Class 12: Definite Integrals – Miscellaneous Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: } \text{If } \int_{0}^{2a}f(x)\,dx = \int_{0}^{a}f(x)\,dx+\int_{0}^{a}f(k-x)\,dx, \text{ then the value of } k \text{ is }$
$\displaystyle \text{(a) } a \quad \text{(b) } 2a \quad \text{(c) independent of } a \quad \text{(d) None of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, } \int_{0}^{2a} f(x)\,dx=\int_{0}^{a} f(x)\,dx+\int_{0}^{a} f(k-x)\,dx \quad (i)$
$\displaystyle \text{We know that, } \int_{0}^{2a} f(x)\,dx=\int_{0}^{a} f(x)\,dx+\int_{0}^{a} f(2a-x)\,dx \quad (ii)$
$\displaystyle \text{On comparing Eqs. (i) and (ii), we get } k=2a$
$\\$

$\displaystyle \textbf{Question 2: } \text{Evaluate } \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx \quad (i)$
$\displaystyle \text{Using } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx$
$\displaystyle I=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\,dx \quad (ii)$
$\displaystyle \text{On adding (i) and (ii), we get}$
$\displaystyle 2I=\int_{2}^{8} \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}}\,dx$
$\displaystyle =\int_{2}^{8} 1\,dx=[x]_{2}^{8}$
$\displaystyle =8-2=6$
$\displaystyle \therefore I=3$
$\\$

$\displaystyle \textbf{Question 3: } \text{Evaluate } \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}}x}{\sin^{\frac{3}{2}}x+\cos^{\frac{3}{2}}x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}}x}{\sin^{\frac{3}{2}}x+\cos^{\frac{3}{2}}x}\,dx \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \frac{\cos^{\frac{3}{2}}x}{\cos^{\frac{3}{2}}x+\sin^{\frac{3}{2}}x}\,dx \quad (ii)$
$\displaystyle \text{On adding (i) and (ii), we get}$
$\displaystyle 2I=\int_{0}^{\frac{\pi}{2}} 1\,dx=\frac{\pi}{2}$
$\displaystyle \therefore I=\frac{\pi}{4}$
$\\$

$\displaystyle \textbf{Question 4: } \text{Evaluate } \int_{-1}^{2}|x^{3}-x|\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{-2}^{2} |x^{3}-x|\,dx$
$\displaystyle x^{3}-x=x(x-1)(x+1)$
$\displaystyle \text{Thus, } x=-1,0,1 \text{ are critical points}$
$\displaystyle I=\int_{-2}^{-1} -(x^{3}-x)\,dx+\int_{-1}^{0} (x^{3}-x)\,dx+\int_{0}^{1} -(x^{3}-x)\,dx+\int_{1}^{2} (x^{3}-x)\,dx$
$\displaystyle =-\int_{-2}^{-1} (x^{3}-x)\,dx+\int_{-1}^{0} (x^{3}-x)\,dx-\int_{0}^{1} (x^{3}-x)\,dx+\int_{1}^{2} (x^{3}-x)\,dx$
$\\$

$\displaystyle \textbf{Question 5: } \text{Evaluate } \int_{1}^{4}|x-2|\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, } |x-2|=\begin{cases} -(x-2), & 1\leq x<2 \\ (x-2), & 2\leq x\leq 4 \end{cases}$
$\displaystyle \text{Let } I=\int_{1}^{4} |x-2|\,dx$
$\displaystyle =\int_{1}^{2} -(x-2)\,dx+\int_{2}^{4} (x-2)\,dx$
$\displaystyle =\left[-\frac{x^{2}}{2}+2x\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2x\right]_{2}^{4}$
$\displaystyle =\left[-2+4-\left(-\frac{1}{2}+2\right)\right]+\left[8-8-\left(2-4\right)\right]$
$\displaystyle =\left[2-\frac{3}{2}\right]+\left[0-(-2)\right]$
$\displaystyle =\frac{1}{2}+2=\frac{5}{2}$
$\\$

$\displaystyle \textbf{Question 6: } \text{Evaluate } \int_{-6}^{3}|x+3|\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{-3}^{3} |x+3|\,dx$
$\displaystyle =\int_{-3}^{0} -(x+3)\,dx+\int_{0}^{3} (x+3)\,dx$
$\displaystyle =\left[-\frac{x^{2}}{2}-3x\right]_{-3}^{0}+\left[\frac{x^{2}}{2}+3x\right]_{0}^{3}$
$\displaystyle =\left[0-0-\left(-\frac{9}{2}+9\right)\right]+\left[\frac{9}{2}+9\right]$
$\displaystyle =\left[-\frac{9}{2}+9\right]+\left[\frac{9}{2}+9\right]$
$\displaystyle =9+9=18$
$\\$

$\displaystyle \textbf{Question 7: } \text{Evaluate } \int_{0}^{\pi} \frac{x\tan x}{\sec x+\tan x}\,dx. \quad$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}\,dx \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\tan(\pi-x)}{\sec(\pi-x)+\tan(\pi-x)}\,dx$
$\displaystyle =\int_{0}^{\pi}\frac{-(\pi-x)\tan x}{-\sec x-\tan x}\,dx$
$\displaystyle =\int_{0}^{\pi}\frac{(\pi-x)\tan x}{\sec x+\tan x}\,dx \quad (ii)$
$\displaystyle \text{On adding Eqs. (i) and (ii), we get}$
$\displaystyle 2I=\pi\int_{0}^{\pi}\frac{\tan x}{\sec x+\tan x}\,dx$
$\displaystyle =\pi\int_{0}^{\pi}\frac{\tan x(\sec x-\tan x)}{(\sec x+\tan x)(\sec x-\tan x)}\,dx$
$\displaystyle =\pi\int_{0}^{\pi}(\tan x\sec x-\tan^{2}x)\,dx$
$\displaystyle =\pi\left[\int_{0}^{\pi}\sec x\tan x\,dx-\int_{0}^{\pi}(\sec^{2}x-1)\,dx\right]$
$\displaystyle =\pi\left[[\sec x]_{0}^{\pi}-[\tan x-x]_{0}^{\pi}\right]$
$\displaystyle =\pi[(-1-1)-\{(0-\pi)-(0-0)\}]$
$\displaystyle =\pi[-2+\pi]=\pi(\pi-2)$
$\displaystyle \therefore I=\frac{\pi}{2}(\pi-2)$
$\\$

$\displaystyle \textbf{Question 8: } \text{Evaluate } \int_{0}^{\frac{\pi}{4}} \log(1+\tan x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{4}}\log(1+\tan x)\,dx \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\log\left[1+\tan\left(\frac{\pi}{4}-x\right)\right]dx$
$\displaystyle =\int_{0}^{\frac{\pi}{4}}\log\left(\frac{2}{1+\tan x}\right)dx \quad (ii)$
$\displaystyle \text{On adding (i) and (ii), we get}$
$\displaystyle 2I=\int_{0}^{\frac{\pi}{4}}\log\left[(1+\tan x)\cdot \frac{2}{1+\tan x}\right]dx$
$\displaystyle =\int_{0}^{\frac{\pi}{4}}\log 2\,dx$
$\displaystyle =(\log 2)\left[\frac{\pi}{4}-0\right]$
$\displaystyle =\frac{\pi}{4}\log 2$
$\displaystyle \therefore I=\frac{\pi}{8}\log 2$
$\\$

$\displaystyle \textbf{Question 9: } \text{Evaluate } \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^{x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{1+e^{x}}\,dx$
$\displaystyle =\int_{-\frac{\pi}{2}}^{0}\frac{\cos x}{1+e^{x}}\,dx+\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{1+e^{x}}\,dx$
$\displaystyle \text{Let } I_{1}=\int_{-\frac{\pi}{2}}^{0}\frac{\cos x}{1+e^{x}}\,dx$
$\displaystyle \text{Put } x=-t \Rightarrow dx=-dt$
$\displaystyle I_{1}=\int_{\frac{\pi}{2}}^{0}\frac{\cos(-t)}{1+e^{-t}}(-dt)$
$\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{\cos t}{1+e^{-t}}dt$
$\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{e^{t}\cos t}{1+e^{t}}dt$
$\displaystyle \therefore I=\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{1+e^{x}}dx+\int_{0}^{\frac{\pi}{2}}\frac{e^{x}\cos x}{1+e^{x}}dx$
$\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{\cos x(1+e^{x})}{1+e^{x}}dx$
$\displaystyle =\int_{0}^{\frac{\pi}{2}}\cos x\,dx$
$\displaystyle =[\sin x]_{0}^{\frac{\pi}{2}}$
$\displaystyle =1-0=1$
$\\$

$\displaystyle \textbf{Question 10: } \text{Evaluate } \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\sin x|\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\sin x|\,dx$
$\displaystyle \text{Since } |\sin x|\text{ is an even function}$
$\displaystyle \therefore I=2\int_{0}^{\frac{\pi}{2}} \sin x\,dx$
$\displaystyle =2[-\cos x]_{0}^{\frac{\pi}{2}}$
$\displaystyle =2[-\cos\frac{\pi}{2}+\cos 0]$
$\displaystyle =2[0+1]=2$
$\\$