Unit 1: Integration and Applications [8 Hours]

This unit introduces integration as the reverse process of differentiation and develops the standard techniques needed to evaluate integrals confidently. It covers standard forms, substitution, integration by parts, and the integration of algebraic, logarithmic, and exponential functions. Definite integrals are then evaluated using the fundamental theorem, and the unit closes with business applications: marginal to total cost and revenue, consumer’s and producer’s surplus, and the area under a curve.

Tested Topics

1. Application of integration in business (marginal to total cost/revenue functions) — 3x (2025, 2025, 2024)
2. Area under a curve — 2x (2025, 2024)
3. Integration by substitution — 2x (2025, 2024)
4. Standard forms of integration — 2x (2025, 2024)
5. Consumer's and producer's surplus — 1x (2024)
6. Definite integral evaluation — 1x (2025)

1. Concept of Integration

Definition

Integration is the process of finding a function \(F(x)\) whose derivative is a given function \(f(x)\). If \(F'(x)=f(x)\), then \(F(x)\) is called an antiderivative or primitive of \(f(x)\), and we write
\(\int f(x)\,dx = F(x)+C\), where \(C\) is the constant of integration.
Intuitively, integration reverses differentiation: it recovers the original function from its rate of change.

Detailed Explanation

Differentiation answers “how fast is \(y\) changing?” Integration answers the reverse question: “what function has this given rate of change?” Because many functions can differ by a constant and still have the same derivative, every indefinite integral carries an arbitrary constant \(C\). The symbol \(\int\) is the integral sign, \(f(x)\) is the integrand, \(dx\) indicates the variable of integration, and \(F(x)+C\) is the general antiderivative.

Integration is linear: \(\int [af(x)+bg(x)]\,dx = a\int f(x)\,dx + b\int g(x)\,dx\). This lets us split sums and pull constants out. The process is not always possible in closed form, but for the functions in this unit the standard rules and techniques are sufficient.

Formula and Conditions

\[ \int f(x)\,dx = F(x)+C \quad \Longleftrightarrow \quad \frac{d}{dx}F(x)=f(x) \]

Linearity:

\[ \int [af(x)+bg(x)]\,dx = a\int f(x)\,dx + b\int g(x)\,dx \]

Conditions: \(f\) must be continuous on the interval of interest for the basic rules to apply. The constant \(C\) must always be added for indefinite integrals. A definite integral \(\int_a^b f(x)\,dx\) is a number and does not need \(C\).

Procedure

1. Identify the integrand and the variable of integration.
2. Use linearity to split sums and pull constants outside the integral.
3. Apply standard integration formulas or a suitable technique.
4. Add the constant \(C\) for an indefinite integral.
5. Check by differentiating the answer.

Numerical Example 1: Reversing differentiation

Q. Differentiate \(x^2+3x\), and hence write \(\int (2x+3)\,dx\).

Solution

Step 1: Differentiate.

\[ \frac{d}{dx}(x^2+3x) = 2x+3 \]

Step 2: Reverse the process.

\[ \int (2x+3)\,dx = x^2+3x+C \]

Check

Differentiate \(x^2+3x+C\): \(2x+3\), which matches the integrand.

Explanation of Each Step

1. Step 1 — The power rule gives \(\frac{d}{dx}x^2=2x\) and \(\frac{d}{dx}3x=3\).
2. Step 2 — Since differentiation of \(x^2+3x\) gives \(2x+3\), an antiderivative is \(x^2+3x\). The constant \(C\) is added because any constant disappears on differentiation.

2. Techniques of Integration

2.1 Standard Forms

Definition

Standard forms are the basic integration formulas obtained directly from known derivatives. They are the building blocks for all later techniques.

Detailed Explanation

Each standard form is the reverse of a differentiation rule. For example, \(\frac{d}{dx}x^{n+1}=(n+1)x^n\) gives \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\) for \(n\neq -1\). The exceptional case \(n=-1\) gives \(\int \frac{1}{x}\,dx=\ln|x|+C\). Exponential and trigonometric functions have their own standard forms. A table of these forms should be memorized because substitution and by parts reduce complicated integrals to these basics.

Formula and Conditions

Procedure

1. Split the integrand into terms using linearity.
2. Match each term to a standard form.
3. Integrate term by term.
4. Add a single constant \(C\).
5. Check by differentiating.

Numerical Example 1: Polynomial, reciprocal, and exponential

Q. Evaluate \(\displaystyle\int \left(6x^2 - \frac{4}{x} + 5e^x\right)dx\).

Solution

Step 1: Split by linearity.

\[ \int \left(6x^2 - \frac{4}{x} + 5e^x\right)dx = 6\int x^2\,dx - 4\int \frac{1}{x}\,dx + 5\int e^x\,dx \]

Step 2: Integrate each term.

\[ 6\int x^2\,dx = 6\cdot \frac{x^3}{3} = 2x^3 \]

\[ -4\int \frac{1}{x}\,dx = -4\ln|x| \]

\[ 5\int e^x\,dx = 5e^x \]

Step 3: Combine and add \(C\).

∴ \(\displaystyle\int \left(6x^2 - \frac{4}{x} + 5e^x\right)dx = 2x^3 - 4\ln|x| + 5e^x + C\)

Check

Differentiate: \(6x^2 - \frac{4}{x} + 5e^x\), which is the original integrand.

Explanation of Each Step

1. Step 1 — The integral of a sum is the sum of integrals, and constants can be moved outside.
2. Step 2 — Use \(\int x^2 dx=x^3/3\), \(\int (1/x)dx=\ln|x|\), and \(\int e^x dx=e^x\).
3. Step 3 — Combine the results and add the single constant \(C\).

Numerical Example 2: Exponential base and power

Q. Evaluate \(\displaystyle\int \left(2^x + x^{-1/2}\right)dx\).

Solution

Step 1: Integrate term by term.

\[ \int 2^x\,dx = \frac{2^x}{\ln 2} \]

\[ \int x^{-1/2}\,dx = \frac{x^{1/2}}{1/2} = 2\sqrt{x} \]

Step 2: Combine.

∴ \(\displaystyle\int \left(2^x + x^{-1/2}\right)dx = \frac{2^x}{\ln 2} + 2\sqrt{x} + C\)

Check

Differentiate: \(\frac{2^x\ln 2}{\ln 2} + 2\cdot\frac{1}{2}x^{-1/2} = 2^x + x^{-1/2}\).

Explanation of Each Step

1. Step 1 — Use \(\int a^x dx=a^x/\ln a\) and the power rule with \(n=-1/2\).
2. Step 2 — Add the constant \(C\).

2.2 Integration by Substitution

Definition

Integration by substitution reverses the chain rule. It replaces a complicated expression by a new variable \(u\), turning the integral into a standard form.

Detailed Explanation

If an integral contains a function and its derivative, substitute \(u=g(x)\). Then \(du=g'(x)\,dx\), and the integral becomes \(\int f(u)\,du\). For definite integrals, the limits must also be changed from \(x\)-values to \(u\)-values. The method works best when the integrand can be written as a product \(f(g(x))g'(x)\).

Formula and Conditions

\[ \int f(g(x))g'(x)\,dx = \int f(u)\,du,\qquad u=g(x),\quad du=g'(x)\,dx \]

For definite integrals:

\[ \int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du \]

Conditions: \(g\) must be differentiable and the substitution must convert the whole integral into \(u\). Always return to the original variable for indefinite integrals.

Procedure

1. Choose \(u\) as the inner function whose derivative also appears.
2. Compute \(du=g'(x)\,dx\).
3. Replace all \(x\)-terms and \(dx\) by \(u\)-terms and \(du\).
4. Integrate in \(u\).
5. Substitute back \(u=g(x)\) and add \(C\). For definite integrals, change limits and evaluate directly.

Numerical Example 1: \(\int x e^{x^2}dx\)

Q. Evaluate \(\displaystyle\int x e^{x^2}\,dx\).

Solution

Step 1: Choose substitution.

Let \(u=x^2\). Then \(du=2x\,dx\), so \(x\,dx=\frac{1}{2}du\).

Step 2: Substitute.

\[ \int x e^{x^2}\,dx = \int e^u\cdot \frac{1}{2}\,du = \frac{1}{2}\int e^u\,du \]

Step 3: Integrate.

\[ \frac{1}{2}e^u + C \]

Step 4: Back-substitute.

∴ \(\displaystyle\int x e^{x^2}\,dx = \frac{1}{2}e^{x^2}+C\)

Check

Differentiate: \(\frac{1}{2}e^{x^2}\cdot 2x = x e^{x^2}\).

Explanation of Each Step

1. Step 1 — The derivative of \(x^2\) is \(2x\), and \(x\) appears in the integrand, so \(u=x^2\) is natural.
2. Step 2 — Replace \(x^2\) by \(u\) and \(x\,dx\) by \(\frac{1}{2}du\).
3. Step 3 — \(\int e^u du=e^u\).
4. Step 4 — Replace \(u\) by \(x^2\) and add \(C\).

Numerical Example 2: \(\int \frac{2x+1}{x^2+x+3}dx\)

Q. Evaluate \(\displaystyle\int \frac{2x+1}{x^2+x+3}\,dx\).

Solution

Step 1: Let \(u=x^2+x+3\).

Then \(du=(2x+1)\,dx\).

Step 2: Substitute.

\[ \int \frac{2x+1}{x^2+x+3}\,dx = \int \frac{1}{u}\,du \]

Step 3: Integrate.

\[ \ln|u|+C \]

Step 4: Back-substitute.

∴ \(\displaystyle\int \frac{2x+1}{x^2+x+3}\,dx = \ln|x^2+x+3|+C\)

Check

Differentiate: \(\frac{2x+1}{x^2+x+3}\).

Explanation of Each Step

1. Step 1 — The derivative of the denominator is \(2x+1\), which is exactly the numerator.
2. Step 2 — The integral becomes \(\int \frac{1}{u}du\).
3. Step 3 — \(\int \frac{1}{u}du=\ln|u|+C\).
4. Step 4 — Replace \(u\) by the original denominator.

Numerical Example 3: Definite integral by substitution

Q. Evaluate \(\displaystyle\int_0^1 x\sqrt{x^2+1}\,dx\).

Solution

Step 1: Let \(u=x^2+1\).

Then \(du=2x\,dx\), so \(x\,dx=\frac{1}{2}du\).

Step 2: Change limits.

When \(x=0\), \(u=0^2+1=1\). When \(x=1\), \(u=1^2+1=2\).

Step 3: Substitute and integrate.

\[ \int_0^1 x\sqrt{x^2+1}\,dx = \frac{1}{2}\int_1^2 \sqrt{u}\,du = \frac{1}{2}\int_1^2 u^{1/2}\,du \]

\[ = \frac{1}{2}\left[\frac{2}{3}u^{3/2}\right]_1^2 = \frac{1}{3}\left[u^{3/2}\right]_1^2 \]

\[ = \frac{1}{3}\left(2^{3/2}-1^{3/2}\right) = \frac{1}{3}(2\sqrt{2}-1) \]

∴ \(\displaystyle\int_0^1 x\sqrt{x^2+1}\,dx = \frac{2\sqrt{2}-1}{3} \approx 0.6095\)

Check

Numerically, \(2\sqrt{2}\approx 2.8284\), so \((2.8284-1)/3=1.8284/3=0.6095\). The integrand is positive on [0,1], so the result is positive.

Explanation of Each Step

1. Step 1 — The derivative of \(x^2+1\) is \(2x\), matching the \(x\) in the integrand.
2. Step 2 — For a definite integral, change the limits to \(u\)-values.
3. Step 3 — Integrate \(u^{1/2}\) to get \(\frac{2}{3}u^{3/2}\), then evaluate from 1 to 2.

2.3 Integration by Parts

Definition

Integration by parts reverses the product rule. It is used for integrals of products such as \(x e^x\), \(x\ln x\), or \(x^2\sin x\).

Detailed Explanation

From the product rule \(\frac{d}{dx}(uv)=u'v+uv'\), rearranging and integrating gives

\[ \int u\,dv = uv - \int v\,du \]

Choose \(u\) to be the part that becomes simpler when differentiated, and \(dv\) to be the part that can be integrated. A useful order for choosing \(u\) is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. The function appearing earlier in this list is usually chosen as \(u\).

Formula and Conditions

\[ \int u\,dv = uv - \int v\,du \]

For definite integrals:

\[ \int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du \]

Conditions: \(u\) and \(v\) must be differentiable on the interval. The method may need to be applied more than once.

Procedure

1. Choose \(u\) and \(dv\) using LIATE.
2. Compute \(du\) and \(v\).
3. Apply \(\int u\,dv = uv - \int v\,du\).
4. Evaluate the remaining integral.
5. Add \(C\) for indefinite integrals.

Numerical Example 1: \(\int x e^{2x}dx\)

Q. Evaluate \(\displaystyle\int x e^{2x}\,dx\).

Solution

Step 1: Choose \(u=x\), \(dv=e^{2x}dx\).

Then \(du=dx\) and \(v=\frac{e^{2x}}{2}\).

Step 2: Apply integration by parts.

\[ \int x e^{2x}\,dx = x\cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2}\,dx \]

Step 3: Integrate the remaining term.

\[ = \frac{x e^{2x}}{2} - \frac{1}{2}\cdot \frac{e^{2x}}{2} + C \]

\[ = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C \]

∴ \(\displaystyle\int x e^{2x}\,dx = e^{2x}\left(\frac{x}{2}-\frac{1}{4}\right)+C\)

Check

Differentiate: \(\frac{d}{dx}\left[e^{2x}\left(\frac{x}{2}-\frac{1}{4}\right)\right] = 2e^{2x}\left(\frac{x}{2}-\frac{1}{4}\right)+e^{2x}\cdot\frac{1}{2} = x e^{2x} - \frac{1}{2}e^{2x}+\frac{1}{2}e^{2x}=x e^{2x}\).

Explanation of Each Step

1. Step 1 — \(x\) is algebraic, so it is chosen as \(u\); \(e^{2x}dx\) is \(dv\).
2. Step 2 — Apply the formula.
3. Step 3 — Integrate \(e^{2x}/2\) to get \(e^{2x}/4\), then combine.

Numerical Example 2: \(\int x^2\ln x\,dx\)

Q. Evaluate \(\displaystyle\int x^2\ln x\,dx\).

Solution

Step 1: Choose \(u=\ln x\), \(dv=x^2dx\).

Then \(du=\frac{1}{x}dx\) and \(v=\frac{x^3}{3}\).

Step 2: Apply integration by parts.

\[ \int x^2\ln x\,dx = \ln x\cdot \frac{x^3}{3} - \int \frac{x^3}{3}\cdot\frac{1}{x}\,dx \]

Step 3: Simplify and integrate.

\[ = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx \]

\[ = \frac{x^3}{3}\ln x - \frac{1}{3}\cdot\frac{x^3}{3} + C \]

∴ \(\displaystyle\int x^2\ln x\,dx = \frac{x^3}{3}\ln x - \frac{x^3}{9}+C\)

Check

Differentiate: \(\frac{d}{dx}\left(\frac{x^3}{3}\ln x - \frac{x^3}{9}\right) = x^2\ln x + \frac{x^3}{3}\cdot\frac{1}{x} - \frac{x^2}{3} = x^2\ln x\).

Explanation of Each Step

1. Step 1 — Logarithmic comes before algebraic in LIATE, so \(u=\ln x\).
2. Step 2 — Apply the formula.
3. Step 3 — Simplify \(x^3/x=x^2\), integrate, and add \(C\).

3. Integration of Algebraic, Logarithmic and Exponential Functions

Definition

This topic collects the standard methods for integrating three important families: algebraic functions (powers, roots, rational expressions), logarithmic functions (containing \(\ln x\)), and exponential functions (\(e^x\), \(a^x\)).

Detailed Explanation

Algebraic functions are integrated using the power rule, by simplifying rational expressions through division or partial fractions, or by substitution. Logarithmic functions often appear with \(1/x\), so substitution \(u=\ln x\) is common. Exponential functions integrate to themselves, with a constant factor when the exponent is \(ax+b\). The key is to rewrite the integrand into a standard form before integrating.

Formula and Conditions

\[ \int x^n\,dx = \frac{x^{n+1}}{n+1}+C \quad (n\neq -1) \]

\[ \int \frac{1}{x}\,dx = \ln|x|+C \]

\[ \int e^{ax+b}\,dx = \frac{1}{a}e^{ax+b}+C \quad (a\neq 0) \]

\[ \int a^x\,dx = \frac{a^x}{\ln a}+C \quad (a>0,\ a\neq 1) \]

For rational functions, first divide if the degree of the numerator is not less than the degree of the denominator.

Procedure

1. Simplify the integrand: divide, factor, or rewrite roots as powers.
2. Identify the family: algebraic, logarithmic, or exponential.
3. Apply the appropriate standard form or substitution.
4. Add \(C\) and check by differentiation.

Numerical Example 1: Rational algebraic function

Q. Evaluate \(\displaystyle\int \frac{x+3}{x-2}\,dx\).

Solution

Step 1: Divide the numerator by the denominator.

\[ \frac{x+3}{x-2} = 1 + \frac{5}{x-2} \]

Step 2: Integrate term by term.

\[ \int \left(1 + \frac{5}{x-2}\right)dx = x + 5\ln|x-2| + C \]

∴ \(\displaystyle\int \frac{x+3}{x-2}\,dx = x + 5\ln|x-2| + C\)

Check

Differentiate: \(1+\frac{5}{x-2} = \frac{x-2+5}{x-2} = \frac{x+3}{x-2}\).

Explanation of Each Step

1. Step 1 — Since the numerator and denominator have the same degree, divide: \(x+3 = 1(x-2)+5\).
2. Step 2 — Integrate 1 to get \(x\), and \(5/(x-2)\) to get \(5\ln|x-2|\).

Numerical Example 2: Logarithmic function

Q. Evaluate \(\displaystyle\int \frac{(\ln x)^2}{x}\,dx\).

Solution

Step 1: Let \(u=\ln x\).

Then \(du=\frac{1}{x}dx\).

Step 2: Substitute.

\[ \int \frac{(\ln x)^2}{x}\,dx = \int u^2\,du \]

Step 3: Integrate.

\[ = \frac{u^3}{3}+C \]

Step 4: Back-substitute.

∴ \(\displaystyle\int \frac{(\ln x)^2}{x}\,dx = \frac{(\ln x)^3}{3}+C\)

Check

Differentiate: \(\frac{3(\ln x)^2}{3}\cdot\frac{1}{x} = \frac{(\ln x)^2}{x}\).

Explanation of Each Step

1. Step 1 — The derivative of \(\ln x\) is \(1/x\), which appears in the integrand.
2. Step 2 — Replace \(\ln x\) by \(u\) and \(dx/x\) by \(du\).
3. Step 3 — Use the power rule.
4. Step 4 — Replace \(u\) by \(\ln x\).

Numerical Example 3: Exponential function

Q. Evaluate \(\displaystyle\int \frac{e^{2x}+1}{e^x}\,dx\).

Solution

Step 1: Simplify.

\[ \frac{e^{2x}+1}{e^x} = e^x + e^{-x} \]

Step 2: Integrate.

\[ \int (e^x + e^{-x})\,dx = e^x - e^{-x} + C \]

∴ \(\displaystyle\int \frac{e^{2x}+1}{e^x}\,dx = e^x - e^{-x} + C\)

Check

Differentiate: \(e^x + e^{-x} = \frac{e^{2x}+1}{e^x}\).

Explanation of Each Step

1. Step 1 — Split the fraction: \(e^{2x}/e^x = e^x\), and \(1/e^x=e^{-x}\).
2. Step 2 — Integrate \(e^x\) to \(e^x\) and \(e^{-x}\) to \(-e^{-x}\).

4. Methods of Evaluating Definite Integrals

Definition

A definite integral \(\int_a^b f(x)\,dx\) represents the signed area between the curve \(y=f(x)\) and the \(x\)-axis from \(x=a\) to \(x=b\). It is evaluated using the fundamental theorem of calculus.

Detailed Explanation

The fundamental theorem states that if \(F\) is any antiderivative of \(f\), then \(\int_a^b f(x)\,dx = F(b)-F(a)\). For substitution in a definite integral, change the limits from \(x\)-values to \(u\)-values and never substitute back. For integration by parts, evaluate the product term at the limits. Definite integrals can be positive, negative, or zero depending on the area above and below the axis.

Formula and Conditions

\[ \int_a^b f(x)\,dx = [F(x)]_a^b = F(b)-F(a) \]

Substitution:

\[ \int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du \]

Properties:

\[ \int_a^b f(x)\,dx = -\int_b^a f(x)\,dx,\qquad \int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx \]

Conditions: \(f\) must be continuous on \([a,b]\).

Procedure

1. Find an antiderivative \(F(x)\) by any valid method.
2. Write \([F(x)]_a^b\).
3. Substitute the upper limit, then the lower limit, and subtract.
4. For substitution, change the limits before integrating.
5. Simplify and state the final value.

Numerical Example 1: Polynomial

Q. Evaluate \(\displaystyle\int_0^1 (3x^2+2x)\,dx\).

Solution

Step 1: Find antiderivative.

\[ \int (3x^2+2x)\,dx = x^3+x^2 \]

Step 2: Evaluate from 0 to 1.

\[ [x^3+x^2]_0^1 = (1^3+1^2) - (0^3+0^2) = 2 \]

∴ \(\displaystyle\int_0^1 (3x^2+2x)\,dx = 2\)

Check

The integrand is positive on [0,1], so the area is positive. Differentiating \(x^3+x^2\) gives \(3x^2+2x\).

Explanation of Each Step

1. Step 1 — Use the power rule term by term.
2. Step 2 — Apply the fundamental theorem: substitute 1, then 0, and subtract.

Numerical Example 2: Rational term

Q. Evaluate \(\displaystyle\int_1^2 \left(x+\frac{1}{x}\right)dx\).

Solution

Step 1: Antiderivative.

\[ \int \left(x+\frac{1}{x}\right)dx = \frac{x^2}{2}+\ln|x| \]

Step 2: Evaluate.

\[ \left[\frac{x^2}{2}+\ln x\right]_1^2 = \left(\frac{4}{2}+\ln 2\right) - \left(\frac{1}{2}+\ln 1\right) \]

\[ = (2+\ln 2) - \left(\frac{1}{2}+0\right) = \frac{3}{2}+\ln 2 \]

∴ \(\displaystyle\int_1^2 \left(x+\frac{1}{x}\right)dx = \frac{3}{2}+\ln 2 \approx 2.1931\)

Check

\(\ln 2\approx 0.6931\), so \(1.5+0.6931=2.1931\). The integrand is positive on [1,2].

Explanation of Each Step

1. Step 1 — Integrate \(x\) to \(x^2/2\) and \(1/x\) to \(\ln|x|\).
2. Step 2 — Substitute limits and simplify.

Numerical Example 3: Definite substitution

Q. Evaluate \(\displaystyle\int_0^1 x e^{x^2}\,dx\).

Solution

Step 1: Let \(u=x^2\).

Then \(du=2x\,dx\), so \(x\,dx=\frac{1}{2}du\).

Step 2: Change limits.

\(x=0\Rightarrow u=0\), \(x=1\Rightarrow u=1\).

Step 3: Integrate.

\[ \int_0^1 x e^{x^2}\,dx = \frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}[e^u]_0^1 \]

\[ = \frac{1}{2}(e-1) \]

∴ \(\displaystyle\int_0^1 x e^{x^2}\,dx = \frac{e-1}{2} \approx 0.8591\)

Check

\(e\approx 2.7183\), so \((2.7183-1)/2=0.8591\).

Explanation of Each Step

1. Step 1 — The derivative of \(x^2\) is \(2x\), so \(u=x^2\).
2. Step 2 — Change limits to \(u\)-values.
3. Step 3 — Integrate \(e^u\) and evaluate from 0 to 1.

5. Application of Integration in Business and Economics

5.1 Marginal to Total Cost/Revenue Functions

Definition

Marginal cost \(MC\) is the derivative of total cost \(TC\) with respect to quantity \(x\). Marginal revenue \(MR\) is the derivative of total revenue \(TR\). Integration recovers total functions from marginal functions.

Detailed Explanation

If \(MC=\frac{dTC}{dx}\), then \(TC=\int MC\,dx + C\). The constant \(C\) is the fixed cost, found from a given total cost at some quantity (often \(x=0\)). Similarly, \(TR=\int MR\,dx + C\), and if no sales give zero revenue, \(C=0\). Average cost is \(AC=TC/x\).

Formula and Conditions

\[ TC = \int MC\,dx + C,\qquad TR = \int MR\,dx + C \]

\[ AC = \frac{TC}{x} \]

Conditions: \(x\ge 0\). The constant is determined by a given initial value, such as fixed cost.

Procedure

1. Write the given marginal function.
2. Integrate to get the total function plus \(C\).
3. Use the given total cost/revenue at a specific quantity to find \(C\).
4. Write the total function.
5. If asked, divide by \(x\) to get average cost.

Numerical Example 1: Marginal revenue

Q. If \(MR=10-4x\), find the total revenue function and the total revenue when \(x=2\).

Solution

Step 1: Integrate \(MR\).

\[ TR = \int (10-4x)\,dx = 10x - 2x^2 + C \]

Step 2: Find \(C\).

When \(x=0\), \(TR=0\). So \(0=0-0+C\Rightarrow C=0\).

Step 3: Write \(TR\).

\[ TR = 10x - 2x^2 \]

Step 4: Evaluate at \(x=2\).

\[ TR(2) = 10(2) - 2(2^2) = 20 - 8 = 12 \]

∴ \(TR=10x-2x^2\), and \(TR(2)=12\)

Check

Differentiate \(TR\): \(10-4x=MR\). Also \(TR(0)=0\).

Explanation of Each Step

1. Step 1 — Integrate \(10\) to \(10x\) and \(-4x\) to \(-2x^2\).
2. Step 2 — No sales means no revenue, so \(C=0\).
3. Step 3 — Write the total revenue function.
4. Step 4 — Substitute \(x=2\).

Numerical Example 2: Marginal cost and fixed cost

Q. If \(MC=6x+5\) and fixed cost is Rs 20, find \(TC\) and \(AC\).

Solution

Step 1: Integrate \(MC\).

\[ TC = \int (6x+5)\,dx = 3x^2+5x+C \]

Step 2: Use fixed cost.

Fixed cost is \(TC(0)=C=20\).

Step 3: Write \(TC\).

\[ TC = 3x^2+5x+20 \]

Step 4: Find \(AC\).

\[ AC = \frac{TC}{x} = 3x+5+\frac{20}{x} \]

∴ \(TC=3x^2+5x+20\), and \(AC=3x+5+\frac{20}{x}\)

Check

Differentiate \(TC\): \(6x+5=MC\). \(TC(0)=20\).

Explanation of Each Step

1. Step 1 — Integrate term by term.
2. Step 2 — Fixed cost is the cost when \(x=0\).
3. Step 3 — Include the constant.
4. Step 4 — Divide by \(x\) for average cost.

5.2 Consumer's Surplus and Producer's Surplus

Definition

Consumer’s surplus is the difference between what consumers are willing to pay and what they actually pay. Producer’s surplus is the difference between what producers receive and the minimum they would accept.

Detailed Explanation

At market equilibrium, demand price \(P_d\) equals supply price \(P_s\). If equilibrium quantity is \(x_0\) and price is \(P_0\), then

\[ CS = \int_0^{x_0} P_d(x)\,dx - P_0 x_0 \]

\[ PS = P_0 x_0 - \int_0^{x_0} P_s(x)\,dx \]

These integrals measure the area between the demand curve and the equilibrium price line, and between the equilibrium price line and the supply curve.

Formula and Conditions

\[ CS = \int_0^{x_0} P_d\,dx - P_0 x_0,\qquad PS = P_0 x_0 - \int_0^{x_0} P_s\,dx \]

Conditions: \(P_d\) and \(P_s\) are functions of quantity \(x\). Equilibrium is found by solving \(P_d=P_s\).

Procedure

1. Find equilibrium: solve \(P_d=P_s\) for \(x_0\), then find \(P_0\).
2. Compute \(\int_0^{x_0} P_d\,dx\).
3. Compute \(P_0 x_0\).
4. \(CS = \int P_d - P_0 x_0\).
5. Compute \(\int_0^{x_0} P_s\,dx\).
6. \(PS = P_0 x_0 - \int P_s\).

Numerical Example 1: Consumer’s and producer’s surplus

Q. If \(P_d=18-2x\) and \(P_s=2x+2\), find the consumer’s surplus and producer’s surplus.

Solution

Step 1: Find equilibrium.

\[ 18-2x = 2x+2 \Rightarrow 16=4x \Rightarrow x_0=4 \]

\[ P_0 = 18-2(4)=10 \]

Step 2: Consumer’s surplus.

\[ \int_0^4 (18-2x)\,dx = [18x-x^2]_0^4 = (72-16)-0 = 56 \]

\[ P_0 x_0 = 10(4)=40 \]

\[ CS = 56-40 = 16 \]

Step 3: Producer’s surplus.

\[ \int_0^4 (2x+2)\,dx = [x^2+2x]_0^4 = (16+8)-0 = 24 \]

\[ PS = 40-24 = 16 \]

∴ \(CS=16\), \(PS=16\)

Check

At \(x=4\), both prices equal 10. The areas are positive and equal in this symmetric example.

Explanation of Each Step

1. Step 1 — Set demand equal to supply to find equilibrium quantity and price.
2. Step 2 — Integrate demand from 0 to 4, subtract rectangle \(P_0x_0\).
3. Step 3 — Subtract integral of supply from the rectangle \(P_0x_0\).

6. Area Under a Curve

Definition

The area under a curve \(y=f(x)\) between \(x=a\) and \(x=b\) is the definite integral of the positive parts of \(f\). If \(f(x)\ge 0\) on \([a,b]\), the area is \(\int_a^b f(x)\,dx\). If the curve crosses the \(x\)-axis, split the interval and take absolute values.

Detailed Explanation

Definite integrals give signed area. Area is always positive. When the curve lies below the \(x\)-axis, the integral is negative, so we take its absolute value. The standard method is to find where the curve cuts the axis, split the interval at those points, integrate each piece, and add the absolute values.

Formula and Conditions

\[ \text{Area} = \int_a^b |f(x)|\,dx \]

If \(f(x)\ge 0\) on \([a,b]\), then \(\text{Area}=\int_a^b f(x)\,dx\). If \(f\) changes sign, split at the zeros.

Procedure

1. Sketch or examine the curve to find where it cuts the \(x\)-axis.
2. Split the interval at those zeros.
3. Integrate each piece.
4. Take absolute values where the integral is negative.
5. Add the areas.

Numerical Example 1: Curve above axis

Q. Find the area bounded by \(y=x^2\), the \(x\)-axis, and the ordinates \(x=1\) and \(x=3\).

Solution

Step 1: Set up integral.

Since \(x^2\ge 0\) on [1,3],

\[ \text{Area} = \int_1^3 x^2\,dx \]

Step 2: Integrate.

\[ \int_1^3 x^2\,dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3}-\frac{1}{3} = \frac{26}{3} \]

∴ \(\text{Area} = \frac{26}{3} \approx 8.6667\) square units

Check

The integrand is positive, so the area is positive. \(\frac{26}{3}\approx 8.6667\).

Explanation of Each Step

1. Step 1 — The curve is above the axis on [1,3].
2. Step 2 — Use the power rule and evaluate from 1 to 3.

Numerical Example 2: Curve crossing axis

Q. Find the area bounded by \(y=x^2-4x+3\), the \(x\)-axis, and the ordinates \(x=0\) and \(x=4\).

Solution

Step 1: Find zeros.

\[ x^2-4x+3=(x-1)(x-3) \]

Zeros at \(x=1\) and \(x=3\).

Step 2: Determine signs.

On [0,1], \(y>0\). On [1,3], \(y<0\). On [3,4], \(y>0\).

Step 3: Antiderivative.

\[ F(x)=\frac{x^3}{3}-2x^2+3x \]

Step 4: Evaluate each piece.

\[ F(0)=0,\quad F(1)=\frac{1}{3}-2+3=\frac{4}{3} \]

\[ F(3)=\frac{27}{3}-18+9=0 \]

\[ F(4)=\frac{64}{3}-32+12=\frac{4}{3} \]

Area from 0 to 1: \(\frac{4}{3}-0=\frac{4}{3}\).

Area from 1 to 3: \(\left|0-\frac{4}{3}\right|=\frac{4}{3}\).

Area from 3 to 4: \(\frac{4}{3}-0=\frac{4}{3}\).

Step 5: Total area.

\[ \text{Area}=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4 \]

∴ \(\text{Area}=4\) square units

Check

The three regions are equal in area here, giving a total of 4.

Explanation of Each Step

1. Step 1 — Factor to find where the curve crosses the \(x\)-axis.
2. Step 2 — Test a point in each interval to determine sign.
3. Step 3 — Integrate the polynomial.
4. Step 4 — Evaluate the antiderivative at the split points.
5. Step 5 — Take absolute values and add.

Exam Questions

1. Integrate: \(\int \left(3x^4 + 2x^2 + 2x + \frac{1}{x}\right) dx\) (2025)
2. Find the area bounded by the curve \(y = x^2 - 3x + 2\) and \(x\)-axis and two ordinates at \(x = 1\) and \(x = 4\). (2025)
3. Integrate the following:
   (a) \(\int_0^2 2x\sqrt{2x+1}\, dx\)
   (b) \(\int \dfrac{dx}{x + x\ln x}\) (2025)
4. If the marginal revenue function \(MR = 3x^2 + 2x + 10\), find the total revenue when \(x = 5\) and \(x = 10\). (2025)
5. The marginal cost function is \(MC = 12 + 10x - 6x^2\). The total cost of producing the first 4 units is Rs 64. Find the total cost function. (2025)
6. Integrate: \(\int \frac{x+2}{x-5} dx\) (2024)
7. Find the area of the region bounded by the curve \(y \= \frac{3}{2}x^2\), \(x\) axis and the two ordinates \(x = 0\) and \(x = 2\). (2024)
8. The marginal cost function of manufacturing \(x\) units of a product is given by MC = \(3x^2 - 10x + 3\). The total cost of producing one unit of the product is Rs. 7. Find average cost function. (2024)
9. Integrate the following:
   (a) \( \int \frac{\log x + 5}{x} dx \)
   (b) \( \int (2x + 1)\sqrt{5x + 2} dx \) (2024)
10. The demand function for a product is \(P_d = 8 - 2x\) and the supply function is \(P_s = 2x\) under pure competition. Find the consumer's surplus and producer's surplus. (2024)

Whole Chapter Summary

Concept and standard forms: Integration reverses differentiation and is linear. Standard forms are the base for all techniques.

Substitution and by parts: Substitution reverses the chain rule; by parts reverses the product rule. Choose \(u\) by LIATE for by parts.

Algebraic, logarithmic, exponential: Simplify rational functions by division; use \(u=\ln x\) for logarithmic integrals; exponential integrals gain a factor \(1/a\).

Definite integrals: Evaluate with \(F(b)-F(a)\); change limits for substitution.

Business applications: Integrate marginal functions to get total functions; use initial conditions for the constant. Consumer’s surplus is demand area minus rectangle; producer’s surplus is rectangle minus supply area.

Area under a curve: Use \(\int |f(x)|dx\), splitting at zeros when the curve crosses the axis.

Key Formulas and Points Sheet

Last-Minute Revision Sheet