — Area under the curve as total pollutant: The total pollutant is the area under the rate-of-pollution curve from t=0 to t=6.

📐 Formula: The area of triangle OABC is ( |OA| \times |OC| ), where OA is the rate at t=6 (( \sqrt{6} \approx 2.45 )) and OC is time (( t=6 )). Area of triangle ( = (2.45 \times 6) = 14.7 )

📐 Formula: The area under the curve is ( \int_0^6 \sqrt{t} , dt = \int_0^6 t^{1/2} , dt = \left[ \frac{t^{3/2}}{3/2} \right]_0^6 = \frac{2}{3} t^{3/2} \bigg|_0^6 ) ( = \frac{2}{3} (6^{3/2}) = \frac{2}{3} (14.7) = 9.8 )

📌 Example: The pollutant volume (the triangle minus the area under the curve) is: ( \triangle OBC = 14.7 - 9.8 = 4.9 ) Since units are in hundreds of gallons: ( \triangle OBC = 4.9 \times 100 = 490 ) gallons. 💡 Why this matters: This shows how to find the total accumulated pollutant, which is the difference between a hypothetical constant rate and the actual decreasing rate.

TOPIC 042: CROP FARMING ANALYSIS USING INTEGRALS

It is estimated that ( t ) days from now a farmer’s crop will be increasing at the rate of ( 0.3t^2 + 0.6t + 1 ) bushels per day. The market price remains fixed at $3 per bushel.

🔑 Definition — Rate of change of crop: ( \frac{dC(t)}{dt} = 0.3t^2 + 0.6t + 1 ) bushels per day.

To find how much the value of the crop increases during the next 5 days, we first find the total crop function, ( C(t) ).

📐 Formula: ( C(t) = \int_0^5 (0.3t^2 + 0.6t + 1) , dt ) ( C(t) = \left[ 0.3 \times \frac{t^3}{3} + 0.6 \times \frac{t^2}{2} + t \right]_0^5 = \left[ 0.1t^3 + 0.3t^2 + t \right]_0^5 ) ( C(t) = [0.1(5^3 - 0^3) + 0.3(5^2 - 0^2) + (5 - 0)] = [0.1(125) + 0.3(25) + 5] ) ( C(t) = 12.5 + 7.5 + 5 = 25 ) bushels.

📌 Example: The value of the crop is ( V_C(t) = C(t) \times P = 25 \times $3 = $75 ). Therefore, ( V_C(t) = 0.3t^3 + 0.9t^2 + 3t \bigg|_0^5 = $75 ). Interpretation: The value of the crop will be $75, after 5 months, if the price is $3 per bushel.

TOPIC 043: MOBILE PRODUCTION ANALYSIS USING INTEGRALS

Bejax Corporation's rate of production of new mobiles is ( \frac{dP}{dt} = 1500 \left(2 - \frac{t}{2t+5}\right) ) units/month.

🔑 Definition — Rate of production: The instantaneous rate at which mobile phones are being produced.

To find how many telephones are produced during the third month, we integrate the rate from t=2 to t=3.

📐 Formula: ( P(t) = \int 1500 \left(2 - \frac{t}{2t+5}\right) dt = 1500 \left( \int 2 , dt - \int \frac{t}{2t+5} dt \right) )

The integral ( \int \frac{t}{2t+5} dt ) requires integration by substitution. Let ( u = 2t + 5 ), so ( t = \frac{u-5}{2} ) and ( du = 2 dt ) or ( \frac{du}{2} = dt ). ( \int \frac{t}{2t+5} dt = \int \frac{(u-5)/2}{u} \cdot \frac{du}{2} = \frac{1}{4} \int \frac{u-5}{u} du = \frac{1}{4} \int (1 - \frac{5}{u}) du ) ( = \frac{1}{4} (u - 5 \ln|u|) + c = \frac{1}{4} (2t+5 - 5 \ln|2t+5|) + c' )

Therefore, ( P(t) = 1500 \left( 2t - \frac{1}{4} (2t+5 - 5 \ln|2t+5|) \right) ) ( P(t) = 1500 \left( 2t - \frac{2t+5}{4} + \frac{5}{4} \ln|2t+5| \right) )

📌 Example: For the third month, we evaluate ( \int_2^3 P(t) dt ). Substituting lower limit ( t=2 ) and upper limit ( t=3 ): ( P(3) - P(2) = 1500 [ 2(3) - 2(2) - \left( \frac{2(3)+5}{4} - \frac{2(2)+5}{4} \right) + \frac{5}{4} ( \ln|2(3)+5| - \ln|2(2)+5| ) ] ) ( = 1500 [ 6 - 4 - \left( \frac{11}{4} - \frac{9}{4} \right) + \frac{5}{4} ( \ln 11 - \ln 9 ) ] ) ( = 1500 [ 2 - \frac{2}{4} + \frac{5}{4} \ln \left( \frac{11}{9} \right) ] = 1500 [ \frac{3}{2} + \frac{5}{4} \ln(1.222) ] ) ( = 1500 [ 1.5 + 0.2504 ] = 1500 \times 1.7504 = 2625.6 ) Interpretation: Approximately 2626 telephones will be produced during the third month.

TOPIC 044: FIRM'S AVERAGE PROFIT ANALYSIS USING INTEGRALS

The firm’s average profit when output ( Q ) varies between ( a ) and ( b ) units is given by a definite integral.

🔑 Definition — Average profit: ( \bar{\pi} = \frac{1}{b-a} \int_a^b f(Q) dQ ), where ( f(Q) = TR(Q) - TC(Q) ).

Calculate a firm’s average profit when ( TR = 100(1 - e^{-0.1Q}) ), ( TC = 0.1Q^2 + 2Q + 1 ), and output varies from 3 to 8 units.

📐 Formula: ( \pi = f(Q) = TR - TC = 100 - 100e^{-0.1Q} - 0.1Q^2 - 2Q - 1 ) ( \pi = 99 - 100e^{-0.1Q} - 2Q - 0.1Q^2 )

📌 Example: ( \bar{\pi} = \frac{1}{8-3} \int_3^8 (99 - 100e^{-0.1Q} - 2Q - 0.1Q^2) dQ ) ( \bar{\pi} = \frac{1}{5} [ 99Q - 100 \left( \frac{e^{-0.1Q}}{-0.1} \right) - Q^2 - \frac{0.1}{3} Q^3 ]_3^8 ) ( \bar{\pi} = \frac{1}{5} [ 99Q + 1000 e^{-0.1Q} - Q^2 - 0.03Q^3 ]_3^8 ) ( \bar{\pi} = \frac{1}{5} [ (99(8-3)) + 1000(e^{-0.8} - e^{-0.3}) - (8^2 - 3^2) - 0.03(8^3 - 3^3) ] ) ( \bar{\pi} = \frac{1}{5} [ 495 + 1000(0.449 - 0.741) - 55 - 0.03(485) ] ) ( \bar{\pi} = \frac{1}{5} [ 495 - 292 - 55 - 14.55 ] = \frac{1}{5} [ 133.45 ] = 26.69 ) Interpretation: Between 3 and 8 units of output, the firm makes 26.46 units of average profit.

TOPIC 045: FIRM'S AVERAGE COST ANALYSIS USING INTEGRALS

The cost of producing ( x ) units of a new product is ( C(Q) = 3Q^2 - 2 ) hundred dollars.

🔑 Definition — Average cost: The average cost of producing ( Q ) units from ( a ) to ( b ) is ( AC_{avg} = \frac{1}{b-a} \int_a^b C(Q) dQ ).

📌 Example: Find the average cost of producing 3 to 7 units. ( AC(Q) = \frac{1}{7-3} \int_3^7 (3Q^2 - 2) dQ ) ( AC(Q) = \frac{1}{4} [ 3 \int_3^7 Q^2 dQ - 2 \int_3^7 1 dQ ] ) ( AC(Q) = \frac{1}{4} [ 3 \left( \frac{Q^3}{3} \bigg|_3^7 \right) - 2 (Q|_3^7) ] ) ( AC(Q) = \frac{1}{4} [ (7^3 - 3^3) - 2(7 - 3) ] = \frac{1}{4} [ (343 - 27) - 2(4) ] ) ( AC(Q) = \frac{1}{4} [ 316 - 8 ] = \frac{1}{4} \times 308 = 77 ) Interpretation: The firm will incur 77 units as average cost if we consider production of 3 to 7 units.

⭐ Key Takeaways

The lecture demonstrates four core applications of definite integrals in firm analysis: (1) calculating total accumulated pollutant from a given rate function requires finding the area under the curve within specific time limits; (2) to find the increase in crop or production value over time, integrate the rate of change function and multiply by price; (3) for production during a specific month, integrate the production rate function between the start and end of that month; and (4) average values (profit, cost) over an output range are found by dividing the definite integral of the function by the range width.

🧠 Quick Revision Questions

1. In the pollution analysis, what does the area of triangle OABC represent versus the area under the curve?
2. What is the first step to find the value of crop increase if you are given a rate of change of crop function?
3. For the mobile production problem, what are the lower and upper limits of integration for finding production during the third month?
4. Write the general formula for a firm's average profit when output varies between ( a ) and ( b ) units.
5. In the average cost example, what does the 4 in the denominator ( \frac{1}{4} ) represent?

📘 Lecture 12 — FIRM’S ANALYSIS USING INTEGRALS (CONTINUED 2)

📖 Overview: This lecture continues the application of definite integrals to firm analysis, focusing on how to compute average values of economic functions over specific time intervals. It covers average price, production, and inventory analysis, and introduces the important concepts of net savings and net excess profit, demonstrating how the area between curves can quantify accumulated economic benefits.

🗂️ Topics Covered

The lecture covers five main topics: average food price analysis using integrals, average production analysis using integrals, average inventory analysis using integrals, net savings analysis using integrals, and net excess profit analysis using integrals. Each topic presents a real-world economic scenario and demonstrates the step-by-step application of the definite integral formula for the average value of a function.

📝 Lecture Summary

TOPIC 046: AVERAGE FOOD PRICE ANALYSIS USING INTEGRALS

This topic demonstrates how to find the average price of a commodity over time using the definite integral formula for the average value of a function. Records show that t months after the beginning of the year, the price of ground beef is (P(t) = 0.09t^2 - 0.2t + 4) dollars per pound. The average price during the first 3 months is found by integrating the price function over the interval ([0, 3]).

🔑 Definition — Average Value of a Function: (f_{avg}(x) = \frac{1}{b-a} \int_a^b f(x) dx) 📐 Formula: (P_{avg}(t) = \frac{1}{3-0} \int_0^3 (0.09t^2 - 0.2t + 4) dt) 📌 Example: (\int_0^3 (0.09t^2 - 0.2t + 4) dt = [0.03t^3 - 0.1t^2 + 4t]0^3 = 0.03(27) - 0.1(9) + 4(3) = 0.81 - 0.9 + 12 = 11.91). Then (P{avg}(t) = \frac{1}{3} \times 11.91 = 3.97) dollars per pound. 💡 Why this matters: This technique allows economists to determine the typical price level over a specified period, smoothing out fluctuations.

TOPIC 047: AVERAGE PRODUCTION ANALYSIS USING INTEGRALS

This section applies the same average value formula to a production function, (Q(L) = 500L^{2/3}), where Q is units produced and L is worker-hours. The goal is to find the average product of labor (AP_L) as labor varies from 1,000 to 2,000 worker-hours.

🔑 Definition — Average Product of Labor (AP_L): The average output per unit of labor input over a specific range. 📐 Formula: (Q_{avg}(L) = \frac{1}{2000-1000} \int_{1000}^{2000} (500L^{2/3}) dL) 📌 Example: The integral is evaluated as ( \frac{1}{1000} \times 500 \times \frac{3}{5} [L^{5/3}]_{1000}^{2000} = \frac{3}{10} [(2000)^{5/3} - (1000)^{5/3}] = \frac{3}{10} [2,174,80.21 - 100,000] = \frac{3}{10} [2,074,80.21]). This results in (AP_L = 65,244.06) units.

TOPIC 048: AVERAGE INVENTORY ANALYSIS USING INTEGRALS

This topic shows how to calculate the average inventory level over a year. An inventory of 60,000 kg is used at a constant rate and is exhausted after 1 year. The inventory function is given as (y(t) = -60,000(t-1)). The average inventory (AI_t) over the interval [0,1] is found using the same average value formula.

🔑 Definition — Average Inventory (AI): The mean level of stock held over a given time period. 📐 Formula: (I_{avg}(t) = \frac{1}{1-0} \int_0^1 {-60,000(t-1)} dt) 📌 Example: (AI_t = -60,000 \int_0^1 (t-1) dt = -60,000 [\frac{t^2}{2} - t]_0^1 = -60,000 [(\frac{1}{2} - 0) - (1-0)] = -60,000 [-\frac{1}{2}] = 30,000) kg. 💡 Why this matters: This is crucial for inventory management, helping to determine the typical stock level and related holding costs.

TOPIC 049: NET SAVINGS ANALYSIS USING INTEGRALS

This problem analyzes a hospital's decision to install a solar power system. The savings function is (S(t) = 0.3t + 45.6) and the cost function is (C(t) = 0.5t^2 + 2t). The steps involve: (a) Sketching the graphs of both functions. (b) Finding when savings exceed costs by solving (S(t) = C(t)): (0.5t^2 + 1.7t - 45.6 = 0), which solves to (t = 8) years (ignoring the negative root). The net savings is the area between the curves. (c) Calculating the net savings as the difference of the integrals: (\int_0^8 S(t) dt - \int_0^8 C(t) dt).

🔑 Definition — Net Savings: The total financial benefit, calculated as the integral of savings minus the integral of costs over a period. 📐 Formula: (\Delta OEA = \int_0^8 (0.3t + 45.6) dt - \int_0^8 (0.5t^2 + 2t) dt) 📌 Example: (= 0.3(32) + 45.6(8) - [0.5(170.67) + 2(32)] = 9.6 + 364.8 - [85.34 + 64] = 374.4 - 149.34 = 225.06). Since the data is in $1000s, the net savings is (225.06 \times 1000 = $225,065).

TOPIC 050: NET EXCESS PROFIT ANALYSIS USING INTEGRALS

This topic introduces the concept of net excess profit (NE). It measures the extra profit generated by one investment plan (Plan 2) over another (Plan 1) during the time when Plan 2's rate of profitability is higher. The formula is the definite integral of the difference between the two rates of profitability, (P'_2(t) - P'_1(t)), over the interval where it is positive.

🔑 Definition — Net Excess Profit (NE): The total accumulated extra profit of one investment over another, calculated over the period of its advantage. 📐 Formula: (NE = \int_0^N {P'_2(t) - P'_1(t)} dt) 📌 Example: Given (P'_1(t) = 50 + t^2) and (P'_2(t) = 200 + 5t), the intersection point is found by solving (50 + t^2 = 200 + 5t), which gives (t = 15) years. The net excess profit is: (NE = \int_0^{15} {(200 + 5t) - (50 + t^2)} dt = 1687.5) hundred dollars, or ($168,750). 💡 Why this matters: This is a vital tool for investment appraisal, allowing decision-makers to quantify the comparative advantage of one project over another.

⭐ Key Takeaways

The average value of any function over an interval can be found using the formula (\frac{1}{b-a} \int_a^b f(x) dx), which is applied to price, production, and inventory functions. Net savings and net excess profit are both computed as the area between two curves, which represents the total accumulated benefit or difference over the relevant time period. The critical step in these analyses is first identifying the interval of interest, often found by solving for the intersection points of the two functions. Finally, the result must be interpreted in the context of the problem's units, as demonstrated by converting the net savings from hundreds of dollars to actual dollars.

🧠 Quick Revision Questions

1. What is the formula for the average value of a function (f(x)) over the interval ([a, b])?
2. If the price of a good is given by (P(t) = 2t + 5), what is the average price from (t=0) to (t=4)?
3. How do you find the time period during which net savings are positive in a cost-benefit analysis?
4. What is the graphical interpretation of net excess profit between two investment plans?
5. If the net excess profit is calculated as 2500 hundred dollars, what is the value in actual dollars?

📘 Lecture 13 — FIRM’S ANALYSIS USING INTEGRALS (CONTINUED 3)

📖 Overview: This lecture continues the application of definite integrals to firm analysis. It covers the Gini Index as a measure of income inequality using Lorenz curves, and then applies integration to analyze average monthly sales, employee efficiency over time, advertising campaign reach, and the net profit over the useful life of a machine. These practical examples demonstrate how calculus is used to make informed business and economic decisions.

🗂️ Topics Covered

This lecture presents five distinct applications of definite integrals in economic analysis: calculating the Gini Index for income distribution comparison; finding average monthly sales after a product launch; determining the average efficiency rate of a new employee; measuring the number of people reached by an advertising campaign over time; and computing the net profit generated by a machine over its useful life, defined by the point where marginal revenue equals marginal cost.

📝 Lecture Summary

TOPIC 051: GINI INDEX USING INTEGRALS

Area plays an important role in the study of Lorentz curves, which are used by economists to measure income inequality. A Lorentz curve is a graph of the function L(x), which plots the cumulative percentage of income against the cumulative percentage of the population. The ratio of the area between the line of perfect equality (y = x) and the Lorentz curve to the total area under the line of perfect equality is used as a measure of inequality. This ratio is called the Gini index, denoted GI (also called the index of income inequality).

🔑 Definition — Gini Index (GI): A measure of income inequality within a population, derived from the Lorentz curve, ranging from 0 (perfect equality) to 1 (perfect inequality).

📐 Formula: $$GI = \frac{\text{area between } y=L(x) \text{ and } y=x}{\text{area under } y=x \text{ over } 0 \le x \le 1} = 2 \int_0^1 [x - L(x)] dx$$ → The Gini index is twice the area between the line of perfect equality (y=x) and the Lorentz curve L(x).

📌 Example: A governmental agency determines that the Lorentz curves for the distribution of income for dentists and contractors in a certain state are given by the functions L₁(x) = x¹·⁷ and L₂(x) = 0.8x² + 0.2x respectively. For which profession is the distribution of income more fairly distributed?

Solution: Calculate and compare the Gini Index for both professions.

Step 1: Gini Index for Dentists (GI_dentists) $$GI_{dentists} = 2 \int_0^1 {x - L_1(x)} dx = 2 \int_0^1 {x - x^{1.7}} dx$$ $$= 2 \left[ \left(\frac{x^2}{2}\right)_0^1 - \left(\frac{x^{2.7}}{2.7}\right)_0^1 \right] = 2 \left[ \left(\frac{1}{2} - 0\right) - \left(\frac{1}{2.7} - 0\right) \right]$$ $$= 2 \left[ \frac{1}{2} - \frac{1}{2.7} \right] = 0.2593$$

Step 2: Gini Index for Contractors (GI_contractors) $$GI_{contractors} = 2 \int_0^1 {x - (0.8x^2 + 0.2x)} dx = 2 \int_0^1 {0.8x - 0.8x^2} dx$$ $$= 1.6 \int_0^1 (x - x^2) dx = 1.6 \left[ \left(\frac{x^2}{2}\right)_0^1 - \left(\frac{x^3}{3}\right)_0^1 \right]$$ $$= 1.6 \left[ \frac{1}{2} - \frac{1}{3} \right] = 0.2667$$

Interpretation: As GI_dentists = 0.2593 and GI_contractors = 0.2667, and it is evident that GI_dentists < GI_contractors. It is straightforward to suggest that income inequality is higher among contractors as measured through the Gini Index. 💡 Why this matters: A lower Gini index indicates a more equal distribution of income, which is crucial for assessing economic fairness and social welfare.

TOPIC 052: SALES ANALYSIS USING INTEGRALS

A manufacturer determines that t months after introducing a new product, the company’s sales will be S(t) thousand dollars, where $$S(t) = \frac{750t}{\sqrt{4t^2 + 25}}$$ What are the average monthly sales of the company over the first 6 months after the introduction of the new product?

Solution: The time span for the average is the first 6 months, so the interval is [0, 6]. The average value of a function over an interval [a, b] is defined as: $$f_{avg}(x) = \frac{1}{b - a} \int_a^b f(x) dx$$

$$S_{avg}(t) = \frac{1}{6 - 0} \int_0^6 \frac{750t}{\sqrt{4t^2 + 25}} dt = \frac{1}{6} \times 750 \int_0^6 \frac{t}{\sqrt{4t^2 + 25}} dt$$ Let X = 750 × this integral, then S_avg(t) = (1/6) × X. Solve X using integration by substitution:

• Let u = 4t² + 25. Then du/dt = 8t, so t dt = du / 8.
• Change limits: When t = 0 ⇒ u = 4(0)² + 25 = 25; When t = 6 ⇒ u = 4(6)² + 25 = 169.

$$X = 750 \int_{25}^{169} \frac{du/8}{\sqrt{u}} = \frac{750}{8} \int_{25}^{169} u^{-1/2} du = 93.75 \left[ \frac{u^{1/2}}{1/2} \right]_{25}^{169}$$ $$X = 187.5 \left[ \sqrt{169} - \sqrt{25} \right] = 187.5 \times (13 - 5) = 1500$$

Now, substitute X back into the average formula: $$S_{avg}(t) = \frac{1}{6} \times (1500) = 250 \text{ thousand dollars}$$

Interpretation: Thus, for the 6-month period immediately after the introduction of the new product, the company’s sales average $250,000 per month.

TOPIC 053: EFFICIENCY ANALYSIS USING INTEGRALS

After t months on the job, a postal clerk can sort Q(t) = 700 - 400e^(-0.5t) letters per hour. What is the average rate at which the clerk sorts mail during the first 3 months on the job?

Solution: The sorting-efficiency function is given. We find the average efficiency over the interval [0, 3] using the formula for the average of a function.

$$Q_{avg}(t) = \frac{1}{3 - 0} \int_0^3 (700 - 400e^{-0.5t}) dt$$ $$= \frac{1}{3} \left{ 700 \int_0^3 1 dt - 400 \int_0^3 e^{-0.5t} dt \right}$$ $$= \frac{1}{3} \left{ 700(t]_0^3) - 400 \left(\frac{e^{-0.5t}}{-0.5}\right)_0^3 \right}$$ $$= \frac{1}{3} \left{ 700(3-0) + 800(e^{-0.5(3)} - e^{-0.5(0)}) \right}$$

$$= \frac{1}{3} \left{ 2100 + 800(e^{-1.5} - e^0) \right} = \frac{1}{3} \left{ 2100 + 800(0.223 - 1) \right}$$ $$= \frac{1}{3} \left{ 2100 + 800(-0.777) \right} = \frac{1}{3} (2100 - 621.6) = \frac{1}{3} (1478.4)$$

$$Q_{avg}(t) = 492.83$$

Interpretation: The average efficiency of the clerk in sorting mail during the first 3 months is 492.83 letters per hour.

TOPIC 054: ADVERTISEMENT ANALYSIS USING INTEGRALS

An advertising agency begins a campaign to promote a new product and determines that t days later, the number of people N(t) who have heard about the product is changing at a rate given by N'(t) = 5t² - (0.04t) / (t²+3) people per day. How many people learn about the product during the first week? During the second week?

Solution: The rate of change of people hearing the product is N'(t). For the first week (days 0 to 7), we integrate N'(t) over the interval [0, 7].

$$N(t)_{first\ week} = \int_0^7 \left(5t^2 - \frac{0.04t}{t^2 + 3}\right) dt = 5 \int_0^7 t^2 dt - 0.04 \int_0^7 \frac{t}{t^2 + 3} dt$$ Applying the rule ∫ f'(x) / f(x) dx = ln|f(x)|, we note that the derivative of the denominator (t²+3) is 2t. So we multiply the numerator by 2 to match.

$$N(t) = 5 \left(\frac{t^3}{3}\right)_0^7 - 0.02 \int_0^7 \frac{2t}{t^2+3} dt$$ $$= 5 \left(\frac{7^3 - 0}{3}\right) - 0.02 \left( \ln|t^2+3| \right)_0^7$$ $$= \frac{5}{3}(343) - 0.02 \left{ \ln(7^2+3) - \ln(0^2+3) \right}$$ $$= 571.67 - 0.02 (\ln(52) - \ln(3)) = 571.67 - 0.02(3.95 - 1.10)$$ $$= 571.67 - 0.02(2.85) = 571.67 - 0.057 \approx 572$$

Interpretation: During the first week (seven days), approximately 572 people will learn about the new product.

For the second week, the interval is [7, 14]. $$N(t)_{second\ week} = \int_7^{14} \left(5t^2 - \frac{0.04t}{t^2 + 3}\right) dt = \frac{5}{3}(14^3 - 7^3) - 0.02(\ln|14^2+3| - \ln|7^2+3|)$$ $$= \frac{5}{3}(2744 - 343) - 0.02(\ln(199) - \ln(52)) = 4001.67 - 0.02(5.29 - 3.95)$$ $$= 4001.67 - 0.0268 \approx 4002$$

All figures are approximate. N(t)_second week ≈ 4002.

Percentage increase: $$%\Delta(N(t)) = \frac{4002 - 572}{572} \times 100 = \frac{3430}{572} \times 100 \approx 600%$$

Interpretation: It is noticeable that there is a rapid increase in the number of people getting aware of the product during the 2nd week. The advertisement campaign appears a success.

TOPIC 055: PROFIT OVER USEFUL LIFE OF MACHINE USING INTEGRALS

Suppose that t years after being put into use, a machine has generated total revenue R(t) and total cost C(t). Total profit is P(t) = R(t) - C(t). Profit declines when C'(t) > R'(t). The optimal time to dispose of the machine, T, is when C'(T) = R'(T). The time period 0 ≤ t ≤ T is called the useful life of the machine.

📌 Example: A machine generates revenue at a rate R'(t) = 5,000 - 20t² dollars per year and operating costs accumulate at a rate C'(t) = 2,000 + 10t² dollars per year. a) What is the useful life of this machine? b) Compute the net profit generated by the machine over its period of useful life.

Part (a) Solution: The useful life, T, is found where R'(t) = C'(t). $$5,000 - 20t^2 = 2,000 + 10t^2$$ $$\implies 30t^2 = 3,000 \implies t^2 = 100$$ $$t = 10 \text{ years (since t ≥ 0)}$$

Interpretation: The useful life of the machine is 10 years. This gives the interval [0, 10].

Part (b) Solution: Find the net profit over the interval [0, 10]. Net profit is the integral of the profit function (Revenue - Cost). $$\int_0^{10} \pi(t) dt = \int_0^{10} {R'(t) - C'(t)} dt$$ $$= \int_0^{10} {(5,000 - 20t^2) - (2,000 + 10t^2)} dt$$ $$= \int_0^{10} (3,000 - 30t^2) dt = 3,000 \int_0^{10} 1 dt - 30 \int_0^{10} t^2 dt$$ $$= 3,000 (10) - 30 \left( \frac{10^3}{3} \right) = 30,000 - 10,000$$

$$\int_0^{10} \pi(t) dt = 20,000$$

Interpretation: The profit generated by the machine over its useful life (10 years) is $20,000.

⭐ Key Takeaways

1. The Gini Index is a crucial measure of income inequality, calculated as GI = 2∫[x - L(x)]dx; a lower value indicates a more equal distribution of income.
2. The average value of a function over an interval [a, b] is found using the formula f_avg = 1/(b-a) ∫ f(x) dx, which is used to find average sales, efficiency, etc., over time.
3. To find the total number of people reached by an advertisement, you must integrate the rate of change function, N'(t) over the desired time interval.
4. The useful life of a machine is the time T when the marginal revenue equals marginal cost R'(T) = C'(T); the total net profit over that useful life is the integral of the difference between these rates from 0 to T.
5. Complex integrands often require integration by substitution to evaluate, such as when the numerator is a constant multiple of the derivative of the denominator.

🧠 Quick Revision Questions

1. What is the formula for calculating the Gini index, and what does its value represent?
2. How is the "useful life" of a machine defined in terms of marginal revenue and marginal cost?
3. A company's sales are given by S(t) = 100t. What is the formula to find the average monthly sales over the first 12 months?
4. In the advertisement problem, what mathematical operation is applied to the rate function N'(t) to find the number of people who hear about the product during a specific week?
5. If a Lorentz curve is L(x) = x, what is the value of the Gini index, and what does this imply about income distribution?

📘 Lecture 14 — FIRM’S ANALYSIS USING INTEGRALS (CONTINUED 4)

📖 Overview: This lecture applies definite integrals to solve five distinct economic problems: fundraising net earnings analysis, retirement annuity valuation, wage differential calculation, energy resource depletion timing, and present value of a gold mine. Each application demonstrates how integration quantifies total accumulated amounts over time from given rate functions.

🗂️ Topics Covered

The lecture covers five economic applications of integrals: Fund-Raising Net Earnings Analysis (Topic 056), Retirement Annuity Analysis (Topic 057), Wage Differential Using Integrals (Topic 058), Depletion of Energy Resources Analysis (Topic 059), and Present Value of Gold Mine Using Integrals (Topic 060). Each topic involves setting up and evaluating definite integrals with exponential functions, often combined with natural logarithms to solve for time variables.

📝 Lecture Summary

TOPIC 056: FUND-RAISING NET EARNINGS ANALYSIS USING INTEGRALS

It is estimated that t weeks from now, contributions in response to a fundraising campaign will be coming in at the rate of R'(t) = 6,537e⁻⁰·³ᵗ dollars per week, while campaign expenses accumulate at the constant rate of $593 per week. The analysis finds how many weeks the rate of revenue exceeds the rate of cost, and the net earnings generated during that period.

🔑 Definition — Net Earnings (NE): The difference between total revenue and total cost over a time period, calculated as the area between the revenue rate curve and the cost rate curve. 📐 Formula: NE(t) = R(t) − C(t) over [0, t*] → ⊿ABC = ∫₀ᵗ [R'(t) − C'(t)] dt* 📌 Example: Given R'(t) = 6,537e⁻⁰·³ᵗ and C'(t) = 593:

1. Find t*: Equate 6,537e⁻⁰·³ᵗ = 593 → e⁻⁰·³ᵗ = 593/6,537 = 0.0907 → ln(e⁻⁰·³ᵗ) = ln(0.0907) → -0.3t = -2.4 → t = 8 weeks*
2. Net earnings: ⊿ABC = ∫₀⁸ (6,537e⁻⁰·³ᵗ − 593) dt = 6,537∫₀⁸ e⁻⁰·³ᵗ dt − 593∫₀⁸ dt = 6,537[e⁻⁰·³ᵗ/(-0.3)]₀⁸ − 593[t]₀⁸ = -21,790[e⁻²·⁴ − e⁰] − 4,744 = -21,790(0.0907 − 1) − 4,744 = $15,069.647 💡 Why this matters: The rate of revenue exceeds cost for exactly 8 weeks, generating net earnings of $15,069.65.

TOPIC 057: RETIREMENT ANNUITY ANALYSIS USING INTEGRALS

Money transferred continuously into an account over time 0 ≤ t ≤ T at rate f(t), earning interest at annual rate r compounded continuously, has a future value (FV) given by a definite integral.

🔑 Definition — Future Value (FV) of an Income Stream: The total accumulated value at time T of continuous payments earning continuously compounded interest. 📐 Formula: FV = ∫₀ᵀ f(t)eʳ⁽ᵀ⁻ᵗ⁾ dt = eʳᵀ ∫₀ᵀ f(t)e⁻ʳᵗ dt 📌 Example: Tom, age 25, deposits $2,500/year into an IRA at 5% annual rate compounded continuously. Retirement at age 60 means T = 35 years, r = 0.05, f(t) = 2,500: FV = e⁰·⁰⁵ˣ³⁵ ∫₀³⁵ 2,500e⁻⁰·⁰⁵ᵗ dt = e¹·⁷⁵ × 2,500 ∫₀³⁵ e⁻⁰·⁰⁵ᵗ dt = 5.7546 × 2,500[e⁻⁰·⁰⁵ᵗ/(-0.05)]₀³⁵ = -287,730[e⁻¹·⁷⁵ − e⁰] = -287,730(0.1738 − 1) = $237,730.02 💡 Why this matters: Tom will have $237,730.02 in his IRA at age 60 from continuous deposits and compounding.

TOPIC 058: WAGE DIFFERENTIAL USING INTEGRALS

An individual chooses between high school education (salary: g(t) = 8,000e⁰·⁰⁵ᵗ) and undergraduate education (salary: f(t) = 12,000e⁽⁰·⁰⁵⁺⁰·⁰²⁾ᵗ). The difference in total salary over 25 years is calculated.

🔑 Definition — Wage Differential (WD): The difference in total earnings from two different education paths over a career. 📐 Formula: WD = ∫₀²⁵ f(t) dt − ∫₀²⁵ g(t) dt 📌 Example: g(t) = 8,000e⁰·⁰⁵ᵗ, f(t) = 12,000e⁰·⁰⁷ᵗ for 25 years:

1. High school total: ∫₀²⁵ 8,000e⁰·⁰⁵ᵗ dt = 160,000[e⁰·⁰⁵ᵗ]₀²⁵ = 160,000(e¹·²⁵ − 1) = 160,000(3.49 − 1) = $398,400
2. Undergraduate total: ∫₀²⁵ 12,000e⁰·⁰⁷ᵗ dt = 171,428.57[e⁰·⁰⁷ᵗ]₀²⁵ = 171,428.57(e¹·⁷⁵ − 1) = 171,428.57(5.75 − 1) = $814,285.71
3. WD = 814,285.71 − 398,400 = $415,885.71 💡 Why this matters: An undergraduate degree earns an additional $415,885.71 over 25 years compared to high school only.

TOPIC 059: DEPLETION OF ENERGY RESOURCES ANALYSIS USING INTEGRALS

Oil is pumped at rate P'(t) = 1.3e⁰·⁰⁴ᵗ billion barrels/year from a field with 20 billion barrels reserve. The analysis finds P(t), oil pumped in first 3 years, and time T when the field runs dry.

📌 Example: Rate P'(t) = 1.3e⁰·⁰⁴ᵗ, reserve = 20 billion barrels:

1. Oil in first 3 years: ∫₀³ 1.3e⁰·⁰⁴ᵗ dt = 32.5[e⁰·⁰⁴ᵗ]₀³ = 32.5(e⁰·¹² − 1) = 32.5(1.127 − 1) = 4.13 billion barrels
2. Find P(t): P(t) = ∫ P'(t) dt = 32.5e⁰·⁰⁴ᵗ + c. With P(0) = 0: 0 = 32.5(1) + c → c = -32.5, so P(t) = 32.5e⁰·⁰⁴ᵗ − 32.5
3. Time to run dry: Set P(t) = 20: 32.5e⁰·⁰⁴ᵗ − 32.5 = 20 → 32.5e⁰·⁰⁴ᵗ = 52.5 → e⁰·⁰⁴ᵗ = 1.615 → ln(e⁰·⁰⁴ᵗ) = ln(1.615) → 0.04t = 0.479 → t = 11.975 ≈ 12 years 💡 Why this matters: The oil field will be depleted in about 12 years, pumping 4.13 billion barrels in the first 3 years.

TOPIC 060: PRESENT VALUE OF GOLD MINE USING INTEGRALS

Given income P(t) = 600e⁻⁰·⁵ᵗ from a gold mine, with money discounted at 10% per year over 20 years, the present value (PV) is calculated.

🔑 Definition — Present Value (PV) of an Income Stream: The current worth of future income discounted at a continuous rate r over time. 📐 Formula: PV = ∫₀ᵀ R(t)e⁻ʳᵗ dt 📌 Example: R(t) = 600e⁻⁰·⁵ᵗ, r = 0.1, T = 20: PV = ∫₀²⁰ 600e⁻⁰·⁵ᵗ × e⁻⁰·¹ᵗ dt = 600 ∫₀²⁰ e⁻⁰·⁶ᵗ dt = 600[e⁻⁰·⁶ᵗ/(-0.6)]₀²⁰ = -1,000(e⁻¹² − e⁰) = -1,000(6.14×10⁻⁶ − 1) = -1,000(-0.99999386) = $999.99 thousand ≈ $1,000 thousand ($1,000,000) 💡 Why this matters: The gold mine is worth approximately $1,000,000 in present value terms over 20 years of operation.

⭐ Key Takeaways

The central concept is that definite integrals quantify total accumulation from rate functions in economic contexts. For fundraising, equate revenue and cost rates to find the crossover time, then integrate the difference. Future value of continuous annuities uses the formula FV = eʳᵀ∫₀ᵀ f(t)e⁻ʳᵗ dt, while present value follows PV = ∫₀ᵀ R(t)e⁻ʳᵗ dt. Wage differentials are found by subtracting integrals of salary functions over a career. Resource depletion requires first finding the accumulation function P(t) from P'(t) using initial conditions, then solving for T when P(T) equals the reserve. All solutions involve integration of exponential functions and often require natural logarithms to solve for time variables.

🧠 Quick Revision Questions

1. What is the formula for net earnings from a fundraising campaign given revenue rate R'(t) and cost rate C'(t) over period [0, t*]?
2. In the retirement annuity formula FV = eʳᵀ∫₀ᵀ f(t)e⁻ʳᵗ dt, what does each term represent?
3. How do you find the time when an oil field runs dry given the pumping rate P'(t) and total reserve?
4. What is the wage differential formula when comparing two salary functions f(t) and g(t) over T years?
5. In the present value of a gold mine calculation, why is the exponent in the integrand the sum of the income rate exponent and the discount rate?

📘 Lecture 15 — FIRM’S ANALYSIS USING INTEGRALS (CONTINUED 5)

📖 Overview: This lecture demonstrates how definite integrals are applied to various real-world economic problems, including lottery payout valuation, trade deficit analysis, ticket sales forecasting, oil demand comparison, and total utility calculation from marginal utility. Understanding these applications is crucial for economic modeling and decision-making where accumulated changes over time need to be quantified.

🗂️ Topics Covered

This lecture covers five distinct applications of integral calculus in economics: analyzing lottery payout options using present value of continuous income streams, calculating the change in a country's trade deficit by integrating rates of imports and exports, determining ticket sales at a county fair by integrating visitor entry rates, comparing oil demand between two different years using integration of a demand rate function, and finding the increase in total utility from marginal utility when consumption changes from 16 to 100 units.

📝 Lecture Summary

TOPIC 061: LOTTERY PAYOUT ANALYSIS USING INTEGRALS

The winner of a state lottery is offered a choice between receiving $10 million now as a lump sum or receiving A dollars per year for the next 6 years as a continuous income stream. With an annual interest rate of 5% compounded continuously, the problem finds A such that both options have equal present value. The equation $10 = \int_{0}^{6} A e^{-0.05t} dt$ is set up and solved. Evaluating the integral gives $-0.5 = A(e^{-0.3} - 1)$, which simplifies to $-0.5 = A(-0.259)$, yielding A = $1.931 million.

💡 Why this matters: This analysis shows how to determine an equivalent continuous payment stream that makes a lottery winner indifferent between a lump sum and periodic payments.

🔑 Definition — Present Value of Continuous Income Stream: The present value of a continuous income stream of $A$ per year paid for $T$ years at interest rate $r$ compounded continuously is $PV = \int_{0}^{T} A e^{-rt} dt$.

📐 Formula: $\int e^{kt} dt = \frac{e^{kt}}{k}$ → Used to integrate the exponential discount factor.

📌 Example: $10 = A \int_{0}^{6} e^{-0.05t} dt$. Calculation: $10 = A[\frac{e^{-0.05t}}{-0.05}]_{0}^{6} = \frac{A}{-0.05}(e^{-0.3} - 1)$. Then $-0.5 = A(0.741 - 1) = A(-0.259)$, so $A = 1.931$ million. The winner should receive $1.931 million annually for 6 years.

TOPIC 062: BALANCE OF TRADE ANALYSIS USING INTEGRALS

The government estimates that t years from now, imports will increase at rate $I'(t) = 12.5e^{0.2t}$ and exports at rate $E'(t) = 1.7t + 3$, both in billions of dollars per year. The trade deficit is $D(t) = I(t) - E(t)$. To find the change in trade deficit over the next 5 years, the rates are integrated from $t=0$ to $t=5$. The import integral: $\int_{0}^{5} 12.5e^{0.2t} dt = 62.5(e^{1} - e^{0}) = 62.5(2.718 - 1) = 62.5(1.718) = 107.375$ billion. The export integral: $\int_{0}^{5} (1.7t + 3) dt = 1.7(\frac{25}{2}) + 3(5) = 1.7(12.5) + 15 = 21.25 + 15 = 36.25$ billion. The trade deficit is $D(t) = 107.375 - 36.25 = 71.125$ billion.

💡 Why this matters: This demonstrates how integrating rates of change over time gives the total accumulated change in economic aggregates like trade deficits.

🔑 Definition — Trade Deficit: The difference between a country's total imports and total exports over a period, calculated as $D(t) = I(t) - E(t)$.

📐 Formula: $\int e^{kt} dt = \frac{e^{kt}}{k}$ and $\int (at + b) dt = a\frac{t^{2}}{2} + bt + C$ → Used to integrate import and export rate functions.

📌 Example: Imports: $\int_{0}^{5} 12.5e^{0.2t} dt = 12.5[\frac{e^{0.2t}}{0.2}]{0}^{5} = 62.5(e^{1} - 1) = 62.5 \times 1.718 = 107.375$. Exports: $\int{0}^{5} (1.7t + 3) dt = 1.7[\frac{t^{2}}{2}]{0}^{5} + 3[t]{0}^{5} = 1.7(12.5) + 15 = 36.25$. Deficit: $107.375 - 36.25 = 71.125$ billion dollars.

TOPIC 063: TICKET SALES ANALYSIS USING INTEGRALS

The promoters estimate that t hours after gates open at 9:00 A.M., visitors enter at the rate of $-4(t+2)^{3} + 54(t+2)^{2}$ people per hour. To find how many people enter between 10:00 A.M. and noon, the time mapping is: 9:00 A.M. ($t=0$), 10:00 A.M. ($t=1$), noon ($t=3$). The ticket sales function $TS'(t) = VE'(t) = -4(t+2)^{3} + 54(t+2)^{2}$ is integrated from $t=1$ to $t=3$. The integral is solved by substitution: $\int_{1}^{3} -4(t+2)^{3} dt + \int_{1}^{3} 54(t+2)^{2} dt = -4[\frac{(t+2)^{4}}{4}]{1}^{3} + 54[\frac{(t+2)^{3}}{3}]{1}^{3} = -[(5)^{4} - (3)^{4}] + 18[(5)^{3} - (3)^{3}] = -(625 - 81) + 18(125 - 27) = -544 + 18(98) = -544 + 1764 = 1220$ people.

💡 Why this matters: This shows how integrals accumulate instantaneous rates to find total quantities over time intervals, applicable to any flow variable.

🔑 Definition — Rate of Ticket Sales: The function $TS'(t)$ representing the number of visitors entering per hour at time $t$, which when integrated gives total tickets sold.

📐 Formula: $\int (t + a)^{n} dt = \frac{(t + a)^{n+1}}{n+1} + C$ → Used for polynomial rate functions.

📌 Example: $TS = \int_{1}^{3} [-4(t+2)^{3} + 54(t+2)^{2}] dt = -4[\frac{(t+2)^{4}}{4}]{1}^{3} + 54[\frac{(t+2)^{3}}{3}]{1}^{3} = -[(5)^{4} - (3)^{4}] + 18[(5)^{3} - (3)^{3}] = -(625 - 81) + 18(125 - 27) = -544 + 1764 = 1220$.

TOPIC 064: DEMAND ANALYSIS OF OIL USING INTEGRALS

It is estimated that t years from the beginning of 2005, the demand for oil changes at rate $D'(t) = (1 + 2t)^{-1}$ billion barrels per year. To compare demand in 2006 ($t=1$ to $t=2$) versus 2009 ($t=4$ to $t=5$), the integrals are computed. For 2006: $\int_{1}^{2} \frac{1}{1+2t} dt = \frac{1}{2}[\ln|1+2t|]{1}^{2} = \frac{1}{2}(\ln 5 - \ln 3) = \frac{1}{2}(1.6094 - 1.0986) = 0.2554$ billion barrels (255.4 million barrels). For 2009: $\int{4}^{5} \frac{1}{1+2t} dt = \frac{1}{2}[\ln|1+2t|]_{4}^{5} = \frac{1}{2}(\ln 11 - \ln 9) = \frac{1}{2}(2.3979 - 2.1972) = 0.1003$ billion barrels (100.3 million barrels). Comparative analysis: 2006 demand (255.4 million) is greater than 2009 demand (100.3 million).

💡 Why this matters: This demonstrates how integrals can compare cumulative quantities between different time periods for resource planning.

🔑 Definition — Demand Rate Function: $D'(t) = (1 + 2t)^{-1}$ representing the instantaneous rate of change of oil demand at time $t$.

📐 Formula: $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C$ → Used for integration of rational functions where numerator is derivative of denominator.

📌 Example: For 2006: $\int_{1}^{2} \frac{1}{1+2t} dt$. Multiply and divide by 2: $\frac{1}{2}\int_{1}^{2} \frac{2}{1+2t} dt = \frac{1}{2}[\ln|1+2t|]_{1}^{2} = \frac{1}{2}(\ln 5 - \ln 3) = \frac{1}{2}(0.5108) = 0.2554$ billion barrels.

TOPIC 065: MARGINAL TO TOTAL UTILITY USING INTEGRALS

Given the marginal utility function $MU_x = \frac{3x^{2} + 6\sqrt{x} + 20}{x}$, where $x$ is the number of units consumed, we find the increase in total utility when $x$ goes from 16 to 100 units. The integral is: $\int_{16}^{100} \frac{3x^{2} + 6\sqrt{x} + 20}{x} dx = \int_{16}^{100} (3x + 6x^{-\frac{1}{2}} + \frac{20}{x}) dx$. This separates into three integrals: $3\int_{16}^{100} x dx = 3[\frac{x^{2}}{2}]{16}^{100} = 3(5000 - 128) = 3 \times 4872 = 14616$. The second: $6\int{16}^{100} x^{-\frac{1}{2}} dx = 6[2x^{\frac{1}{2}}]{16}^{100} = 12(10 - 4) = 12 \times 6 = 72$. The third: $20\int{16}^{100} \frac{1}{x} dx = 20[\ln x]_{16}^{100} = 20(4.6052 - 2.7726) = 20 \times 1.8326 = 36.652$. Total: $14616 + 72 + 36.652 = 14724.652$.

💡 Why this matters: This illustrates how marginal utility can be integrated to find the total utility change over a consumption range, a fundamental tool in microeconomic analysis.

🔑 Definition — Marginal Utility: The additional utility derived from consuming one more unit of a good, given by $MU_x$, which when integrated yields total utility $TU(x)$.

📐 Formulas: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, $\int \frac{1}{x} dx = \ln|x| + C$, $\int x^{-\frac{1}{2}} dx = 2x^{\frac{1}{2}} + C$ → Used for each term of the decomposed function.

📌 Example: $\int_{16}^{100} \frac{3x^{2} + 6\sqrt{x} + 20}{x} dx = 3[\frac{x^{2}}{2}]{16}^{100} + 6[2\sqrt{x}]{16}^{100} + 20[\ln x]_{16}^{100} = 3(5000 - 128) + 12(10 - 4) + 20(4.6052 - 2.7726) = 14616 + 72 + 36.652 = 14724.652$.

⭐ Key Takeaways

The five applications in this lecture demonstrate that definite integrals are essential for converting rates of change into total accumulated values across various economic contexts — from present value of income streams to trade deficits, ticket sales, oil demand forecasting, and utility analysis. The critical skills include setting up correct integration limits based on time mappings, applying appropriate integration rules (exponential, polynomial, and logarithmic), and interpreting results in economic terms. The trade deficit, ticket sales, and demand examples all require carefully converting problem time frames into integration limits. The utility example shows how to decompose complex marginal utility functions into simpler integrable components. Students must remember to multiply or divide by constants when using the logarithmic integration rule $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)|$ to adjust for non-unit derivatives in the denominator.

🧠 Quick Revision Questions

1. In the lottery payout problem, why is the discount factor $e^{-0.05t}$ and what does the integral $\int_{0}^{6} A e^{-0.05t} dt$ represent?
2. How do you determine whether the trade deficit in Topic 062 increased or decreased over the 5-year period?
3. In the ticket sales problem, what time mapping converts 9:00 A.M. to noon into integration limits, and why is the lower limit 1 instead of 0?
4. Why is the demand for oil higher in 2006 than in 2009 when the rate function $D'(t) = (1+2t)^{-1}$ is decreasing over time?
5. In the utility problem, how do you decompose $\frac{3x^{2} + 6\sqrt{x} + 20}{x}$ into three separate integrals, and what rule is used to integrate the $\frac{20}{x}$ term?

📘 Lecture 16 — FIRM’S ANALYSIS USING INTEGRALS (CONTINUED 6)

📖 Overview: This lecture explores how integrals can be applied to real-world business and economic problems. It focuses on using integration to analyze unsold housing inventory over time, calculate total labor costs from marginal cost functions, measure asset depreciation rates, and compare the long-term cost-effectiveness of different investment options. These applications demonstrate the power of calculus in managerial decision-making.

🗂️ Topics Covered

The lecture covers four distinct applications of integrals: analyzing the total number of unsold homes in a real estate market where existing and new homes are added at different rates; deriving the total labor cost function from a marginal labor cost function and calculating the cost of employing a range of labor units; determining the depreciation of an industrial machine over a specific year using its rate of value change; and comparing the total cost of two alternative heating systems over a ten-year period to determine which is more cost-effective.

📝 Lecture Summary

TOPIC 066: REAL ESTATE ANALYSIS USING INTEGRALS

In a certain community, the fraction of homes that remain unsold for at least t weeks is approximately f(t) = e^(-0.2t). If 200 homes are currently on the market, the function for existing unsold homes is g(t) = 200e^(-0.2t). Additionally, new homes are placed on the market at a rate of 8 per week, so the function for unsold new homes is h(t) = 8e^(-0.2(10-t)). The goal is to find the total number of unsold homes after 10 weeks.

The total unsold homes U(t) is the sum of the pre-existing unsold homes and the new unsold homes. For pre-existing homes, we simply evaluate g(10) without integration, as they are already on the market. For new homes, we must integrate h(t) from 0 to 10 because they are added as a stream over time. The calculation is as follows:

U(t) = g(10) + ∫₀¹⁰ h(t) dt U(t) = 200e^(-0.2*10) + ∫₀¹⁰ {8e^(-0.2(10-t))} dt U(t) = 200e^(-2) + 8 ∫₀¹⁰ (e^(-2+0.2t)) dt U(t) = 200e^(-2) + 8 [e^(-2+0.2t) / 0.2]₀¹⁰ U(t) = 200e^(-2) + 40 [e^(-2+0.2t)]₀¹⁰ U(t) = 200e^(-2) + 40(e^(0) - e^(-2)) U(t) = 200(0.135) + 40(1 - 0.135) U(t) = 27 + 40(0.865) U(t) = 27 + 34.6 = 61.6

The result shows that after 10 weeks, approximately 27 of the original 200 homes will remain unsold, and about 34.6 of the new homes added will be unsold, for a total of roughly 62 unsold homes.

🔑 Definition — Pre-existing unsold: Homes already on the market at the start of the period; their number is found by evaluating the function at the end time. 🔑 Definition — New unsold: Homes added to the market over time as a stream; their total is found by integrating the rate function over the time period. 📐 Formula: U(t) = g(T) + ∫₀ᵀ h(t) dt → The total number of an item (unsold homes) after time T is the sum of the initial stock's value at T plus the accumulated new stock added over the interval. 📌 Example: With g(10) = 27 and ∫₀¹⁰ h(t) dt = 34.6, the total unsold U(10) = 27 + 34.6 = 61.6. This means after 10 weeks, about 62 homes are unsold.

TOPIC 067: DYNAMICS OF LABOR COST USING INTEGRALS

The marginal labor cost (MLC) function is given by MLC = 3 + 4L. The total labor cost function is the integral of the marginal cost. To find the specific function, we integrate and use the initial condition that total labor cost is zero when L = 0. The total labor cost function is found by indefinite integration: ∫ (3 + 4L) dL = 3L + 2L² + C. With L=0 and cost = 0, we find C=0. Therefore, the total labor cost function is TLC = 3L + 2L².

To calculate the cost of employing successive labor units from L = 1 to L = 7, we use a definite integral: ∫₁⁷ MLC(L) dL = ∫₁⁷ (3 + 4L) dL = 3∫₁⁷ dL + 4∫₁⁷ L dL = 3[L]₁⁷ + 2[L²]₁⁷ = 3(7 - 1) + 2(49 - 1) = 18 + 96 = 114

The cost of employing the 1st through 7th labor units sums to 114.

💡 Why this matters: This shows how to move from a marginal function (rate of change) back to the total function (level), a fundamental concept in economics. 🔑 Definition — Marginal Labor Cost (MLC): The additional cost incurred by hiring one more unit of labor. 📐 Formula: Total Labor Cost = ∫ MLC(L) dL → The integral of the marginal cost function yields the total cost function. 📌 Example: If MLC = 3 + 4L, then total cost from L=1 to L=7 is ∫₁⁷ (3 + 4L) dL = 114.

TOPIC 068: DEPRECIATION ANALYSIS USING INTEGRALS

The resale value of an industrial machine decreases over a 10-year period. When the machine is x years old, the rate at which its value is changing is V'(x) = 220(x - 10) dollars per year. To find how much the machine depreciates during the second year (from x=1 to x=2), we integrate the rate of change over that interval.

The change in value V(2) - V(1) is given by: ∫₁² V'(x) dx = ∫₁² {220(x - 10)} dx = 220 [∫₁² x dx - 10∫₁² dx] = 220 [ (x²/2) |₁² - 10(x)|₁² ] = 220 [ (4/2 - 1/2) - 10(2 - 1) ] = 220 [ (1.5) - 10 ] = 220 * (-8.5) = -1870

The change in value is -1870, meaning the machine's value declined by $1870 during the second year. Therefore, the machine depreciates by $1870 in the second year.

🔑 Definition — Rate of Depreciation: The instantaneous rate at which an asset's value is decreasing, often given as V'(x). 📐 Formula: Depreciation over interval [a, b] = ∫ₐᵇ V'(x) dx → The total change in value over a time period is the integral of the rate of change. 📌 Example: If V'(x) = 220(x - 10), the depreciation in the second year (x=1 to x=2) is ∫₁² 220(x - 10) dx = -1870, a decline of $1870.

TOPIC 069: COST EFFECTIVENESS DECISION USING DIFFERENTIAL EQUATIONS

A hospital considers two alternative heating systems. The rates of increase in costs are given by the differential equations dC₁/dt = 0.01t and dC₂/dt = t^(0.25). The initial cost for each system at t=0 is £50,000.

To find the total cost functions, we integrate the rate equations. For the total cost over the next 10 years, we find the additional cost incurred between t=0 and t=10. The total cost at any time t is the initial cost plus the integral of the rate from 0 to t.

For System 1: C₁(t) = 50000 + ∫₀ᵗ 0.01u du = 50000 + 0.005t² The additional cost over 10 years is ∫₀¹⁰ 0.01t dt = 0.005 * 100 = 0.5.

For System 2: C₂(t) = 50000 + ∫₀ᵗ u^(0.25) du = 50000 + (t^(1.25)/1.25) The additional cost over 10 years is ∫₀¹⁰ t^(0.25) dt = (10^(1.25) - 0) / 1.25 = 17.7828 / 1.25 = 14.226.

Since the additional cost for System 1 (£0.5) is less than for System 2 (£14.226), the 1st heating system is more cost-effective over the next 10 years. The graph of the cumulative costs would visually confirm this.

🔑 Definition — Differential Equation: An equation that relates a function to its derivatives, often used to model rates of change in economic contexts. 📐 Formula: Total Cost = Initial Cost + ∫₀ᵀ (Rate of Cost Increase) dt → The total cost at time T is the initial cost plus the accumulation of cost over time. 📌 Example: System 1: ∫₀¹⁰ 0.01t dt = 0.5; System 2: ∫₀¹⁰ t^(0.25) dt = 14.226. Since 0.5 < 14.226, System 1 is more cost-effective.

⭐ Key Takeaways

The lecture demonstrates that integrals are essential tools for converting marginal functions (rates of change) into total functions (levels). A crucial distinction is made between evaluating an existing stock at a future time and integrating a continuous flow over a time interval to find its total accumulated value. The sign of a definite integral is critical: a negative result for a depreciation function indicates a decline in value. When comparing cost-effectiveness, the system with the lower integral of its cost rate function over the specified time period is superior, assuming equal initial costs.

🧠 Quick Revision Questions

1. In the real estate example, why is the number of pre-existing unsold homes found by evaluating g(10), while the number of new unsold homes is found by integrating h(t) from 0 to 10?
2. If the marginal labor cost function is MLC = 5 + 6L and total labor cost is zero when L=0, what is the total cost of employing 2 to 5 units of labor?
3. A machine's value changes at a rate of V'(x) = 100(5 - x). What is the total depreciation during the third year (from x=2 to x=3)?
4. Two cost rate functions are C₁'(t) = 2t and C₂'(t) = 3t². Which system is more cost-effective over the period t=0 to t=5 if both have an initial cost of £100, and why?
5. What is the key difference in methodology between calculating the total unsold from a one-time stock of homes versus a continuous stream of new homes being added?

📘 Lecture 17 — Domar Growth Model

📖 Overview: This lecture introduces the Domar Growth Model, which analyzes the dual effect of investment on both aggregate demand and productive capacity. It develops a mathematical framework to derive the time path of investment required for an economy to maintain full utilization of its productive capacity over time.

🗂️ Topics Covered

The lecture is structured into three topics: Topic 070 establishes the framework of the Domar Growth Model, explaining how changes in investment affect both aggregate demand (through the Keynesian investment multiplier) and productive capacity (through the capacity-capital ratio). Topic 071 solves the model to derive the time path of investment. Topic 072 provides numerical examples and graphical analysis of investment time paths under different savings rates.

📝 Lecture Summary

DOMAR GROWTH MODEL: FRAMEWORK

The Domar Growth Model is based on the premise that a change in the rate of investment affects both aggregate demand and the productive capacity of an economy. The purpose is to find a time path of investment ( I(t) ) along which an economy can grow while maintaining full utilization of its productive capacity ( \kappa ).

Change in investment causes a dual change: ( \Delta Y ) (Change in Income/Aggregate Demand) and ( \Delta \kappa ) (Change in Productive Capacity). From Keynesian Investment Multiplier Theory: ( \frac{dY}{dI} = K = \frac{1}{s} ), where ( s ) is the Marginal Propensity to Save. Rearranging: ( dY = \frac{1}{s} \times dI ). For dynamic analysis, differentiating with respect to time ( t ): ( \frac{dY}{dt} = \frac{1}{s} \frac{dI}{dt} ).

🔑 Definition — Productive Capacity-Capital Ratio ( \rho ): ( \rho = \frac{\kappa}{K} ), where ( \kappa ) is productive capacity and ( K ) is capital. For example, if ( \rho = 5 ), each unit of capital produces 5 units of output. ( \rho ) is assumed constant as technology is constant.

📐 Formula: ( \kappa = \rho K ) → Productive capacity equals the capacity-capital ratio multiplied by capital stock.

Differentiating ( \kappa = \rho K ) with respect to ( t ): ( \frac{d\kappa}{dt} = \rho \frac{dK}{dt} ). Since the rate of change of capital is investment ( I ): ( \frac{d\kappa}{dt} = \rho I ).

In static equilibrium: ( Y = \kappa ) (current income equals potential production). In dynamic equilibrium: ( \frac{dY}{dt} = \frac{d\kappa}{dt} ), meaning the rate of change of income over time must equal the rate of change of productive capacity over time.

DOMAR GROWTH MODEL: SOLUTION

Starting from the dynamic equilibrium condition ( \frac{dY}{dt} = \frac{d\kappa}{dt} ), substitute the derived values: ( \frac{1}{s} \frac{dI}{dt} = \rho I ).

Rearranging for ( I ) in terms of ( t ): ( \frac{dI}{I} = \rho s , dt ). Introducing integrals on both sides: ( \int \frac{dI}{I} = \int \rho s , dt ).

This gives: ( \ln|I| + c_I = \rho s t + c_t ). Simplifying: ( \ln|I| = \rho s t + c ), where ( c = c_t - c_I ) is another constant.

Raising both sides to the natural exponent ( e ): ( e^{\ln|I|} = e^{\rho s t + c} ), so ( I = e^{\rho s t} e^c ). Let ( e^c = A ) (another constant): ( I(t) = A e^{\rho s t} ). This is the indeterminate/general time path of investment due to constant ( A ).

To definitize the arbitrary constant, introduce the initial condition ( t = 0 ): ( I(0) = A e^{\rho s (0)} = A ). Substituting back: ( I(t) = I(0) e^{\rho s t} ).

📐 Formula: ( I(t) = I(0) e^{\rho s t} ) → Definite solution for the time path of investment, where ( I(0) ) is initial investment, ( s ) is marginal propensity to save (constant), and ( \rho ) is the productive capacity-capital ratio (constant).

DOMAR GROWTH MODEL: NUMERICAL

The solution ( I(t) = I(0) e^{\rho s t} ) is an exponential function. Comparing with the standard form ( e^{rt} ), we get ( r = \rho s ). So, the growth of investment depends on the expression ( \rho s ).

Assuming ( \rho = 2 ) and varying ( s ) from 0.01 to 0.5, with initial condition ( I(0) = 1 ):

Observations: The composite graph shows 7 time paths generated by alternative values of Marginal Propensity to Save (( s )). All time paths share a common initial condition ( I(0) = 1 ). An increase in the value of ( s ) uplifts the time path. Specifically, an increase in marginal propensity to save increases the rate at which investment takes place. As increased savings lead to increased investment, the rate of growth of investment also increases.

💡 Why this matters: The Domar model shows that a higher savings rate produces a steeper exponential growth path for investment, demonstrating the trade-off between current consumption and future growth potential.

⭐ Key Takeaways

The Domar Growth Model demonstrates the dual effect of investment on both aggregate demand (through the Keynesian multiplier) and productive capacity (through the capacity-capital ratio). Dynamic equilibrium requires that the rate of change of income equals the rate of change of productive capacity (( dY/dt = d\kappa/dt )). The definite solution for the time path of investment is ( I(t) = I(0)e^{\rho s t} ), an exponential function where ( r = \rho s ). The growth rate of investment depends on both the productive capacity-capital ratio (( \rho )) and the marginal propensity to save (( s )). A higher marginal propensity to save leads to a higher rate of investment growth, as increased savings fuel increased investment.

🧠 Quick Revision Questions

1. What are the two effects of a change in investment in the Domar Growth Model?
2. Write the dynamic equilibrium condition for the Domar Growth Model and explain what it means.
3. What is the productive capacity-capital ratio (( \rho )) and why is it assumed to be constant?
4. Derive the definite solution for the time path of investment ( I(t) ) from the equilibrium condition ( \frac{1}{s} \frac{dI}{dt} = \rho I ).
5. If the marginal propensity to save increases from 0.1 to 0.3 with ( \rho = 2 ), what happens to the growth rate of investment?

📘 Lecture 18 — First Order Differential Equations

📖 Overview: This lecture introduces first-order differential equations, their classification by order and degree, and the standard form for solving them. It covers three cases—homogeneous and two non-homogeneous—showing how to find general and definite solutions for dynamic economic models.

🗂️ Topics Covered

The lecture begins with definitions of differential equations, order, and degree, then presents the standard form dy/dt + u(t)·y = w(t). It systematically covers three cases: the homogeneous case (a ≠ 0, b = 0), non-homogeneous case I (a ≠ 0, b ≠ 0), and non-homogeneous case II (a = 0, b ≠ 0), each with derivations of complementary functions and particular solutions. Verification by differentiation is demonstrated, and three numerical examples with graphs illustrate all cases, concluding with a summary table.

📝 Lecture Summary

FIRST-ORDER DIFFERENTIAL EQUATIONS: INTRODUCTION

A differential equation expresses an explicit or implicit relationship between a function y = f(t) and one or more of its derivatives or differentials. For example, dy/dx = 10x is a first-order differential equation because the highest derivative is a first-order derivative. y' = 12y is also first-order, while y'' - 2y' + 19 = 0 is a second-order differential equation as it contains y''.

In addition to order (the highest derivative present), degree is the power of the highest order derivative in the equation.

🔑 Definition — Order: the highest derivative present in the differential equation 🔑 Definition — Degree: the power of the highest order derivative in the equation

Solution: The solution of a differential equation will be y = f(x) with no derivative term present.

A standard form of First-Order Differential Equation is: dy/dt + u(t) · y = w(t)

• dy/dt: derivative of dependent variable y w.r.t independent variable t
• u(t): coefficient of y, a function in terms of t
• w(t): term, a function in terms of t

When u(t) and w(t) are constant, using symbols u(t) = a and w(t) = b, the standard form becomes: dy/dt + a · y = b

FIRST-ORDER DIFFERENTIAL EQUATIONS: HOMOGENEOUS CASE

Standard Form: dy/dt + a · y = b

Depending on the values of a and b, there are 3 possible cases:

• Case I: a ≠ 0, b = 0 [Homogeneous Case]
• Case II: a ≠ 0, b ≠ 0 [Non-Homogeneous Case I]
• Case III: a = 0, b ≠ 0 [Non-Homogeneous Case II]

Case I: Homogeneous Case (a ≠ 0, b = 0) The standard form dy/dt + ay = b becomes: dy/dt + ay = 0

Rearranging: dy/dt = −ay → (1/y)(dy/dt) = −a → dy/y = −a dt

Integrating w.r.t t: ∫ dy/y = −∫ a dt → ln|y| + c₁ = −at + c₂ → ln|y| = −at + c (where c = c₂ − c₁)

Raising exponent on both sides: e^(ln|y|) = e^(−at + c) → y(t) = Ae^(−at) (where A = e^c) — General Solution

Introducing the initial condition (t = 0) to definitize the arbitrary constant: y(0) = Ae^(−a·0) = Ae⁰ = A, so A = y(0) → y(t) = y(0)e^(−at) — Definite Solution

FIRST-ORDER DIFFERENTIAL EQUATIONS: NONHOMOGENEOUS CASE – I

Case II: Non-Homogeneous Case I (a ≠ 0, b ≠ 0) The standard form dy/dt + ay = b remains unchanged. Since no term equals zero (b ≠ 0), this is also called the Complete Equation.

In the complete equation of time path, there is an equilibrium value and the deviation from it. For example, in a graph where 50 is the equilibrium value and ±40e^(−2t) is the deviation, the time path converges towards equilibrium (50) in either case—whether deviation is positive or negative.

The solution has two components:

• Complementary Function (y_c): ±40e^(−2t) — complements the equilibrium value
• Particular Solution (y_p): 50 — one particular value of y out of infinite values on the whole time path

Complementary Function (y_c) is equivalent to the general solution of the homogeneous case: (y_c)_{NHC} = {y(t)}_HC = Ae^(−at)

So: y(t) = Ae^(−at) + y_p

Particular Solution (y_p): For a ≠ 0, y_p can be a constant k: Let y = k, then dy/dt = 0. Substituting into dy/dt + ay = b: 0 + a·k = b → k = b/a So: y_p = b/a

Therefore: y(t) = Ae^(−at) + b/a — General Solution of Non-Homogeneous Case with a ≠ 0

For Definite Solution: Put t = 0: y(0) = Ae⁰ + b/a = A + b/a → A = y(0) − b/a → y(t) = {y(0) − b/a} e^(−at) + b/a — Definite Solution of Non-Homogeneous Case with a ≠ 0

📌 Example (from graph): For dy/dt + 4y = 12, a = 4, b = 12 General Solution: y(t) = Ae^(−4t) + 3 Definite Solution: y(t) = {y(0) − 3}e^(−4t) + 3

FIRST-ORDER DIFFERENTIAL EQUATIONS: NONHOMOGENEOUS CASE – II

Case III: Non-Homogeneous Case II (a = 0, b ≠ 0) The standard form dy/dt + ay = b with a = 0 becomes: dy/dt + (0)y = b → dy/dt = b

Solution y(t) has two components: y(t) = y_c + y_p

Complementary Function (y_c): Equivalent to the homogeneous case: (y_c)_{NHC} = {y(t)}_HC = Ae^(−at) Since a = 0: y_c = Ae^(−(0)t) = Ae⁰ = A So: y(t) = A + y_p

Particular Solution (y_p): If y is considered a constant k (as in Case II), dy/dt = 0 → 0 = b, which is false since b ≠ 0. So we experiment with y = kt: y = kt → dy/dt = k From the differential equation dy/dt = b, we get k = b So: y_p = bt

Therefore: y(t) = A + bt — General Solution of NHC with a = 0

Definite Solution: Put t = 0: y(0) = A + b(0) = A → A = y(0) → y(t) = y(0) + bt — Definite Solution of Non-Homogeneous Case with a = 0

📌 Example (from graph): For dy/dt = 23, a = 0, b = 23 General Solution: y(t) = A + 23t Definite Solution: y(t) = y(0) + 23t

VERIFICATION OF SOLUTION OF FIRST-ORDER DIFFERENTIAL EQUATIONS

The validity of all solutions can be checked by differentiation. Consider the definite solution for a ≠ 0: y(t) = {y(0) − b/a} e^(−at) + b/a

Differentiating w.r.t t: dy/dt = −a {y(0) − b/a} e^(−at)

Substituting into the standard form dy/dt + ay = b: −a{y(0) − b/a}e^(−at) + a[{y(0) − b/a}e^(−at) + b/a] = −a{y(0) − b/a}e^(−at) + a{y(0) − b/a}e^(−at) + a·(b/a) = 0 + b = b ✓ (LHS = RHS)

📌 Numerical Example: Verify if y(t) = 4(1 − e^(−t)) is the definite solution of dy/dt + y = 4. Differentiating: dy/dt = 4[0 − (−1)e^(−t)] = −4e^(−t) Substituting into dy/dt + y = 4: −4e^(−t) + 4(1 − e^(−t)) = −4e^(−t) + 4 − 4e^(−t) = 4 ✓

FIRST-ORDER DIFFERENTIAL EQUATIONS: SUMMARY & NUMERICALS

1. dy/dt − 2y = 0: A Homogeneous Case Comparing with dy/dt + ay = b: a = −2, b = 0 General Solution: y(t) = Ae^(−(−2)t) = Ae^(2t) Definite Solution (with t = 0): A = y(0) → y(t) = y(0)e^(2t) Interpretation: There is exponential growth (e^(2t)) in the solution.

2. dy/dt + 4y = 12: Non-Homogeneous Case (a ≠ 0, b ≠ 0) a = 4, b = 12 General Solution: y(t) = Ae^(−4t) + 12/4 = Ae^(−4t) + 3 Definite Solution (with t = 0): A = y(0) − 3 → y(t) = {y(0) − 3}e^(−4t) + 3 Interpretation: The particular integral y_p = 3 is the equilibrium. The complementary function has exponential decay converging to equilibrium.

3. dy/dt = 23: Non-Homogeneous Case (a = 0, b ≠ 0) Rewriting: dy/dt + (0)y = 23 → a = 0, b = 23 General Solution: y(t) = A + 23t Definite Solution (with t = 0): A = y(0) → y(t) = y(0) + 23t

Summary Table of All Cases:

⭐ Key Takeaways

First-order differential equations have the standard form dy/dt + ay = b, and their solution always consists of a complementary function (y_c = Ae^(−at)) plus a particular solution (y_p). The three cases depend on whether the constant term b and coefficient a are zero or non-zero: the homogeneous case (b=0) has only y_c; non-homogeneous case I (a≠0, b≠0) has y_p = b/a representing the equilibrium; and non-homogeneous case II (a=0, b≠0) has y_p = bt, giving a linear time path. The arbitrary constant A is definitized using the initial condition y(0). All solutions can be verified by differentiating and substituting back into the original equation.

🧠 Quick Revision Questions

1. What is the standard form of a first-order differential equation with constant coefficients?
2. How do you find the complementary function for any first-order linear differential equation?
3. In the non-homogeneous case with a ≠ 0, how is the particular solution y_p determined?
4. Why does the non-homogeneous case with a = 0 require a non-constant particular solution (y = kt) instead of a constant?
5. How do you verify that y(t) = {y(0) − b/a}e^(−at) + b/a is indeed the correct solution of dy/dt + ay = b?

📘 Lecture 19 — Dynamics of Market Price

📖 Overview: This lecture extends the static supply-demand model into a dynamic framework, analyzing how market prices adjust over time when the market is not in equilibrium. It introduces the concept of a time-path for price, derived from a first-order differential equation, and examines the conditions for dynamic stability, supported by a numerical example.

🗂️ Topics Covered

The lecture begins by defining the static equilibrium of the market model and then transitions to a dynamic sense where price follows a time-path. It introduces an adjustment equation linking price changes to excess demand, solves the resulting first-order differential equation to find a definite time-path for price, analyzes the dynamic stability condition requiring the complementary function to decay to zero, and concludes with a numerical example to illustrate the convergence process.

📝 Lecture Summary

TOPIC 079: DYNAMICS OF MARKET PRICE: FRAMEWORK

The standard microeconomic model begins with demand and supply equations: quantity demanded is Qd = a - bP and quantity supplied is Qs = -c + dP, where all parameters (a, b, c, d) are positive constants. In a static sense, equilibrium is found by setting Qd = Qs. Solving this gives the static equilibrium price P* = (a + c) / (b + d), which is a positive constant due to the positive parameters. In a dynamic sense, the price P(t) follows a time-path rather than being a single point. This time-path is expressed as P(t) = Pc + Pp, where the particular solution Pp is the static equilibrium price P*, and Pc is the complementary function representing deviation from equilibrium. The initial condition P(0) can start above, at, or below the equilibrium price P*. Cases where P(0) > P* or P(0) < P* require adjustment over time.

TOPIC 080: DYNAMICS OF MARKET PRICE: THE TIME PATH

Price adjustments over time are measured by the rate of change dP/dt. This rate is proportional to excess demand, which is defined as (Qd - Qs). A shortage (excess demand > 0) or surplus (excess demand < 0) drives price changes. The adjustment equation is formalized as dP/dt = j(Qd - Qs), where j > 0 is the adjustment coefficient. At equilibrium, dP/dt = 0 and Qd - Qs = 0. Substituting Qd and Qs into the adjustment equation yields:

dP/dt = j[(a + c) - (b + d)P]. This is rearranged to the standard first-order differential equation form: dP/dt + j(b+d)P = j(a+c). By comparing this to the form dy/dt + ay = b, we have y = P, a = j(b+d) (a positive constant, thus a ≠ 0), and b = j(a+c) (a positive constant, thus b ≠ 0). This is a non-homogeneous case with a ≠ 0.

The definite solution for the time-path of this case is: y(t) = [y(0) - b/a] e^(-at) + b/a. Substituting the values for this model, we get:

P(t) = [P(0) - (a+c)/(b+d)] e^(-j(b+d)t) + (a+c)/(b+d)

Since (a+c)/(b+d) = P*, and letting k = j(b+d), the final definite time-path for market price is:

🔑 Definition — Definite Time-Path: P(t) = [P(0) - P*] e^(-kt) + P*. This equation describes the path of price P at any time t, starting from its initial value P(0) and converging towards its equilibrium value P* at a rate determined by k.

TOPIC 081: DYNAMICS OF MARKET PRICE: DYNAMIC STABILITY

For a time-path to be dynamically stable, it must converge to its equilibrium value as time goes to infinity. This means the deviation from equilibrium, represented by the complementary function Pc, must fade to zero. In the definite solution P(t) = [P(0) - P*] e^(-kt) + P*, the term Pc = [P(0) - P*] e^(-kt). For stability, we require Pc → 0 as t → ∞. The term e^(-kt) will approach zero as t → ∞ only if the exponent is negative, which requires k > 0. Since k = j(b+d) and all parameters j, b, d are positive, k is always positive. Therefore, regardless of whether the initial deviation [P(0) - P*] is positive (starting above equilibrium) or negative (starting below equilibrium), e^(-kt) decays to zero. This causes Pc to vanish, and the time-path P(t) to converge to P*. The system is thus proven to be dynamically stable.

📌 Example: Consider P(t) = ±40e^(-4t) + 50. The equilibrium price is P* = 50. If P(0) = 90, the deviation is +40, and the time-path slides down from 90 towards 50. If P(0) = 10, the deviation is -40, and the time-path rises from 10 towards 50. In both cases, as t increases, the term 40e^(-4t) decays to zero, and the price converges to 50.

TOPIC 082: DYNAMICS OF MARKET PRICE: NUMERICAL

Consider the market model: Qs = 3P - 4, Qd = -5P + 20, and the adjustment equation dP/dt = 0.2(Qd - Qs), with the initial condition P(0) = 2. The goal is to find P(t), Qs(t), and Qd(t), and determine stability.

First, substitute Qd and Qs into the adjustment equation: dP/dt = 0.2[(-5P+20) - (3P-4)]. This simplifies to dP/dt + 1.6P = 4.8. Comparing to dy/dt + ay = b, we have a = 1.6 and b = 4.8. Since a ≠ 0 and b ≠ 0, this is a non-homogeneous case with a ≠ 0. Using the definite solution formula:

P(t) = [P(0) - b/a] e^(-at) + b/a = [2 - (4.8/1.6)] e^(-1.6t) + (4.8/1.6)

This gives the definite time-path for market price: P(t) = (-1) e^(-1.6t) + 3.

Dynamic Stability: The equilibrium price P* is b/a = 3. The complementary function is Pc = -e^(-1.6t), which is a negative deviation. As time t increases, -e^(-1.6t) decays towards zero, causing P(t) to converge to P* = 3 from below. The system is dynamically stable.

🔑 Definition — Dynamic Stability: A time-path is dynamically stable if the complementary function Pc approaches zero as time approaches infinity (t → ∞), ensuring the path converges to the equilibrium P*.

📌 Example (continuing): The table shows the convergence of P(t) = -e^(-1.6t) + 3.

The graph would show the time-path P(t) (green line) starting at 2 and rising asymptotically towards the equilibrium price line P* = 3 (red line), while the deviation Pc decays to zero.

💡 Why this matters: The model shows that a simple price adjustment mechanism based on excess demand guarantees convergence to the static equilibrium, a foundational result in dynamic economic analysis.

⭐ Key Takeaways

A fundamental extension of the supply-demand model introduces dynamics via an adjustment equation, dP/dt = j(Qd - Qs), which links price changes to excess demand. Solving this first-order differential equation yields a definite time-path for price: P(t) = [P(0) - P*] e^(-kt) + P*. For dynamic stability, the time-path must converge to the static equilibrium P*, which requires that the complementary function [P(0)-P*]e^(-kt) decays to zero as t → ∞. This decay is guaranteed because k = j(b+d) > 0, so regardless of whether price starts above or below equilibrium, it will converge, confirming that the market is dynamically stable.

🧠 Quick Revision Questions

1. What is the adjustment equation that links the change in price over time to market conditions?
2. What is the standard form of the first-order differential equation derived from the adjustment equation, and what are a and b in terms of the model's parameters?
3. What is the general form of the definite solution for the time-path of price, P(t)?
4. What is the condition for dynamic stability of the price time-path, and why is it always satisfied given the model's assumptions?
5. In the numerical example (Qs = 3P - 4, Qd = -5P + 20), what is the equilibrium price, and what is the expression for the time-path of price?

📘 Lecture 20 — PRICE DYNAMICS USING DIFFERENTIAL EQUATIONS

📖 Overview: This lecture demonstrates how first-order differential equations are applied to model price dynamics in economic systems. It covers three key applications: calculating the consumer's reservation price from demand elasticity, modeling the time-path of price level in the money market, and analyzing price dynamics using the equation of exchange. Understanding these models is crucial for predicting how prices adjust over time to economic shocks.

🗂️ Topics Covered

This lecture introduces the concept of reservation price as the price at which demand becomes zero, derived from a demand function using integration and elasticity information. It then models the dynamics of price and money market, where the rate of price change depends on excess money supply, deriving a time-path for prices. Finally, it applies the equation of exchange to model price dynamics when transaction demand for money is considered constant, analyzing the dynamic stability of the resulting time-path.

📝 Lecture Summary

TOPIC 083: CONSUMER'S RESERVATION PRICE USING DIFFERENTIAL EQUATIONS

Given the direct demand function, the own price elasticity of demand is used to derive the demand function. The elasticity formula is εd = (dQ/dP) × (P/Q). The given elasticity expression εd = −(3P+2P²)/Q is rewritten to isolate dQ/dP.

The differential equation is set up: dQ/dP + 2P = −3. Since the dependent variable is Q, an alternative approach is used. The derivative term is kept on the left-hand side and both sides are integrated with respect to P:

∫ (dQ/dP) dP = ∫ (−2P − 3) dP

This yields the General Solution: Q(P) = −P² − 3P + c, where c is an integration constant.

Using the given information that at equilibrium Q = 50 when P = 10, the constant c is found: 50 = −(10)² − 3(10) + c 50 = −100 − 30 + c c = 180

Substituting c back gives the Definite Solution: Q(P) = −P² − 3P + 180.

The Reservation Price (PR) is defined as the price at which demand for a good or service does not start. Mathematically, it is the price when Q = 0: 0 = −P² − 3P + 180 P² + 3P − 180 = 0

Using the quadratic formula: P = [−3 ± √(3² − 4(1)(−180))] / 2(1) P = [−3 ± √729] / 2 P = (−3 ± 27) / 2

This gives two values: P = 12 or P = −15. The unrealistic negative value is discarded.

🔑 Definition — Reservation Price (PR): The price at which no demand for a good or service exists, i.e., the price at which Q = 0.

📐 Formula: Quadratic Formula x = [−b ± √(b² − 4ac)] / 2a → used to solve for the roots of a quadratic equation of the form ax² + bx + c = 0.

📌 Example: Given εd = −(3P+2P²)/Q, Q=50, P=10. The demand function is derived as Q(P) = −P² − 3P + 180. Setting Q=0 gives PR = 12. Verification: Q(12) = −144 − 36 + 180 = 0.

💡 Why this matters: The reservation price represents the maximum price consumers are willing to pay, making it a critical concept for pricing strategy and market analysis.

TOPIC 084: DYNAMICS OF PRICE & MONEY MARKET USING DIFFERENTIAL EQUATIONS

In this model, money supply (Mˢ) is fixed by the government. The transaction demand for money (L) is positively related to price level P(t) and real output Y: L = aPY, where a > 0 is the constant of proportionality.

Inflation (dP/dt), the increase in the overall price level over time, depends on excess money supply, which is the difference between money supply and transaction demand for money. The model is: dP/dt = j(Mˢ − L), where j > 0 implies price level rises with excess money supply.

Substituting L gives: dP/dt = j(Mˢ − aPY) dP/dt = jMˢ − ajYP dP/dt + ajYP = jMˢ

Comparing with the standard form dy/dt + ay = b: dy/dt = dP/dt, y = P, a = ajY, b = jMˢ

This is a non-homogeneous case with a ≠ 0 since both a and b are non-zero positive constants.

The definite solution for the time-path is: P(t) = {P(0) − (jMˢ)/(ajY)} e^(−ajY)t + (jMˢ)/(ajY) P(t) = {P(0) − Mˢ/(aY)} e^(−ajY)t + Mˢ/(aY)

This is the Definite Time-path for Market price in Dynamic Sense.

Interpretation: Mˢ appears in numerator and has a positive relationship with price level. Y is in the denominator and has a negative relationship with price level. If money supply increases, prices increase; if output increases, prices decrease due to increased supply of goods/services.

Dynamic Stability: For the time-path to be dynamically stable, it must converge towards equilibrium. The complementary function (Pc) represents the deviation from equilibrium, and for stability, Pc → 0 as t → ∞. Pc = {P(0) − Mˢ/(aY)} e^(−ajY)t

Since a, j, Mˢ, and Y are positive, the exponent e^(−ajY)t → 0 as t → ∞. Therefore, regardless of whether P(0) is greater or smaller than the equilibrium price (P*), the time-path converges. Hence, the time-path is dynamically stable.

🔑 Definition — Excess Money Supply: The difference between money supply (Mˢ) and transaction demand for money (L), driving price level changes.

🔑 Definition — Dynamic Stability: Property of a time-path where deviations from equilibrium fade to zero as time approaches infinity (Pc → 0 as t → ∞).

📐 Formula: Differential Equation for Price Dynamics dP/dt = j(Mˢ − aPY) → the rate of price change depends on the adjustment coefficient (j) and excess money supply.

📌 Example: With Mˢ = 500, Y = 100, a = 0.2, j = 0.5, P(0) = 20. The equilibrium price P* = Mˢ/(aY) = 500/20 = 25. P(t) = {20 − 25} e^(−0.5×0.2×100)t + 25 = −5e^(−10)t + 25. As t→∞, e^(−10t)→0, so P(t)→25, showing convergence to equilibrium.

💡 Why this matters: This model demonstrates that monetary policy (changing Mˢ) and real output growth (changing Y) have predictable effects on the price level over time, and the system naturally returns to equilibrium.

TOPIC 085: PRICE DYNAMICS & EQUATION OF EXCHANGE USING DIFFERENTIAL EQUATIONS

The equation of exchange is: Mˢ × V = P × Y, where Mˢ is money supply, V is velocity of money, P is general price level, and Y is aggregate output.

Rewriting for Mˢ: Mˢ = (P × Y) / V

The transaction demand for money (L) is assumed to have only a transactive component and is constant. Price adjustments occur due to differences between demand and supply of money: dP/dt = j(Mˢ − L)

Substituting Mˢ: dP/dt = j((PY)/V − L) dP/dt = j(Y/V)P − jL dP/dt − j(Y/V)P = −jL

Comparing with standard form dy/dt + ay = b: dy/dt = dP/dt, y = P, a = −j(Y/V), b = −jL

This is a non-homogeneous case with a ≠ 0.

Using the definite solution formula: P(t) = {P(0) − (−jL)/(−jY/V)} e^{−(−jY/V)t} + (−jL)/(−jY/V) P(t) = {P(0) − VL/Y} e^{(Y/V)t} + VL/Y

This is the Definite Time-path for Market price in Dynamic Sense.

Interpretation: V (velocity of money) and L (transaction demand for money) appear in the numerator, having positive relationships with price level. Y (aggregate output) is in the denominator, having a negative relationship with price level. Faster money circulation or higher transaction demand increases prices; greater aggregate output reduces prices.

Dynamic Stability: For stability, the complementary function must approach zero as t → ∞. Pc = {P(0) − VL/Y} e^{(Y/V)t}

Since Y and V are positive, the exponent (Y/V) is positive, making e^{(Y/V)t} → ∞ as t → ∞. Therefore, regardless of whether P(0) is greater or smaller than equilibrium price (P*), the deviation grows without bound. Hence, the time-path is dynamically unstable.

🔑 Definition — Equation of Exchange: Mˢ × V = P × Y, showing the relationship between money supply, velocity of money, price level, and aggregate output.

📐 Formula: Price Time-path from Equation of Exchange P(t) = {P(0) − VL/Y} e^{(Y/V)t} + VL/Y → the time-path of price level when transaction demand for money is constant.

📌 Example: With V = 5, L = 100, Y = 200, P(0) = 2. The equilibrium price P* = VL/Y = 500/200 = 2.5. P(t) = {2 − 2.5} e^{(200/5)t} + 2.5 = −0.5e^(40t) + 2.5. As t→∞, e^(40t)→∞, so P(t)→∞, showing divergence from equilibrium.

💡 Why this matters: Unlike the previous stable money market model, this specification with constant transaction demand leads to unstable price dynamics, illustrating how different modeling assumptions critically affect economic predictions about price stability.

⭐ Key Takeaways

The consumer's reservation price can be derived by integrating the elasticity of demand function to obtain the demand function, then solving for the price where quantity demanded is zero, discarding any negative price values. In the money market model where transaction demand depends on price and output, the price time-path is dynamically stable, converging to equilibrium regardless of initial conditions, provided all parameters are positive. However, when using the equation of exchange with constant transaction demand, the exponent becomes positive, causing the time-path to diverge away from equilibrium and become dynamically unstable. The key difference between the two models is whether the transaction demand for money varies with price (stable) or is constant (unstable). Understanding the conditions for dynamic stability is essential for predicting whether an economic system will return to equilibrium after a shock.

🧠 Quick Revision Questions

1. What is the reservation price and how is it mathematically determined from a demand function?
2. In the money market model (Topic 084), what condition must hold for the price time-path to be dynamically stable?
3. Why does the equation of exchange model (Topic 085) yield a dynamically unstable time-path while the money market model (Topic 084) yields a stable one?
4. How does an increase in aggregate output (Y) affect the price level in each of the three models presented in this lecture?
5. In the equation of exchange model, what are the implications of a positive exponent (Y/V) on the time-path of prices as time approaches infinity?

📘 Lecture 21 — INCOME DYNAMICS USING DIFFERENTIAL EQUATIONS

📖 Overview: This lecture applies first-order differential equations to analyze the dynamic behavior of national income under different economic conditions. It examines how aggregate income adjusts over time to maintain equilibrium in the money market, how income taxation affects the stability of the national income path, and how an individual's consumption evolves dynamically. Understanding these dynamics is crucial for predicting whether economies converge to or diverge from equilibrium.

🗂️ Topics Covered

The lecture covers three main applications: first, income dynamics and the money market, where the time path of national income ensuring money market equilibrium is derived and its dynamic stability is assessed. Second, income dynamics with income tax, where a national income model including taxation is solved for its time path and stability. Third, dynamics of consumption, where a simple differential equation for an individual's consumption growth is solved and analyzed for convergence to equilibrium.

📝 Lecture Summary

TOPIC 086: INCOME DYNAMICS & MONEY MARKET USING DIFFERENTIAL EQUATIONS

On the money market, money supply (M^s) is fixed by the government. The transaction demand for money (L) depends on income (Y) and is positively related to its incremental increase (\frac{dY}{dt}). The demand function is given by (L = aY + l \frac{dY}{dt}), where (0 < a < 1) and (l > 0). To find the time path of aggregate national income that ensures equilibrium on the money market, we set demand equal to supply: (L = M^s). This gives the first-order differential equation (aY + l \frac{dY}{dt} = M^s). Rearranging to standard form (\frac{dy}{dt} + ay = b) yields (\frac{dY}{dt} + \frac{a}{l} Y = \frac{M^s}{l}). Here, (\frac{dy}{dt} = \frac{dY}{dt}), (y = Y), (a = \frac{a}{l}), and (b = \frac{M^s}{l}).

This is a Non-Homogeneous case with (a \neq 0) because both (b) and (a) are non-zero positive constants. The definite solution for the time path is: [ Y(t) = \left{ Y(0) - \frac{M^s}{a} \right} e^{-\frac{a}{l}t} + \frac{M^s}{a} ] This is the Definite Time-path for National Income in Dynamic Sense. The interpretation is that since (M^s) appears in the numerator, it has a positive relationship with aggregate national income (Y).

💡 Why this matters: For dynamic stability, the time path must converge to equilibrium. The equilibrium is the particular solution (Y_p = \frac{M^s}{a}). Convergence requires the complementary function (Y_c = \left{ Y(0) - \frac{M^s}{a} \right} e^{-\frac{a}{l}t}) to approach zero as (t \to \infty). Since (a, l, M^s > 0), the exponential term (e^{-\frac{a}{l}t}) decays to zero. Regardless of whether initial income (Y(0)) is greater or smaller than equilibrium (Y^*), the deviation fades, ensuring the time path of national income converges towards the equilibrium income level. Hence, the time path is dynamically stable.

🔑 Definition — Dynamic Stability: The property of a time path such that the deviation from equilibrium (complementary function) approaches zero as time approaches infinity, causing the variable to converge to equilibrium. 📌 Example: If (Y(0) = 100), (M^s = 200), (a = 0.5), and (l = 10), then (Y_p = 400). The deviation component is ((100 - 400)e^{-0.05t} = -300e^{-0.05t}). As (t \to \infty), this deviation decays to zero, and (Y(t) \to 400).

TOPIC 087: INCOME DYNAMICS & INCOME TAX USING DIFFERENTIAL EQUATIONS

A national income model is given by the system: (Y = C + I), with consumption (C = a + bY) (where (a > 0) is autonomous consumption and (0 < b < 1) is the marginal propensity to consume), and investment (I = \alpha + \beta \frac{dY}{dt}) (where (\alpha > 0) and (0 < \beta < 1)). When an income tax with rate (r < 1) is introduced, the consumption function becomes (C = a + b(1 - r)Y). Substituting into (Y = C + I) and rearranging to a first-order differential equation gives: [ \frac{dY}{dt} + \left( \frac{1 - b + br - \beta}{-\beta} \right) Y = \left( -\frac{a}{\beta} \right) ] This is a Non-Homogeneous case with (a \neq 0). The definite solution for the time path is: [ Y(t) = \left{ Y(0) - \left( \frac{a}{1 - b + br - \beta} \right) \right} e^{\left( \frac{1 - b + br - \beta}{\beta} \right)t} + \left( \frac{a}{1 - b + br - \beta} \right) ] This is the Definite Time-path for National Income in Dynamic Sense. The interpretation is that national income (Y) is negatively related to the tax rate (r), as (r) appears in the denominator.

💡 Why this matters: For dynamic stability, we examine the complementary function (Y_c = \left{ Y(0) - \left( \frac{a}{1 - b + br - \beta} \right) \right} e^{\left( \frac{1 - b + br - \beta}{\beta} \right)t}). With parameters (a, b, \beta, r > 0) and (b < 1, r < 1), the exponent (\left( \frac{1 - b + br - \beta}{\beta} \right)) is likely positive, causing the deviation to grow without bound. Regardless of initial national income (Y(0)) relative to equilibrium (Y^*), the time-path of national income diverges away from the equilibrium level. Hence, the time-path is dynamically unstable.

🔑 Definition — Dynamically unstable: A time path that does not converge to equilibrium; instead, the deviation from equilibrium grows as time passes, causing the variable to move away from its equilibrium value. 📌 Example: If (a=10, b=0.8, \beta=0.2, r=0.25), then (1 - b + br - \beta = 1 - 0.8 + 0.2 - 0.2 = 0.2), so the exponent is (0.2/0.2 = 1). Any non-zero deviation will grow exponentially as (e^t), so (Y(t)) diverges.

TOPIC 088: DYNAMICS OF CONSUMPTION USING DIFFERENTIAL EQUATIONS

The consumption function of an individual grows with time at a rate given by the equation: (\frac{dC}{dt} = -2C + 100). Given that (C(0) = 10), we find the time path. Rearranging to the standard form (\frac{dy}{dt} + ay = b) gives (\frac{dC}{dt} + 2C = 100), where (a = 2) and (b = 100). This is a Non-Homogeneous case with (a \neq 0). The definite solution is: [ C(t) = \left{ C(0) - \frac{b}{a} \right} e^{-at} + \frac{b}{a} ] Substituting values: (C(0) = 10), (b/a = 50), so: [ C(t) = (10 - 50)e^{-2t} + 50 = -40e^{-2t} + 50 ] This is the Definite Time-Path of Consumption of the individual. The expression has an exponential decay term (-40e^{-2t}) (the complementary function (C_c)) and the particular solution (C_p = 50).

💡 Why this matters: For dynamic stability, as (t \to \infty), (e^{-2t} \to 0), so the deviation (C_c \to 0). The time path of consumption (C(t)) rises from its initial value of 10 and converges to the equilibrium consumption (C^* = 50). This is confirmed by the table and graph, which show the negative deviation decaying over time and the time path converging to equilibrium. The time path is dynamically stable.

🔑 Definition — Complementary Function ((C_c)): The part of the solution to a differential equation that represents the deviation from equilibrium; it decays to zero for a dynamically stable system. 📐 Formula: (C(t) = C_c + C_p) → (C(t) = \underbrace{-40e^{-2t}}{\text{deviation}} + \underbrace{50}{\text{equilibrium}}) 📌 Example: At (t=0), (C(0) = -40 + 50 = 10). At (t=2), (C(2) = -40e^{-4} + 50 \approx -0.7326 + 50 = 49.2674). The consumption is close to its equilibrium value of 50.

⭐ Key Takeaways

The most critical takeaway is the concept of dynamic stability, which is determined by the exponent in the exponential term of the complementary function. For the money market, the exponent (-\frac{a}{l}t) is negative, ensuring convergence and dynamic stability. However, when income tax is introduced, the exponent becomes positive, causing the national income path to diverge, making it dynamically unstable. This shows how economic policies like taxation can fundamentally alter the stability of an economy. The third example, on consumption dynamics, reinforces that a negative coefficient in the differential equation leads to exponential decay and convergence to equilibrium. Students must be able to derive the differential equation, identify the case (non-homogeneous with (a \neq 0)), apply the correct solution formula, and evaluate the sign of the exponent to determine stability.

🧠 Quick Revision Questions

1. What is the condition for a first-order differential equation solution (Y(t) = {Y(0) - b/a}e^{-at} + b/a) to be dynamically stable?
2. In the money market model, how does the money supply (M^s) affect the equilibrium level of national income (Y^*)?
3. Why does the introduction of an income tax rate (r) cause the time path of national income to become dynamically unstable in the model presented?
4. If an individual's consumption follows (\frac{dC}{dt} = -kC + m) with (k > 0), what is the long-run equilibrium consumption level (C^*)?
5. Explain why a negative deviation from equilibrium in the consumption example ((C(t) = -40e^{-2t} + 50)) decreases over time rather than increases.

📘 Lecture 22 — National Income Determination Using Differential Equations

📖 Overview: This lecture demonstrates how first-order differential equations are used to model and analyze the dynamic behavior of national income over time. It explores the conditions for dynamic stability in two-sector and three-sector economies, including the effects of induced government spending and investment dynamics on the time path of key economic variables.

🗂️ Topics Covered

This lecture covers five main topics: national income determination using differential equations with consumption and investment functions, a numerical example of national income time-path calculation, income dynamics with induced government spending as a third sector, dynamics of investment using differential equations, and dynamics of interest rate determination in the capital market using investment and savings functions with parameter conditions for stability.

📝 Lecture Summary

Topic 089: National Income Determination Using Differential Equations

The national income model is given by the system: Y = C + I, C = a + bY (where a is autonomous consumption and b is the marginal propensity to consume), and I = αY + β(dY/dt) (with α > 0, β < 1). Substituting gives a first-order differential equation: dY/dt + ((1 - b - α)/(-β))Y = -a/β. This is a Non-Homogenous Case with a ≠ 0 (assuming 1 - b - α ≠ 0). The definite solution for the time path of national income is:

Y(t) = {Y(0) - (a/(1 - b - α))} * e^(((1 - b - α)/β)t) + (a/(1 - b - α))

🔑 Definite Time-path: The specific solution showing national income as a function of time, consisting of a complementary function (deviation from equilibrium) and a particular integral (equilibrium value).

📐 Formula: Y(t) = {Y(0) - Y_p} * e^(((1-b-α)/β)t) + Y_p, where Y_p = a/(1-b-α) is the equilibrium national income.

📌 Example: For given a > 0, 0 < b < 1, α > 0, β < 1, regardless of whether initial income Y(0) is above or below equilibrium, since (1 - b - α)/β > 0 when b > 0, α < 1, β > 0, the exponential term grows over time, making the time path diverge from equilibrium — hence dynamically unstable.

Topic 090: National Income Determination: Numerical

Given: dY/dt = 0.5(C + I - Y), C = 0.8Y + 400, I = 600, with Y(0) = 7000. Substituting yields: dY/dt + 0.1Y = 500. This is a Non-Homogenous Case with a = 0.1 ≠ 0, b = 500 ≠ 0. The definite solution is:

Y(t) = {7000 - 500/0.1} * e^(-0.1t) + 500/0.1 = 2000 * e^(-0.1t) + 5000

🔑 Complementary Function (Deviation): The component Y_c = 2000 * e^(-0.1t) that measures the deviation from equilibrium and decays exponentially over time.

📐 Formula: Y(t) = Y_c + Y_p = 2000e^(-0.1t) + 5000, where Y_p = 5000 is the equilibrium income.

📌 Example: At t = 0, Y(0) = 7000; at t = 20, Y(20) ≈ 5270.67; at t = 100, Y(100) ≈ 5000.09. As t → ∞, Y_c → 0 and Y(t) → Y* = 5000, showing convergence to equilibrium — dynamically stable.

Topic 091: Income Dynamics and Induced Government Spending Using Differential Equations

Adding government spending G = gY (where g < 1) to the two-sector model gives: Y = C + I + G. Substituting yields: dY/dt + ((1 - b - α - g)/(-β))Y = -a/β. The definite time path is:

Y(t) = {Y(0) - (a/(1 - b - α - g))} * e^(((1 - b - α - g)/β)t) + (a/(1 - b - α - g))

🔑 Induced Government Spending: Government expenditure proportional to national income (G = gY), where g < 1, which adds a negative term in the denominator of the equilibrium expression, raising the equilibrium income level.

📐 Formula: Y(t) = {Y(0) - Y_p} * e^(((1-b-α-g)/β)t) + Y_p, where Y_p = a/(1 - b - α - g).

📌 Example: Since 1 - b - α - g > 0 when b, α, g are positive and sum < 1, the exponential exponent (1 - b - α - g)/β is positive (β > 0). Thus, regardless of initial income, the time path diverges from equilibrium — dynamically unstable. The government spending g has a positive effect on equilibrium income levels.

Topic 092: Dynamics of Investment Using Differential Equations

A principal of $60 is invested, and the investment value I(t) satisfies: dI/dt = 0.002I + 5, with I(0) = 2000. Rearranging: dI/dt - 0.002I = 5. This is a Non-Homogenous Case with a = -0.002, b = 5. The definite solution is:

I(t) = {2000 - 5/(-0.002)} * e^(0.002t) + 5/(-0.002) = -500 * e^(-0.002t) + 2500

🔑 Deviation Time-Path: The complementary function I_c = -500 * e^(-0.002t) shows a negative deviation (since initial investment is below equilibrium) that decays exponentially over time.

📐 Formula: I(t) = -500e^(-0.002t) + 2500, where I_p = 2500 is the equilibrium investment.

📌 Example: At t = 0, I(0) = 2000; at t = 10, I(10) ≈ 2009.90; at t = 100, I(100) ≈ 2498.15. As t → ∞, I_c → 0 and I(t) → I* = 2500 — dynamically stable.

💡 Why this matters: The investment dynamics show that even with an initial level below equilibrium, the investment time path converges smoothly to the equilibrium level due to the negative exponential decay.

Topic 093: Dynamics of Interest Rate in Capital Market Using Differential Equations

Investment: I = a - bi + ρ(di/dt), Savings: S = -c + di + σ(di/dt). Equilibrium requires I = S. Substituting yields: di/dt - {(b + d)/(ρ - σ)}i = -{(a + c)/(ρ - σ)}. The definite time path is:

i(t) = {i(0) - ((a + c)/(b + d))} * e^(((b + d)/(ρ - σ))t) + ((a + c)/(b + d))

🔑 Dynamic Stability Condition: For stability, the exponent (b + d)/(ρ - σ) must be negative. Since b + d > 0 (positive parameters), we need ρ - σ < 0, i.e., ρ < σ (the investment adjustment parameter must be less than the savings adjustment parameter).

📐 Formula: i(t) = {i(0) - i_p} * e^(((b+d)/(ρ-σ))t) + i_p, where i_p = (a + c)/(b + d) is the equilibrium interest rate.

📌 Example: Given a, b, c, d > 0, if ρ < σ, then (b + d)/(ρ - σ) < 0, and the exponential term decays to zero. As t → ∞, i(t) → i* = (a + c)/(b + d). Regardless of whether initial interest rate is above or below equilibrium, convergence occurs — dynamically stable.

⭐ Key Takeaways

The lecture demonstrates that first-order differential equations are essential for modeling the dynamic adjustment paths of national income, investment, and interest rates over time. The critical factor determining dynamic stability is the sign of the coefficient in the exponential term: a negative coefficient ensures convergence to equilibrium (dynamically stable), while a positive coefficient causes divergence (dynamically unstable). For national income models without government or with induced government spending, the time path is dynamically unstable when the marginal propensity to consume and investment parameters are positive and their sum is less than one. In contrast, investment dynamics and interest rate determination can exhibit dynamic stability when the relevant adjustment parameters satisfy specific conditions (negative exponential coefficient). The equilibrium level (particular integral) always equals the autonomous spending divided by the marginal propensity to save or leakages, regardless of stability properties.

🧠 Quick Revision Questions

1. What is the condition for dynamic stability in the interest rate model, and why does the sign of (ρ - σ) matter?

2. Derive the definite time path for national income when induced government spending G = gY is included, and explain how g affects the equilibrium income level.

3. In the numerical example with Y(0) = 7000, why does the national income time path converge to 5000 despite starting above equilibrium?

4. For the investment dynamics problem, explain why the complementary function is negative (-500) and how this affects the path of investment over time.

5. What distinguishes a dynamically stable time path from an unstable one in the context of first-order differential equations, using the exponential term's coefficient as the deciding factor?