Some Properties of Definite Integrals

PROPERTIES OF DEFINITE INTEGRATION

1. Change of variable of integration is immaterial so long as limits of integration remain the same i.e. $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(t)dt$

2. $\underset{a}{\overset{b}{\int }}f(x)dx=-\underset{b}{\overset{a}{\int }}f(x)dx$

3. $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{c}{\int }}f(x)dx+\underset{c}{\overset{b}{\int }}f(x)dx$.

Generally we break the limit first at the points where f(x) is discontinuous and second at the points where definition of f(x) changes.

Illustration 1: Evaluate $\underset{0}{\overset{\frac{5\pi }{12}}{\int }}\left[\tan x\right]dx$, where [.] is the greatest integer function.

Solution: Let I =$\underset{0}{\overset{\frac{5\pi }{12}}{\int }}\left[\tan x\right]dx$

Value of tan x at x = $\frac{5\pi }{12}$ is 2 + $\sqrt{3}$

Value of tan x at x = 0 is 0

Integers between 0 and 2 + $\sqrt{3}$ are 1, 2, 3

$\therefore$ tan x = 1, tan x = 2, tan x = 3

$\Rightarrow$ x = tan^-1 1, x = tan^-1 2, x = tan^-1 3

$\therefore$ I = $\underset{0}{\overset{{\tan }^{-1}1}{\int }}\left[\tan x\right]dx+\underset{{\tan }^{-1}1}{\overset{{\tan }^{-1}2}{\int }}\left[\tan x\right]dx+\underset{{\tan }^{-1}2}{\overset{{\tan }^{-1}3}{\int }}\left[\tan x\right]dx$+ $\underset{{\tan }^{-1}3}{\overset{\frac{5\pi }{12}}{\int }}\left[\tan x\right]dx$

= $\underset{0}{\overset{{\tan }^{-1}1}{\int }}0dx+\underset{{\tan }^{-1}1}{\overset{{\tan }^{-1}2}{\int }}1dx+\underset{{\tan }^{-1}2}{\overset{{\tan }^{-1}3}{\int }}2dx$+ $\underset{{\tan }^{-1}3}{\overset{\frac{5\pi }{12}}{\int }}3dx$

= 0 + $\left({\tan }^{-1}2-{\tan }^{-1}1\right)+2\left({\tan }^{-1}3-{\tan }^{-1}2\right)+3\left(\frac{5\pi }{12}-{\tan }^{-1}3\right)$

= $\frac{5\pi }{4}-\frac{\pi }{4}-{\tan }^{-1}3-{\tan }^{-1}2$

= $\pi$ $-\left[{\tan }^{-1}\left(\frac{3+2}{1-6}\right)+\pi \right]$= -tan^-1 (-1)

= $\frac{\pi }{4}$.

4. $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$. In particular $\underset{0}{\overset{a}{\int }}f(x)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$.

Illustration 2: If f, g, h be continuous function on [0, a] such that

f(a - x) = f(x), g(a - x) = - g(x) and 3h(x) - 4h(a - x) = 5, then prove that ${\int }_0^{a}f\left(x\right)\text{ g}\left(\text{x}\right)\text{ h}\left(\text{x}\right)dx=0$.

Solution: I = $\underset{0}{\overset{a}{\int }}f(x)g(x)h(x)dx$ = $\underset{0}{\overset{a}{\int }}f(a-x)g(a-x)h(a-x)dx$

= –$\underset{0}{\overset{a}{\int }}f(x)g(x)h(a-x)dx$

7I = 3I + 4I

= $\underset{0}{\overset{a}{\int }}f(x)g(x)\{3h(x)-4h(a-x)\}dx$

= 5$\underset{0}{\overset{a}{\int }}f(x)g(x)dx$ = 0, since f (a – x) g (a – x) = –f (x) g (x)

$\Rightarrow$ I = 0

If f (x) is a periodic function with period T, then

$\underset{a}{\overset{a+nT}{\int }}f(x)dx=n\underset{0}{\overset{T}{\int }}f(x)dxwheren\in I$,

In particular,

(i) if a = 0, $\underset{0}{\overset{nT}{\int }}f(x)dx=n\underset{0}{\overset{T}{\int }}f(x)dx$ where n $\in$ I

(ii) If n = 1, $\underset{a}{\overset{a+T}{\int }}f(x)dx=\underset{0}{\overset{T}{\int }}f(x)dx$

Illustration 3: Evaluate $\underset{0}{\overset{10\pi }{\int }}\left|\sin x\right|dx$.

Solution: Let I =$\underset{0}{\overset{10\pi }{\int }}\left|\sin x\right|dx$

We know that |sinx| is a periodic function with period $\pi$

Hence I = $10\underset{0}{\overset{\pi }{\int }}\left|\sin x\right|dx$