Engneering Mathematics

3. Integral Calculus

Definition of Integration

Integration is the mathematical process of finding the accumulation of a quantity. It is the inverse operation of differentiation. The integral of a function represents the total accumulation over an interval, such as area under a curve or total distance traveled.

There are two main types of integrals:

• Indefinite Integral: Represents a family of functions and includes a constant of integration$C$C.
• Definite Integral: Evaluates the net accumulation over a specific interval$[a, b]$[a,b].

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2. Indefinite and Definite Integrals

a) Indefinite Integral

An indefinite integral finds a general function whose derivative is the given function. It is written as:

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

∫f(x)dx=F(x)+C

where

$C$

C is the constant of integration.

Example:

$\int 2x \,dx = x^2 + C$

∫2xdx=x2+C

b) Definite Integral

A definite integral calculates the total accumulation over an interval

$[a, b]$

[a,b] and is written as:

$\int_{a}^{b} f(x) \,dx = F(b) - F(a)$

∫ab​f(x)dx=F(b)−F(a)

where

$F(x)$

F(x) is the antiderivative of

$f(x)$

f(x).

Example:

$\int_{1}^{3} (x^2) \,dx = \left[ \frac{x^3}{3} \right]_1^3$

∫13​(x2)dx=[3x3​]13​

$= \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$

=333​−313​=327​−31​=326​

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3. Techniques of Integration

a) Substitution Method

Used when an integral contains a composite function. We set

$u$

u as an inner function to simplify integration.

Example:

$\int x e^{x^2} \,dx$

∫xex2dx

Let

$u = x^2$

u=x2, so

$du = 2x \,dx$

du=2xdx.
Rewriting the integral:

$\frac{1}{2} \int e^u \,du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C$

21​∫eudu=21​eu+C=21​ex2+C

b) Integration by Parts

Used for integrating the product of two functions, based on:

$\int u \,dv = uv - \int v \,du$

∫udv=uv−∫vdu

Example:

$\int x \ln x \,dx$

∫xlnxdx

Let

$u = \ln x$

u=lnx, so

$du = \frac{1}{x}dx$

du=x1​dx, and let

$dv = xdx$

dv=xdx, so

$v = \frac{x^2}{2}$

v=2x2​.
Applying the formula:

$\int x \ln x \,dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx$

∫xlnxdx=2x2​lnx−∫2x2​⋅x1​dx

$= \frac{x^2}{2} \ln x - \int \frac{x}{2} dx$

=2x2​lnx−∫2x​dx

$= \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$

=2x2​lnx−4x2​+C