# Integration

The process in which we find the integral of some function is called as integration process. In other words we can say that the integration is the opposite of the differentiation process. The integration method is used usually for finding areas, volumes and even central points many a times. There are many more applications where the process of integration is used.

Two different types of integrals come in picture when we talk about integrals. First is indefinite integral and the second is definite integral.

Indefinite integral is often called as general integral at times:

The left picture represents an indefinite integral and the right picture is representing a definite integral. Both pictures showcase the area under a curve f(x) which can be determined using integration process. The difference is that in case of first picture there are no fix interval specified while in right picture the area is to determined in the interval [a, b].When the function is integrated over a given interval, say [c, d] then we say it a definite integral and when no interval is specified we say it indefinite integral.

## Indefinite Integral

Indefinite Integrals:
A function H (x) is called the antiderivative of another function h (x) only if in the given domain of h, for all values of ‘x’, we are having the following.

H’(x) = h (x)

$\int h (x) dx = H (x) + C$

where ‘C’ is the constant of integration.

There are some basic formulas or rule that are specified for the integration process just like in differentiation:

1) $\int a dy$ = ay + C

2) $\int y^n dy$ = $\frac{y^(n +1)}{(n+1)}$ + C; n $\neq$ -1

3) $\int sin y\ dy$ = -$cos y + C$

4) $\int cos y\ dy$ = $sin y + C$

5) $\int sec^2 y\ dy$ = $tan y + C$

6) $\int csc^2 y\ dy$ = -$cot y + C$

7) $\int sec y\ tan y\ dy$ = $sec y + C$

8) $\int csc y\ cot y\ dy$ = -$csc y + C$

9) $\int$ $\frac{1}{y}$ $dy$ = $ln |y| + C$

10) $\int e^y dy$ = $e^y + C$

11) $\int a^y dy$ = $\frac{a^y}{ln a}$ + $C$; $a > 0$, a $\neq$ 1

12) $\int$ $\frac{1}{\sqrt {1 – y^2}}$ $dy$ = $sin^{-1} y + C$

13) $\int$ $\frac{1}{(1 + y^2)}$ dy = $tan^(-1) y + C$

14) $\int$ $\frac{1}{(|y| \sqrt {y^2 – 1})}$ $dy$ = $sec^(-1) + C$

15) $\int (f(y) ± g(y)) dy$ = $\int f(y) dy$ ± $\int g(y) dy$

These formulas are used for only simple given functions. For some complex or compound functions we have other ways to solve them like:

1) Integration by parts

2) Quotient rule

3) Substitution method

4) Partial fractions

Definite integrals: In this case, we simply integrate the given function and then we find the value of the function obtained in the interval given. The process is explained below.

Suppose that $h (x) = H’(x)$. Now in order to find the value of $\int h(x)$ in the interval [a, b], we find the value of H (b) - H (a).

## Examples

Few of the examples on integration are given below:
Example 1: Find the integral of the given function: h (x) = $16 x^7 + 6 x + 8$ .

Solution: $\int h (x) dx$ =$\int (16 x^7 + 6 x + 8) dx$

= 16 $\int x^7 dx$ + 6 $\int x dx$ + 8 $\int dx$

= 16 * $\frac{x^8}{8}$ + 6 $\frac{x^2}{2}$ + 8 x + C

= $2 x^8 + 6 x^2 + 8 x + C$

Example 2: Solve $\int_1^2$ ($x^2+xy + 2$) dx

Solution: $\int_1^2$ ($x^2+xy + 2$) dx = $\int_1^2 (x^2)dx + \int_1^2 (xy)dx + \int_1^2(2dx)$

= $\frac{x^3}{3}|_1^2$ + $\frac{x^2y}{2}|_1^2$ + $2x|_1^2$

= $\frac{1}{3}$ $\times$ (8 - 1) + $\frac{y}{2}$ $\times$ (4 - 1) + 2(2 - 1)

= $\frac{7}{3}$ + $\frac{3y}{2}$ + 2

= $\frac{13}{3}$ + $\frac{3y}{2}$

Example 3: Integrate function with respect to x, f(x) = $\frac{4 + x^2}{1+x^2}$

Solution: f(x) = $\frac{4 + x^2}{1+x^2}$

or f(x) = $\frac{1 +3 + x^2}{1+x^2}$

or f(x) = $\frac{1+ x^2}{1+x^2}$ + $\frac{3}{1+x^2}$

or f(x) = 1 + $\frac{3}{1+x^2}$

Now $\int$f(x) = $\int$(1 + $\frac{3}{1+x^2}$)dx

= $\int$ dx + $\int$ $\frac{3}{1+x^2}$dx

= x + 3 tan$^{-1}$x

Therefore integration of f(x) is x + 3 tan$^{-1}$x