Integral Calculus

Integral Calculus is the branch of calculus focused on accumulation, such as finding areas under curves, volumes, and total changes. It involves indefinite integrals (antiderivatives) and definite integrals (numerical values over intervals). Integral calculus essentially reverses differentiation, helping us reconstruct a function from its rate of change. Widely used in physics, engineering, and economics, it allows us to analyze continuous change and solve real-world problems involving motion, growth, and distribution.

1.0What is Integral Calculus?

Integral Calculus is the branch of calculus concerned with accumulation—how small pieces add up to make a whole. If differential calculus is about breaking things down (derivatives), then integral calculus is about putting things back together (integrals).

It primarily deals with:

• Indefinite Integrals: Represent antiderivatives of functions.
• Definite Integrals: Represent actual values, like area or total change.

2.0Indefinite Integrals

An indefinite integral is the reverse process of differentiation. It gives a general form of all antiderivatives of a function and includes an arbitrary constant of integration C, since derivatives of constants are zero.

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

Here:

• f(x) is the integrand,
• F(x) is any function such that F'(x) = f(x),
• C is the constant of integration.

Example:

$\int 2xdx=x^2+C$

Indefinite integrals are functions, not numbers. They're used when you're looking for a general formula rather than a specific value.

3.0Definite Integrals

A definite integral gives the net accumulation of a quantity over an interval [a, b]. It produces a numerical value and has direct geometric meaning—like the area under a curve.

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

Where:

• F(x) is the antiderivative of f(x),
• a and b are the limits of integration.

Key Properties:

${\int }_a^{a}f(x)dx=0$

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

• If f(x) $\ge$ 0, the integral represents area.

Example:

${\int }_0^2{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^2=\frac{8}{3}$

Definite integrals are numbers, often used to compute areas, volumes, and total change.

4.0The Core Idea: Summing Infinitesimally Small Quantities

To find the area under a curve y = f(x), we slice it into thin rectangles and add their areas. As we take more slices, we get closer to the true area. This infinite sum leads us to a definite integral:

${\int }_a^{b}f(x)dx$

Where:

• f(x): the function
• a, b: the limits of integration
• dx: an infinitesimally small width

5.0The Fundamental Theorem of Calculus

This theorem connects differentiation and integration:

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

Here, F(x) is an antiderivative of f(x), meaning F'(x) = f(x).

6.0Common Integration Techniques

1. Substitution (u-substitution):
Use when an integral looks like a chain rule in reverse.

$\int f(g(x))g^{\prime }(x)dx=\int f(u)du$

2. Integration by Parts:
\int u \, dv = uv - \int v \, du

3. Partial Fractions:
Break complex rational expressions into simpler terms.

4. Trigonometric Identities:
Useful for integrating sin²x, cos²x, etc.

5. Special Integrals:
Learn standard forms like:

$\int \frac{1}{1+x^2}dx={\tan }^{-1}x+C$

7.0Applications of Integral Calculus

• Area under curves
• Volume of solids of revolution
• Displacement and distance from velocity
• Center of mass and moment of inertia
• Probability distributions (in stats)

8.0Solved Examples on Integral Calculus

Example 1: Evaluate: $\int \frac{x}{\sqrt{1+x^2}}dx$

Solution:
Let u = $1+x^2\Rightarrow du=2xdx$

So,

$\frac{du}{2}=xdx$

$\int \frac{x}{\sqrt{1+x^2}}dx=\int \frac{1}{\sqrt{u}}\cdot \frac{du}{2}=\frac{1}{2}\int u^{-1/2}du$

$=\frac{1}{2}\cdot \frac{u^{1/2}}{1/2}=\sqrt{u}=\sqrt{1+x^2}+C$

Example 2: Evaluate: ${\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$

Solution:

Use the property:

${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

Let:

$I={\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$

Apply the property:

$I={\int }_0^{\pi }\frac{(\pi -x)\sin x}{1+{\cos }^{2}x}dx$

Add both:

$2I={\int }_0^{\pi }\frac{x\sin x+(\pi -x)\sin x}{1+{\cos }^{2}x}dx={\int }_0^{\pi }\frac{\pi \sin x}{1+{\cos }^{2}x}dx$

$2I=\pi {\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx$

Let $u=\cos x\Rightarrow du=-\sin xdx$

Change limits:
When x = 0 $\Rightarrow u=1,$
When $x=\pi \Rightarrow u=-1$

So:

$2I=-\pi {\int }_1^{-1}\frac{1}{1+u^2}du=\pi {\int }_{-1}^1\frac{1}{1+u^2}du$

This is a standard integral:

$\pi {\left[{\tan }^{-1}u\right]}_{-1}^1=\pi ({\tan }^{-1}1-{\tan }^{-1}(-1))=\pi \left(\frac{\pi }{4}+\frac{\pi }{4}\right)=\pi \cdot \frac{\pi }{2}\Rightarrow 2I=\frac{{\pi }^2}{2}\Rightarrow I=\frac{{\pi }^2}{4}$

Example 3: Evaluate: ${\int }_0^3[x]dx\text{(where [x] is the greatest integer function)}$

Solution:

Break it at integers:

${\int }_0^3[x]dx={\int }_0^{1}0dx+{\int }_1^{2}1dx+{\int }_2^{3}2dx$= 0 + 1 + 2 = 3

Example 4: Evaluate:

$\int x\ln xdx$

Solution:

Use integration by parts:
Let u = $\ln x,dv=xdx$
Then $du=\frac{1}{x}dx,v=\frac{x^2}{2}$

$\int x\ln xdx=\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\cdot \frac{1}{x}dx=\frac{x^2}{2}\ln x-\int \frac{x}{2}dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$

9.0Practice Questions

1. Find the integral: $\int (3x^2+2x+1)dx$

2. Evaluate:${\int }_0^2{x}^{2}dx$

3. Find the area under the curve $y=\sin \text{x from x = 0 to x =}\pi .$

4. Solve using substitution: $\int x\sqrt{x^2+1}dx$

5. Evaluate using integration by parts: $\int x{e}^{x}dx$