This page introduces you to definite integrals. It gives an explanation like how it is defined and different ways how we can use it and calculation can be done on it.

By the time you have gone through this you would know how to use definite integral as the sum, fundamental theorem of calculus and to calculate the area under the curves and be familiar with limit as a sum.

## Definition of Definite Integral

Let F(x) be any anti derivative {integral} of f(x), then for any two values of the independent variable x, say a and b, the difference F(b)-F(a) is called the definite integral of f(x) from a to b and is denoted by $\int_{a}^{b}f{x}dx$. Thus
$\int_{a}^{b}f(x)dx$ = $F(b) – F(a)$,
where $F(x)$ is the anti derivative of $f(x)$. The number $a$ and $b$ are called the limits of integration; $a$ is the lower limit and $b$ is the upper limit. Usually F(b)-F(a) is abbreviated by writing $F(x)|^{b}_{a}$.

Definite integral is also known as Riemann integral.

## Properties of Definite Integrals

There are some definite integral rules as follows:

$\int_{a}^{b}f{x}dx$ = $-\int_{b}^{a}f{x}dx$

$\int_{a}^{a}f{x}dx$ = $0$

$\int_{a}^{b}f{x}dx$ = $\int_{a}^{b}f{y}dy$

$\int_{a}^{b}f{x}dx$ = $\int_{a}^{c}f{x}dx + \int_{c}^{b}f{x}dx$ where a
$\int_{0}^{a}f{x}dx$ = $\int_{0}^{a}f{a-x}dx$

$\int_{0}^{a}\frac{f{x}}{f(x)+f(a-x)}dx$ = $\frac{a}{2}$

$\int_{0}^{2a}f{x}dx$ = $\int_{0}^{a}f{x}dx +\int_{0}^{a}f{2a-x}dx$

$\int_{0}^{2a}f{x}dx$ = 0 , if $f(2a-x)$=$-f(x)$

= $2\int_{0}^{a}f{x}dx$, if $f(2a-x)$= $f(x)$

$\int_{-a}^{a}f{x}dx$ = $2\int_{0}^{a}f{x}dx$, if $f(-x)$ = $f(x)$ i.e. $f(x)$ is even.

= 0, if $f(-x)$ = $-f(x)$ i.e. $f(x)$ is odd.

$\int_{a}^{b}f{x}dx$ = $\int_{a}^{b}f{a+b-x}dx$

$\int_{a}^{b}\frac{f{x}}{f(x)+f(a+b-x)}dx$ = $\frac{b-a}{2}$

If $f(x)\geq0$ on the interval [a,b] then $\int_{a}^{b}f{x}dx\geq 0$

If f(x) is a periodic function of period a, i.e. $f(a+x)$ = $f(x)$, then

$\int_{0}^{na}f{x}dx$ = $n \int_{o}^{a}f{x}dx$
$\int_{a}^{na}f{x}dx$ = $(n-1)\int_{0}^{a}f{x}dx$
$\int_{b}^{b+a}f{x}dx$ is independent of b.

Theorem: $\int_{a}^{b}f(x)dx$ = $\int_{a}^{b}f(z)dz$ (change of variable)

Proof: Let $\int f(x)dx$ = $F(x)$, then $\int f(z)dz$ = $F(z)$

L.H.S = $\int_{a}^{b}f(x)dx$ = $[F(x)]_{a}^{b}$ = $F(b) - F(a)$

R.H.S = $\int_{a}^{b}f(z)dz$ = $[F(z)]_{a}^{b}$ = $F(b) – F(a)$

$\therefore$ L.H.S = R.H.S

Thus, the value of a definite integral is independent of the variable and depends on the lower and upper limits only.

Theorem: $\int_{a}^{b}f(x)dx$ = $-\int_{b}^{a}f(x)dx$ (Interchange of limits)

Proof: Let $\int f(x)dx$ = $F(x)$

L.H.S = $\int_{a}^{b}f(x)dx$ = $[F(x)]_{a}^{b}$ = $F(b) - F(a)$

R.H.S = $-\int_{b}^{a}f(x)dx$ = $-[F(x)]_{b}^{a}$ = $-[F(a) – F(b)]$

=$F(b)-F(a)$

$\therefore$ L.H.S = R.H.S

Thus, an interchange of the lower and upper limits changes the sign of the definite integrals.

Theorem: $\int_{a}^{b}f(x)dx$ = $\int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx$ where aDecomposition of range of integration)

Proof: Let $\int f(x)dx$ = $F(x)$

L.H.S = $\int_{a}^{b}f(x)dx$ = $[F(x)]_{a}^{b}$ = $F(b) – F(a)$

R.H.S = $\int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx$ = $[F(x)]_{a}^{c} + [F(x)]_{c}^{b}$

= $[F(c) – F(a)] + [F(b) - F(c)]$ = $F(b) - F(a)$

$\therefore$ L.H.S = R.H.S

Summation of Definite Integral: If f(x) is a continuous and single valued function define on the interval [a,b], then the definite integral $\int_{-a}^{a}f{x}dx$ is defined as,

$\int_{a}^{b}f{x}dx$ = $\lim_{h\to o}h[f(a+h)+f(a+2h)+.....+f(a+nh)]$

where $nh$ = $b-a$

f(x) is said to be odd function if $f(-x)$ = $-f(x)$
f(x) is said to be even function if $f(-x)$ = $f(x)$

## Applications of Definite Integrals

• The area bounded by the curve $y$ = $f(x)$, x-axis and the ordinates $x$ = $a$ and $x$ = $b$ (where b>a) is given by
$A$ = $\int_{a}^{b}ydx$ = $\int_{a}^{b}f(x)dx$

• The area bounded by the curve x = g(y) is given by,

$A$ = $\int_{c}^{d}xdy$ = $\int_{a}^{b}g(y)dy$

• The area bounded by the curve y = h(x), x-axis and the two ordinates x=a and x= b is given by
$A$ = $\left | \int_{a}^{c}ydx \right |+ \left | \int_{c}^{b}ydx \right |$

• Let y = f(x) and y = g(x) be two curves which is shown below,

The area under the shaded region can be written as

$A$ = $lower\ curve\ area$ – $upper\ curve\ area$

$A$ = $\int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx$

## Derivative of a Definite Integral

The first fundamental theorem of the calculus says that, if a function $f$ is continuous on [a,b] and $x$ is an arbitrary point in [a,b], then the definite integral   $\int_{a}^{x}f(t)dt$ is a function of the variable upper limit x, and its derivative is given by

$\frac{\mathrm{d} }{\mathrm{d} x}\left [ \int_{a}^{x}f(t)dt \right ]$ = $f(x)$
Example;

Find the $\frac{\mathrm{d} }{\mathrm{d} x}\left [ \int_{1}^{x}t^{3}dt \right ]$.

$\int_{1}^{x}t^{3}dt$ = $\left [ \frac{t^{4}}{4} \right ]_{1}^{x}$ = $\frac{x^{4}}{4}-\frac{1}{4}$

$\therefore$ $\frac{\mathrm{d} }{\mathrm{d} x}\left [ \int_{1}^{x}t^{3}dt \right ]$ = $\frac{\mathrm{d} }{\mathrm{d} x}\left [ \frac{x^{4}}{4}-\frac{1}{4} \right ]$ = $x^{3}$

If the variable x is not appearing in either (or both) of the limits of integration, the definite integral result will not involve x, and so the derivative of that definite integral will be zero.

## Definite Integral Examples

The following problems will help us to learn more about Definite integrals.

### Solved Examples

Question 1: Show that $\int_{0}^{\pi /2}\sin 2x \log \tan xdx$ = 0
Solution:

Let I = $\int_{0}^{\pi /2}\sin 2x \log \tan xdx$

we have $\int_{0}^{a}f(x)dx$ = $\int_{0}^{a}f(a-x)dx$

I = $\int_{0}^{\pi /2}\sin 2(\frac{\pi}{2}-x) \log \tan (\frac{\pi}{2}-x)dx$

=  $\int_{0}^{\pi /2}\sin (\pi-2x) \log \cot xdx$

= $\int_{0}^{\pi /2}\sin 2x \log (\tan x)^{-1}dx$

= $-\int_{0}^{\pi /2}\sin 2x \log \tan xdx$ = -I

$\therefore$ 2I = 0, Hence I = 0.

Question 2: Evaluate $\int_{-2}^{2}|y|dy$
Solution:

$|y| = 0$,if y = 0 which lies between -2 and 2

when $-2\leq y\leq 0$, |y| = $-y$

= $\left [ x-x^{2} \right ]_{1/4}^{1/2} + \left [x^{2}-x \right ]_{1/2}^{1}$

= $\left ( \frac{1}{2}-\frac{1}{4}- \frac{1}{4} +\frac{1}{16} \right ) + \left ( 1-1-\frac{1}{4}+\frac{1}{2}\right )$

=$\frac{5}{16}$