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