Integration – Definite, Indefinite, and Applications

Overview

Integration is the mathematical process of finding the accumulation of quantities. In calculus, it is the inverse of differentiation. Two primary forms are used in the syllabus:

The unit is structured into five sections:

1. Antiderivatives and Indefinite Integration
2. Definite Integration and the Fundamental Theorem of Calculus
3. Techniques of Integration
4. Applications of Integration
5. Integration in Real‑World Contexts

1. Antiderivatives and Indefinite Integration

An antiderivative of $f(x)$ is a function $F(x)$ such that $F^{\prime }(x)=f(x)$. The indefinite integral is written as

$$\int f(x)dx=F(x)+C,$$

where $C$ is an arbitrary constant.

Worked Example 1 – Polynomial Antiderivative

Find $\int (3x^2-4x+5)dx$.

$$\begin{array}{rl}\int 3x^{2}dx& =x^3,\\ \int -4xdx& =-2x^2,\\ \int 5dx& =5x.\end{array}$$

Hence

$$\int (3x^2-4x+5)dx=x^3-2x^2+5x+C.$$

2. Definite Integration and the Fundamental Theorem of Calculus

The definite integral gives the net signed area between the curve $y=f(x)$ and the $x$-axis over $[a,b]$.

Fundamental Theorem of Calculus (Part 1)
If $F$ is an antiderivative of $f$ on $[a,b]$, then

$${\int }_a^{b}f(x)dx=F(b)-F(a).$$

Worked Example 2 – Computing a Definite Integral

Evaluate ${\int }_0^{\pi }\sin xdx$.

Antiderivative: $F(x)=-\cos x$.
Apply the theorem:

$${\int }_0^{\pi }\sin xdx=(-\cos \pi )-(-\cos 0)=(-(-1))-(-1)=2.$$

Figure: Graph of $\sin x$ from $0$ to $\pi$ and the area under the curve

3. Techniques of Integration

3.1 Substitution

Let $u=g(x)$. Then $du=g^{\prime }(x)dx$.

Example 3 – Substitution

$$\int 2x\sqrt{1+x^2}dx.$$

Set $u=1+x^2$, $du=2xdx$.

$$\int \sqrt{u}du=\frac{2}{3}{u}^{3/2}+C=\frac{2}{3}(1+x^2{)}^{3/2}+C.$$

3.2 Integration by Parts

$$\int udv=uv-\int vdu.$$

Example 4 – Integration by Parts

$$\int x{e}^{x}dx.$$

Choose $u=x$, $dv=e^{x}dx$. Then $du=dx$, $v=e^x$.

$$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C=e^x(x-1)+C.$$

3.3 Partial Fractions

Decompose $\frac{P(x)}{Q(x)}$ into simpler fractions.

Example 5 – Partial Fractions

$$\int \frac{1}{x^2-1}dx=\int \frac{1}{(x-1)(x+1)}dx.$$

Decompose:

$$\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.$$

Solve: $A=1/2$, $B=-1/2$.

$$\int \frac{1}{x^2-1}dx=\frac{1}{2}\int \frac{dx}{x-1}-\frac{1}{2}\int \frac{dx}{x+1}=\frac{1}{2}\ln |x-1|-\frac{1}{2}\ln |x+1|+C.$$

3.4 Trigonometric Substitution

Use substitutions like $x=a\sin \theta$ to simplify square roots.

Example 6 – Trigonometric Substitution

$$\int \frac{dx}{\sqrt{1-x^2}}.$$

Let $x=\sin \theta$, $dx=\cos \theta d\theta$.

$$\int \frac{\cos \theta d\theta }{\sqrt{1-{\sin }^2\theta }}=\int \frac{\cos \theta d\theta }{\cos \theta }=\int d\theta =\theta +C.$$

Since $x=\sin \theta$, $\theta =\arcsin x$.

$$\int \frac{dx}{\sqrt{1-x^2}}=\arcsin x+C.$$

4. Applications of Integration

Worked Example 7 – Continuous Compounding

A student deposits $10,000 in a savings account with an annual interest rate of 5 % compounded continuously. Find the amount after 3 years.

$$A=P{e}^{rt}=10,000\times e^{0.05\times 3}\approx 10,000\times e^{0.15}\approx 10,000\times 1.161834\approx 11,\mathrm{618.34.}$$

5. Integration in Real‑World Contexts

In the real world

1. eSewa – Continuous Interest on Credit
   eSewa calculates interest on credit balances using continuous compounding. The formula $A=P{e}^{rt}$ is an integral of the instantaneous growth rate $rP{e}^{rx}$.
2. Daraz – Order Queue Modeling
   The average number of orders in the queue over time is modeled by ${\int }_0^T\lambda (t)dt$, where $\lambda (t)$ is the arrival rate. Integration gives total orders processed.
3. Kathmandu Traffic – Travel Time Estimation
   Travel time along a stretch with variable speed $v(x)$ is ${\int }_{x_1}^{x_2}\frac{dx}{v(x)}$. Integration sums infinitesimal time intervals over distance.

Concrete Worked Example – Kathmandu Traffic

Assume speed varies as $v(x)=30-0.5x$ km/h over a 20 km stretch ($x$ in km). Compute travel time.

$$T={\int }_0^{20}\frac{dx}{30-0.5x}.$$

Let $u=30-0.5x$, $du=-0.5dx$, $dx=-2du$.

$$T={\int }_{30}^{20}\frac{-2du}{u}=2{\int }_{20}^{30}\frac{du}{u}=2[\ln u{]}_{20}^{30}=2(\ln 30-\ln 20)\approx 2(3.401-2.996)\approx 0.81\text{hours}\approx 48.6\text{minutes}.$$

Visuals

Figure: Definite Integral of $f(x)=x^2$ from 0 to 3

Figure: Number Line Showing Limits of Integration

Figure: Bar Chart of Integration Techniques

Mermaid Flowchart – Choosing an Integration Technique

flowchart TD
    A["Start"] --> B{"Is integrand a product?"}
    B -->|"Yes"| C["Use Integration by Parts"]
    B -->|"No"| D{"Is integrand a rational function?"}
    D -->|"Yes"| E["Use Partial Fractions"]
    D -->|"No"| F{"Does integrand contain sqrt(a^2 - x^2)?"}
    F -->|"Yes"| G["Use Trig Substitution"]
    F -->|"No"| H["Use Substitution"]

Advantages and Disadvantages of Integration Techniques

Exam tip

• Identify the type of integral: Indefinite vs. definite.
• Select the appropriate technique: Use the flowchart above as a quick reference.
• Check limits carefully: For definite integrals, evaluate the antiderivative at upper and lower limits.
• Simplify before integrating: Factor, expand, or rationalize to reduce complexity.
• Show all steps: Partial credit is awarded for method selection and intermediate steps.
• Practice with real‑world problems: Continuous compounding, area calculations, and work problems are common.

Real pictures

A scientific calculator used for integration calculations (Image: Matti Blume, CC BY-SA 2.0, via Wikimedia Commons) Graph paper for sketching functions and integration limits (Image: Mikus, CC BY-SA 4.0, via Wikimedia Commons)