# DEFINITE INTEGRALS (Excercise Problems with Solutions)

$\underline{PART\ -\ A}$
$1.\ \color {red}{Evaluate\ :\ \int_1^2 (x\ +\ x^2)\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_1^2 ( x + x^2)\ dx = (\frac{x^2}{2} +\frac{x^3}{3}) \Biggr]_{1}^{2}$
$= [(\frac{2^2 }{2}+\frac{2^3}{3}) – (\frac{1^2}{2} +\frac{1^3}{3})]$
$= [(\frac{4}{2} + \frac{8}{3}) – (\frac{1}{2} +\frac{1}{3})]$
$= [\frac{4}{2} + \frac{8}{3} – \frac{1}{2} -\frac{1}{3}]$
$=\frac{ 12 + 16 -3 – 2}{6}$
$=\frac{23}{6}$
$\boxed{\int_1^2 ( x + x^2)\ dx = \frac{23}{6}}$
$2.\ \color {red}{Evaluate\ :\ \int_0^1 (3x^2\ -\ 2x\ +\ 7)\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_0^1 (3x^2\ -\ 2x\ +\ 7)\ dx = 3\ \frac{x^3}{3} \ -2\ \frac{x^2}{2}\ +\ 7x \Biggr]_{0}^{1}$
$= [(1^3 – 1^2 + 7(1))- 0)]$
$= 1\ -\ 1 +\ 7$
$= 7$
$\boxed{\int_0^1 (3x^2\ -\ 2x\ +\ 7)\ dx =\ 7}$
$3.\ \color {red}{Evaluate\ :\ \int_1^3 (4x\ -\ 5x^2)\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_1^3 (4x\ -\ 5x^2)\ dx = 4\ \frac{x^2}{2} \ -\ 5\ \frac{x^3}{3}\ \Biggr]_{1}^{3}$
$= [(2(3^2)\ -\ 5 \frac{3^3}{3})\ -\ (2(1^2)\ -\ 5\ \frac{1^3}{3})]$
$= [(18\ -\ 45)\ -\ (2\ -\ \frac{5}{3})]$
$= [(-27)\ -\ (2\ -\ \frac{5}{3})]$
$= [-\ 29\ +\ \frac{5}{3}]$
$= \frac{-\ 82}{3}$
$\boxed{\int_1^3 (4x\ -\ 5x^2)\ dx =\ \frac{-\ 82}{3}}$
$4.\ \color {red}{Evaluate: \int_0^\frac{\pi}{4} sec^2 x\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_0^\frac{\pi}{4} sec^2 x\ dx =\ tan\ x\ \Biggr]_{0}^{\frac{\pi}{4}}$
$=\ (tan\ \frac{\pi}{4} – 0)$
$=\ (1\ -\ 0)$
$=\ 1$
$\boxed{\int_0^\frac{\pi}{4} sec^2 x\ dx =\ 1}$
$\underline{PART\ -\ B}$
$5.\ \color {red}{Evaluate\ :\ \int_1^2 (x^2\ +\ x\ +\ 1)\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_1^2 ( x^2\ +\ x\ +\ 1)\ dx = (\frac{x^3}{3} +\frac{x^2}{2}\ +\ x) \Biggr]_{1}^{2}$
$= [(\frac{2^3 }{3}+\frac{2^2}{2}\ +\ 2) – (\frac{1^3}{3} +\frac{1^2}{2}\ +\ 1)]$
$= [(\frac{8}{3}\ +\ \frac{4}{2}\ +\ 2) – (\frac{1}{3}\ +\frac{1}{2}\ +\ 1)]$
$= [\frac{8}{3}\ +\ 2\ +\ 2\ -\ \frac{1}{3} – \frac{1}{2} -\ 1]$
$=\ 3\ +\ \frac{8}{3}\ -\ \frac{1}{3} – \frac{1}{2}$
$=\frac{18\ +\ 16-\ 2\ -3}{6}$
$= \frac{29}{6}$
$\boxed{\int_1^2 ( x^2\ +\ x\ +\ 1)\ dx = \frac{29}{6}}$
$6.\ \color {red}{Evaluate\ :\ \int_0^1 \frac{dx}{\sqrt{1 – x^2}}}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_0^1 \frac{dx}{\sqrt{1 – x^2}}\ =\ sin^{-1}\ x\ \Biggr]_{0}^{1}$
$=[sin^{-1}\ (1) – sin^{-1}\ (0)]$
$=\frac{\pi}{2}\ -\ 0$
$=\frac{\pi}{2}$
$\boxed{\int_0^1 \frac{dx}{\sqrt{1 – x^2}}\ =\frac{\pi}{2}}$
$7.\ \color {red}{Evaluate\ :\ \int_0^\frac{\pi}{2} (2 + sin x)^2 cos x \ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u =2 + sin x$
$\frac{du}{dx}= cos x$
$du\ =\ cos x\ dx$
$\int_0^\frac{\pi}{2} (2 + sin x)^2 cos x \ dx = \int_0^\frac{\pi}{2} u^2 \ du$
$=\Biggr[\frac{u^3}{3}\Biggr]_{0}^{\frac{\pi}{2}}$
$=\Biggr[\frac{(2 + sin x)^3}{3}\Biggr]_{0}^{\frac{\pi}{2}}$
$=[\frac{(2 + sin \frac{\pi}{2})^3}{3} – \frac{( 2 + sin\ 0)^3}{3}]$
$=[\frac{(2 + 1)^3}{3} – \frac{( 2 + 0)^3}{3}]$
$=[\frac{(3)^3}{3} – \frac{(2)^3}{3}]$
$=\ \frac{27}{3} – \frac{8}{3}$
$=\ \frac{27- 8}{3}$
$=\ \frac{19}{3}$
$\boxed{\int_0^\frac{\pi}{2} (2 + sin x)^2 cos x\ dx =\ \frac{19}{3}}$
$\underline{PART\ -\ C}$
$8.\ \color{red}{Evaluate: \int_0^\frac{\pi}{2} \frac{sin^2 x}{ 1- cos x} \ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int_0^\frac{\pi}{2} \frac{sin^2 x}{ 1- cos x} \ dx = \int_0^\frac{\pi}{2} \frac{1 – cos^2 x}{ 1- cos x} \ dx$
$= \int_0^\frac{\pi}{2} \frac{(1+ cos x)(1 – cos x)}{1- cos x}\ dx$
$=\int_0^\frac{\pi}{2} ( 1 + cos x)\ dx$
$=\Biggr[(x\ +\ sin x)\Biggr]_{0}^{\frac{\pi}{2}}$
$=[(\frac{\pi}{2}\ – sin \frac{\pi}{2}) – ( 0 + sin\ 0)]$
$=\frac{\pi}{2}\ -\ 1$
$\boxed{\int_0^\frac{\pi}{2} \frac{sin^2 x}{ 1- cos x} \ dx =\frac{\pi}{2} + 1}$
$9.\ \color{red}{Evaluate: \int_0^\frac{\pi}{4} tan\ x\ sec^2\ x \ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$put\ u\ =\ tan\ x$
$\frac{du}{dx}=\ sec^2\ x$
$du\ =\ sec^2\ x\ dx$
$\int_0^\frac{\pi}{4} tan\ x\ sec^2\ x \ dx = \int_0^\frac{\pi}{4} u \ du$
$=\Biggr[\frac{u^2}{2}\Biggr]_{0}^{\frac{\pi}{4}}$
$=\Biggr[\frac{(tan\ x)^2}{2}\Biggr]_{0}^{\frac{\pi}{4}}$
$=\ \frac{(tan \frac{\pi}{4})^2}{2} – \frac{( tan\ 0)^2}{2}$
$=\ \frac{(1)^2}{2} – \frac{(0)^2}{2}$
$=\frac{1}{2}$
$\boxed{\int_0^\frac{\pi}{4} tan\ x\ sec^2\ x \ dx =\frac{1}{2}}$
$10.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int_0^1\ (2x\ +\ 3)^4\ dx\ \hspace{2cm}\ (ii)\ \int_0^\frac{\pi}{2} cos^2 x\ dx}\ \hspace{10cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(i)\ \int_0^1\ (2x\ +\ 3)^4\ dx\ \hspace{10cm}$
$put\ u\ =\ 2x\ +\ 3$
$\frac{du}{dx}\ =\ 2$
$dx\ = \frac{1}{2}\ du$
$\int_0^1\ (2x\ +\ 3)^4\ dx = \frac{1}{2}\ \int_0^1 u^4\ du$
$= \frac{1}{2}[\frac{u^5}{5}] \Biggr]_{0}^{1}$
$= \frac{1}{2}[\frac{(2x\ +\ 3)^5}{5}] \Biggr]_{0}^{1}$
$=\frac{1}{10}[(2(1) + 3)^5\ -\ (2(0) + 3)^5]$
$=\frac{1}{10}[(5)^5\ – (3)^5]$
$=\frac{1}{10}[3125\ – 243]$
$=\frac{2882}{10}$
$=\ 288.2$
$\boxed{\int_0^1\ (2x\ +\ 3)^4 \ dx =\ 288.2}$
$(ii)\ \int_0^\frac{\pi}{2} cos^2 x\ dx\ \hspace{10cm}$
$\int_0^\frac{\pi}{2} cos^2 x\ dx = \int_0^\frac{\pi}{2} [ \frac{1\ +\ cos\ 2x}{2}]\ dx$
$= \frac{1}{2}[x +\ \frac{sin\ 2x}{2}] \Biggr]_{0}^{\frac{\pi}{2}}$
$=\frac{1}{2}[(\frac{\pi}{2}\ +\ 0) – ( 0\ +\ 0)]$
$=\frac{1}{2}[\frac{\pi}{2}]$
$=\frac{\pi}{4}$
$\boxed{\int_0^\frac{\pi}{2} cos^2 x\ dx =\frac{\pi}{4}}$
$11.\ \color{red}{Evaluate: \int_0^\frac{\pi}{2} log(tan\ x)\ dx}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Use\ the\ property\ \int_{0}^{a} f(x)\ dx = \int_{0}^{a} f(a – x)\ dx$
$Let\ I= \int_0^\frac{\pi}{2} log(tan\ x)\ dx —————— (1)$
$= \int_0^\frac{\pi}{2} {log(tan\ (\frac{\pi}{2} – x))}\ dx$
$= \int_0^\frac{\pi}{2} log(cot\ x)\ dx———————-( 2 )$
$Adding\ (1)\ and\ (2)$
$2\ I= \int_0^\frac{\pi}{2} log(tan\ x)\ dx + \int_0^\frac{\pi}{2} log(cot\ x)\ dx$
$= \int_0^\frac{\pi}{2} (log(tan\ x)\ +\ log( cot\ x))\ dx$
$= \int_0^\frac{\pi}{2} (log(tan\ x\ ×\ cot\ x))\ dx$
$= \int_0^\frac{\pi}{2} log\ 1\ dx$
$= \ 0$
$\boxed{\int_0^\frac{\pi}{2} log(tan\ x)\ dx =\ 0}$