Evaluation of Definite Integrals by using their properties

In this section we shall see how the properties of definite integrals are useful to evaluate definite integrals

Example 1

$\displaystyle \int _{ -1 }^{ 1 }{ f\left( x \right) } dx,\quad where\quad f\left( x \right)\begin{matrix} =1-2x\quad when\quad x\le 0 \\ =1+2x\quad when\quad x\ge 0 \end{matrix}$

Solution:

$\displaystyle we\quad have\quad \int _{ -1 }^{ 1 }{ f\left( x \right) } dx\quad =\int _{ -1 }^{ 0 }{ f\left( x \right) } dx+\int _{ 0 }^{ 1 }{ f\left( x \right) } dx$

$\displaystyle =\int _{ -1 }^{ 0 }{ \left( 1-2x \right) } dx+\int _{ 0 }^{ 1 }{ \left( 1+2x \right) } dx\\ ={ \left[ x-{ x }^{ 2 } \right] }_{ -1 }^{ 0 }+{ \left[ x+{ x }^{ 2 } \right] }_{ 0 }^{ 1 }\\ =\left[ 0-\left( -1-1 \right) \right] +\left[ \left( 1+1 \right) -\left( 0 \right) \right] \\ =4$

Example 2

$\displaystyle \int _{ 0 }^{ 1 }{ \left| 5x-3 \right| } dx$

Solution:

$\displaystyle we\quad have\quad \left| 5x-3 \right| \begin{matrix} =-\left( 5x-3 \right) \quad when\quad 5x-3<0\quad i.e.,\quad x<\frac { 3 }{ 5 } \\ =\left( 5x-3 \right) \quad when\quad 5x-3\ge 0\quad i.e.,\quad x\ge \frac { 3 }{ 5 } \end{matrix}$

$\displaystyle \therefore \int _{ 0 }^{ 1 }{ \left| 5x-3 \right| } dx=\int _{ 0 }^{ \frac { 3 }{ 5 } }{ \left| 5x-3 \right| } dx+\int _{ \frac { 3 }{ 5 } }^{ 1 }{ \left| 5x-3 \right| } dx$

$\displaystyle =\int _{ 0 }^{ \frac { 3 }{ 5 } }{ -\left( 5x-3 \right) } dx+\int _{ \frac { 3 }{ 5 } }^{ 1 }{ \left( 5x-3 \right) } dx\\ ={ \left[ 3-\frac { 5{ x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ \frac { 3 }{ 5 } }+{ \left[ \frac { 5{ x }^{ 2 } }{ 2 } -3x \right] }_{ \frac { 3 }{ 5 } }^{ 1 }\\ =\left( \frac { 9 }{ 5 } -\frac { 9 }{ 10 } \right) +\left( -\frac { 1 }{ 2 } +\frac { 9 }{ 10 } \right) \\ =\frac { 13 }{ 10 }$

Example 3

Prove that $\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { x } }{ \sin { x } +\cos { x } } } dx=\frac { \pi }{ 4 }$

Solution:

$\displaystyle Let\quad I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { x } }{ \sin { x } +\cos { x } } } dx\qquad .......\left( i \right) \\ Then\quad I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { \left( \frac { \pi }{ 2 } -x \right) } }{ \sin { \left( \frac { \pi }{ 2 } -x \right) } +\cos { \left( \frac { \pi }{ 2 } -x \right) } } } dx\qquad \left[ Using\quad \int _{ 0 }^{ a }{ f\left( x \right) } dx=\int _{ 0 }^{ a }{ f\left( a-x \right) } dx \right] \\ or\quad I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos { x } }{ \cos { x } +\sin { x } } } dx\qquad .......\left( ii \right)$

Adding (i) and (ii)

$\displaystyle 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { x } }{ \sin { x } +\cos { x } } } dx+\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos { x } }{ \cos { x } +\sin { x } } } dx\\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { x } +\cos { x } }{ \sin { x } +\cos { x } } } dx\\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ 1 } .dx\\ ={ x }_{ 0 }^{ \frac { \pi }{ 2 } }\\ =\frac { \pi }{ 2 } -0\\ I=\frac { \pi }{ 4 } \\ i.e.\quad \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { x } }{ \sin { x } +\cos { x } } } dx=\frac { \pi }{ 4 }$

Example 4

Prove that $\displaystyle \int _{ 0 }^{ 2a }{ f\left( x \right) } dx=\int _{ 0 }^{ 2a }{ f\left( 2a-x \right) } dx$

Solution:

$\displaystyle Let\quad I=\int _{ 0 }^{ 2a }{ f\left( x \right) } dx$ Put 2a-x so that -dx=dt or dx=-dt

Now, x=0, t= 2a-0 and x=a , t=2a-a=a

$\displaystyle \therefore I=\int _{ 2a }^{ 0 }{ f\left( 2a-t \right) } \left( -dt \right) =-\int _{ 2a }^{ 0 }{ f\left( 2a-t \right) } dt\\ =\int _{ 0 }^{ 2a }{ f\left( 2a-t \right) } dt\qquad \left[ \because \int _{ a }^{ b }{ f\left( x \right) } dx=-\int _{ b }^{ a }{ f\left( x \right) } dx \right] \\ =\int _{ 0 }^{ 2a }{ f\left( 2a-x \right) } dx\qquad \left[ \because \int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ b }{ f\left( t \right) } dt \right] \\ Hence,\quad \int _{ 0 }^{ 2a }{ f\left( x \right) } dx=\int _{ 0 }^{ 2a }{ f\left( 2a-x \right) } dx$

Example 5

Prove that $\displaystyle \int _{ 0 }^{ 2a }{ f\left( x \right) } dx=\int _{ 0 }^{ a }{ f\left( x \right) } dx+\int _{ 0 }^{ a }{ f\left( 2a-x \right) } dx$

Solution:

$\displaystyle Let\quad I=\int _{ 0 }^{ 2a }{ f\left( x \right) } dx\quad Then\\ I=\int _{ 0 }^{ a }{ f\left( x \right) } dx+\int _{ 0 }^{ 2a }{ f\left( x \right) } dx\qquad \left[ \because \int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ c }{ f\left( x \right) } dx+\int _{ c }^{ b }{ f\left( x \right) } dx \right] \\ =\int _{ 0 }^{ a }{ f\left( x \right) } dx+{ I }_{ 1 }\quad where\quad { I }_{ 1 }=\int _{ 0 }^{ 2a }{ f\left( x \right) } dx$

Put 2a-t =x, so that dx=-dt. Also x=a, t=a and x=2a, t=0

$\displaystyle \therefore { I }_{ 1 }=\int _{ a }^{ 2a }{ f\left( x \right) } dx=\int _{ a }^{ 0 }{ f\left( 2a-t \right) } \left( -dt \right) =-\int _{ a }^{ 0 }{ f\left( 2a-t \right) } dt\\ =\int _{ 0 }^{ a }{ f\left( 2a-t \right) } dt=\int _{ 0 }^{ a }{ f\left( 2a-x \right) } dx\\ \therefore I=\int _{ 0 }^{ a }{ f\left( x \right) } dx+\int _{ 0 }^{ a }{ f\left( 2a-x \right) } dx$

Example 6

$\displaystyle \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left| \sin { x } \right| } dx$

Solution:

$\displaystyle Let\quad f\left( x \right) =\left| \sin { x } \right| \\ Then\quad f\left( -x \right) =\left| \sin { \left( -x \right) } \right| \\ =\left| -\sin { x } \right| \\ =\left| \sin { x } \right| =f\left( x \right)$ So, f(x) is an even function.

$\displaystyle \therefore \int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left| \sin { x } \right| } dx\\ =2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left| \sin { x } \right| } dx=2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left| \sin { x } \right| } dx\qquad \left[ \because \sin { x } \ge 0\quad for\quad 0\le x\le \frac { \pi }{ 2 } \right] \\ =2{ \left[ -\cos { x } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\\ =2\left( -\cos { \frac { \pi }{ 2 } } +\cos { 0 } \right) \\ =2$