# DEFINITE INTEGRALS

Definition of Definite Integrals:

$\int_{a}^{b} f(x)\ dx = F(b) – F(a)$
$Note:\ \int_{0}^{a} f(x)\ dx = \int_{0}^{a} f(a – x)\ dx$

Example  1:

$\color{red}{Evaluate: \int_1^2 x^2\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_1^2 x^2\ dx = \frac{x^3}{3} \Biggr]_{1}^{2}$
$= [\frac{2^3}{3} – \frac{1^3}{3}]$
$= \frac{8}{3} – \frac{1}{3}$
$= \frac{8 – 1}{3}$
$= \frac{7}{3}$
$\boxed{\int_1^2 x^2\ dx = \frac{7}{3}}$

#### Example  2:

$\color{red}{Evaluate: \int_0^1 (x^2\ -\ 2)\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_0^1 (x^2\ -\ 2)\ dx = \frac{x^3}{3}\ -\ 2x \Biggr]_{0}^{1}$
$= [(\frac{1^3}{3}\ -\ 2(1))\ -\ 0]$
$= \frac{1}{3}\ -\ 2$
$= \frac{1 – 6}{3}$
$= \frac{-5}{3}$
$\boxed{\int_0^1 (x^2\ -\ 2)dx\ = \frac{-5}{3}}$

Example  3:

$\color{red}{Evaluate: \int_1^2 ( x – x^2)\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_1^2 ( x – x^2)\ dx = (\frac{x^2}{2} -\frac{x^3}{3}) \Biggr]_{1}^{2}$
$= [(\frac{2^2 }{2}-\frac{2^3}{3}) – (\frac{1^2}{2} -\frac{1^3}{3})]$
$= [(\frac{4}{2} – \frac{8}{3}) – (\frac{1}{2} -\frac{1}{3})]$
$= [\frac{4}{2} – \frac{8}{3} – \frac{1}{2} +\frac{1}{3}]$
$=\frac{ 12 – 16 -3 + 2}{6}$
$=\frac{-5}{6}$
$\boxed{\int_1^2 ( x – x^2)\ dx = \frac{-5}{6}}$

Example  4:

$\color{red}{Evaluate: \int_1^2 \frac{1}{x}\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_1^2 \frac{1}{x}\ dx = log\ x \Biggr]_{1}^{2}$
$=[log\ 2 – log\ 1]$
$= log\ 2 – 0$
$= log\ 2$
$\boxed{\int_1^2 \frac{1}{x}\ dx = log\ 2}$

Example  5:

$\color{red}{Evaluate: \int_0^1 (x^2+ 2x + 3)(x + 1)\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_0^1 (x^2+ 2x + 3)(x + 1)\ dx = \int_0^1 (x^3+ x^2 + 2x^2 + 2x + 3x + 3)\ dx$
$=\int_0^1 (x^3+ 3x^2 + 5x + 3)\ dx$
$= \frac{x^4}{4} + 3 \frac{x^3}{3} + 5 \frac{x^2}{2} + 3x \Biggr]_{0}^{1}$
$= [(\frac{1^4}{4} + 1^3 + 5 \frac{1^2}{2} + 3(1))- 0)]$
$= \frac{1}{4} + 1 + \frac{5}{2} + 3$
$= \frac{1 + 4 + 10 + 12}{4}$
$= \frac{27}{4}$
$\boxed{\int_0^1 (x^2+ 2x + 3)(x + 1)\ dx =\frac{27}{4}}$

Example  6:

$\color{red}{Evaluate: \int_0^\frac{\pi}{2} sin^2 x\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$W.K.T\ sin^2 x = \frac{1 – cos\ 2x}{2}$
$\int_0^\frac{\pi}{2} sin^2 x\ dx = \int_0^\frac{\pi}{2} [ \frac{1 – cos\ 2x}{2}]\ dx$
$= \frac{1}{2}[x – \frac{sin\ 2x}{2}] \Biggr]_{0}^{\frac{\pi}{2}}$
$=\frac{1}{2}[(\frac{\pi}{2} – 0) – ( 0 – 0)]$
$=\frac{1}{2}[\frac{\pi}{2}]$
$=\frac{\pi}{4}$
$\boxed{\int_0^\frac{\pi}{2} sin^2 x\ dx =\frac{\pi}{4}}$

Example  7:

$\color{red}{Evaluate: \int_0^\frac{\pi}{2} sin^3 x\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$W.K.T\ sin\ 3x = 3 sin\ x – 4sin^3\ x$
$sin^3 x = \frac{1}{4}[3sin\ x\ -\ sin\ 3x]$
$\int_0^\frac{\pi}{2} sin^3 x\ dx = \int_0^\frac{\pi}{2} \frac{1}{4}[3sin\ x\ -\ sin\ 3x]\ dx$
$= \frac{1}{4}[-\ 3\ cos\ x\ +\ \frac{cos\ 3x}{3}]\Biggr]_{0}^{\frac{\pi}{2}}$
$=\frac{1}{4}[(-3 cos \frac{\pi}{2} + \frac{cos\ 3\frac{\pi}{2}}{3})\ -\ ( -3 cos\ 0 + \frac{cos3(0)}{3})]$
$=\frac{1}{4}[0\ – (-3\ +\ \frac{1}{3})]$
$=\frac{1}{4}[3\ -\ \frac{1}{3}]$
$=\frac{1}{4}[\frac{9\ -\ 1}{3}]$
$=\frac{1}{4}[\frac{8}{3}]$
$=\frac{2}{3}$
$\boxed{\int_0^\frac{\pi}{2} sin^3 x\ dx =\frac{2}{3}}$

#### Example  8:

$\color{red}{Evaluate: \int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx}\ \hspace{18cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$W.K.T\ cos A\ cos B = \frac{1}{2}[cos (A + B) + cos ( A – B)]$
$cos\ 3x\ cos x = \frac{1}{2}[cos (3x\ +\ x)\ +\ cos (3x\ -\ x)]$
$= \frac{1}{2}[cos\ 4x\ +\ cos\ 2x]$
$\int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx = \int_0^\frac{\pi}{2} \frac{1}{2}[cos\ 4x\ +\ cos\ 2x]\ dx$
$= \frac{1}{2}[\frac{sin\ 4x}{4}\ +\ \frac{sin\ 2x}{2}]\Biggr]_{0}^{\frac{\pi}{2}}$
$=\frac{1}{2}[(\frac{sin\ 4\frac{\pi}{2}}{4} + \frac{sin\ 2\frac{\pi}{2}}{2})\ -\ (\frac{sin\ 4(0)}{4}\ + \frac{sin\ 2(0)}{2})]$
$=\frac{1}{2}[(\frac{0}{4}\ +\ \frac{0}{2})- (\frac{0}{4}\ +\ \frac{0}{2})]$
$=\frac{1}{2}[0\ -\ 0]$
$\boxed{\int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx\ =\ 0}$

Example  9:

$\color{red}{Evaluate: \int_0^2 x^2\ \sqrt{1 + x^3}\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u =1 + x^3$
$\frac{du}{dx}= 3x^2$
$x^2\ dx = \frac{1}{3}\ du$
$\int_0^2 x^2\ \sqrt{1 + x^3}\ dx = \frac{1}{3}\ \int_0^2 u^\frac{1}{2}\ du$
$= \frac{1}{3}[\frac{u^\frac{3}{2}}{\frac{3}{2}}] \Biggr]_{0}^{2}$
$=\frac{1}{3} × \frac{2}{3} \Biggr[u^\frac{3}{2} \Biggr]_{0}^{2}$
$=\frac{2}{9} \Biggr[(1 + x^3)^\frac{3}{2} \Biggr]_{0}^{2}$
$=\frac{2}{9}[(1 + 2^3)^\frac{3}{2} – (1 + 0^3)^\frac{3}{2}$
$=\frac{2}{9}[9^\frac{3}{2} – 1^\frac{3}{2}]$
$=\frac{2}{9}[27 – 1]$
$\int_0^2 x^2\ \sqrt{1 + x^3}\ dx\ = \frac{52}{9}$
$\boxed{\int_0^2 x^2\ \sqrt{1 + x^3}\ dx =\frac{52}{9}}$

Example  10:

$\color{red}{Evaluate: \int_0^\frac{\pi}{2} \frac{cos^2 x}{ 1- sin x} \ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_0^\frac{\pi}{2} \frac{cos^2 x}{ 1- sin x} \ dx = \int_0^\frac{\pi}{2} \frac{1 – sin^2 x}{ 1- sin x} \ dx$
$= \int_0^\frac{\pi}{2} \frac{(1+ sin x)(1 – sin x)}{1- sin x}\ dx$
$=\int_0^\frac{\pi}{2} ( 1 + sin x)\ dx$
$=\Biggr[(x – cos x)\Biggr]_{0}^{\frac{\pi}{2}}$
$=[(\frac{\pi}{2} – cos \frac{\pi}{2}) – ( 0 – cos\ 0)]$
$=[(\frac{\pi}{2} – 0) – ( 0 – 1)]$
$=\frac{\pi}{2} + 1$
$\boxed{\int_0^\frac{\pi}{2} \frac{cos^2 x}{ 1- sin x}\ dx =\frac{\pi}{2} + 1}$

Example  11:

$\color{red}{Evaluate: \int_0^\frac{\pi}{2} (2 + sin x)^3 cos x \ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u =2 + sin x$
$\frac{du}{dx}= cos x$

du  = cos x dx

$\int_0^\frac{\pi}{2} (2 + sin x)^3 cos x \ dx = \int_0^\frac{\pi}{2} u^3 \ du$
$=\Biggr[\frac{u^4}{4}\Biggr]_{0}^{\frac{\pi}{2}}$
$=\Biggr[\frac{(2 + sin x)^4}{4}\Biggr]_{0}^{\frac{\pi}{2}}$
$=[\frac{(2 + sin \frac{\pi}{2})^4}{4} – \frac{( 2 + sin\ 0)^4}{4}]$
$=[\frac{(2 + 1)^4}{4} – \frac{( 2 + 0)^4}{4}]$
$=[\frac{(3)^4}{4} – \frac{(2)^4}{4}]$
$=\ \frac{81}{4} – \frac{16}{4}$
$=\ \frac{81- 16}{4}$
$=\ \frac{65}{4}$
$\boxed{\int_0^\frac{\pi}{2} (2 + sin x)^3 cos x\ dx =\ \frac{65}{4}}$

#### Example  12:

$\color{red}{Prove\ that \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx = \frac{\pi}{4}}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Use\ the\ property\ \int_{0}^{a} f(x)\ dx = \int_{0}^{a} f(a – x)\ dx$
$Let\ I= \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx —————— (1)$
$= \int_0^\frac{\pi}{2} \frac{sin\ (\frac{\pi}{2} – x)}{sin\ (\frac{\pi}{2} – x) + cos\ (\frac{\pi}{2} – x)}\ dx$
$= \int_0^\frac{\pi}{2} \frac{cos\ x}{cos\ x + sin\ x}\ dx———————-( 2 )$

Adding ( 1 ) &  ( 2 )

$2\ I= \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx + \int_0^\frac{\pi}{2} \frac{cos\ x}{cos\ x + sin\ x}\ dx$
$= \int_0^\frac{\pi}{2} \frac{sin\ x + cos\ x}{sin\ x + cos\ x}\ dx$
$= \int_0^\frac{\pi}{2} 1\ dx$
$=\Biggr[x \Biggr]_{0}^{\frac{\pi}{2}}$
$=\frac{\pi}{2} – 0$
$=\frac{\pi}{2}$
$I =\frac{\pi}{4}$
$\boxed{\int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx = \frac{\pi}{4}}$