Exercise 9.10 - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $9.10$
Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
The value of $\int_0^{\frac{2}{3}} \frac{d x}{\sqrt{4-9 x^2}}$ is
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{4}$
$(d) \pi$
Solution:
(a) $\frac{\pi}{6}$
Hint:
$
\begin{aligned}
\int_0^{2 / 3} \frac{d x}{\sqrt{(2)^2-(3 x)^2}} & =\left[\frac{1}{3} \sin ^{-1}\left(\frac{3 x}{2}\right)\right]_0^{2 / 3} \\
& =\frac{1}{3} \sin ^{-1}\left(\frac{3}{2} \cdot \frac{2}{3}\right)=\frac{1}{3} \sin ^{-1}(1) \\
& =\frac{1}{3}\left(\frac{\pi}{2}\right)=\frac{\pi}{6}
\end{aligned}
$
Question 2.
The value of $\int_{-1}^2|x| d x$ is
(a) $\frac{1}{2}$
(b) $\frac{3}{2}$
(c) $\frac{5}{2}$
(d) $\frac{7}{2}$
Solution:
(c) $\frac{5}{2}$

Hint:
$
\begin{aligned}
\int_{-1}^2|x| d x & =\int_{-1}^0-x d x+\int_0^2 x d x=\left(-\frac{x^2}{2}\right)_{-1}^0+\left(\frac{x^2}{2}\right)_0^2 \\
& =\frac{1}{2}+\frac{4}{2}=\frac{5}{2} .
\end{aligned}
$
Question 3.
For any value of $n \in \mathbb{Z}, \int_0^\pi e^{\cos ^2 x} \cos ^3[(2 n+1) x] d x$ is
(a) $\frac{\pi}{2}$
(b) $\pi$
(c) 0
(d) 2
Solution:
(c) 0
Hint:
$
\begin{aligned}
\mathrm{I} & =\int_0^\pi e^{\cos ^2 x} \cos ^3[(2 n+1) x] d x \\
\int_0^a f(x) d x & =\int_0^a f(a-x) d x \\
\mathrm{I} & =-\int_0^\pi e^{\cos ^2 x} \cos ^3[(2 n+1) x] d x \\
(1)+(2) \Rightarrow 2 \mathrm{I} & =0 \\
\mathrm{I} & =0 .
\end{aligned}
$
Question 4.
The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos x d x$ is.
(a) $\frac{3}{2}$
(b) $\frac{1}{2}$
(c) 0
(d) $\frac{2}{3}$
Solution:
(d) $\frac{2}{3}$
Hint:

It is an even function
$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos x d x=2 \int_0^{\pi / 2} \sin ^2 x \cos x d x=2\left(\frac{\sin ^3 x}{3}\right)_0^{\pi / 2}=\frac{2}{3} .
$
Question 5.
The value of $\int_{-4}^4\left[\tan ^{-1}\left(\frac{x^2}{x^4+1}\right)+\tan ^{-1}\left(\frac{x^4+1}{x^2}\right)\right] d x$ is.
(b) $2 \pi$
(c) $3 \pi$
(d) $4 \pi$
Solution:
(d) $4 \pi$
Hint:
$
\begin{aligned}
& =\int_{-4}^4 \tan ^{-1}\left[\frac{\frac{x^2}{x^4+1}+\frac{x^4+1}{x^2}}{1-\left(\frac{x^2}{x^4+1}\right)\left(\frac{x^4+1}{x^2}\right)}\right] d x \\
& =\int_{-4}^4 \tan ^{-1}[\infty] d x=\frac{\pi}{2}[x]_{-4}^4=\frac{\pi}{2}[4+4]=4 \pi .
\end{aligned}
$

Question 6.
The value of $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{2 x^7-3 x^5+7 x^3-x+1}{\cos ^2 x}\right) d x$ is..
(a) 4
(b) 3
(c) 2
(d) 0
Solution:
(c) 2
Hint:
$
\begin{aligned}
& =\int_{-\pi / 4}^{\pi / 4} \frac{2 x^7}{\cos ^2 x}-\int_{-\pi / 4}^{\pi / 4} \frac{3 x^5}{\cos ^2 x}+\int_{-\pi / 4}^{\pi / 4} \frac{7 x^3}{\cos ^2 x}-\int_{-\pi / 4}^{\pi / 4} \frac{x}{\cos ^2 x}+\int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos ^2 x} d x \\
& =0-0+0-0+2 \int_0^{\pi / 4} \sec ^2 x d x=2[\tan x]_0^{\pi / 4} \\
& =2\left[\tan \frac{\pi}{4}-\tan (0)\right]=2(1)=2 .
\end{aligned}
$
Question 7.
If $f(x)=\int_0^x t \cos t d t$, then $\frac{d f}{d x}=$

(a) $\cos x-x \sin x$
(b) $\sin x+x \cos x$
(c) $x \cos x$
(d) $x \sin x$
Solution:
(c) $x \cos x$
Hint:
$
\begin{aligned}
& =\int_{-\pi / 4}^{\pi / 4} \frac{2 x^7}{\cos ^2 x}-\int_{-\pi / 4}^{\pi / 4} \frac{3 x^5}{\cos ^2 x}+\int_{-\pi / 4}^{\pi / 4} \frac{7 x^3}{\cos ^2 x}-\int_{-\pi / 4}^{\pi / 4} \frac{x}{\cos ^2 x}+\int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos ^2 x} d x \\
& =0-0+0-0+2 \int_0^{\pi / 4} \sec ^2 x d x=2[\tan x]_0^{\pi / 4} \\
& =2\left[\tan \frac{\pi}{4}-\tan (0)\right]=2(1)=2 .
\end{aligned}
$
Question 8.
The area between $y^2=4 x$ and its latus rectum is
(a) $\frac{2}{3}$
(b) $\frac{4}{3}$
(c) $\frac{8}{3}$
(d) $\frac{5}{3}$
Solution:
(c) $\frac{8}{3}$
Hint:
$
\begin{aligned}
& y^2=4 x \\
& 4 a=4 \Rightarrow a=1
\end{aligned}
$
$
\mathrm{A}=2 \int_0^1 y d x=2 \int_0^1 2 \sqrt{x} d x=4\left(\frac{x^{3 / 2}}{3 / 2}\right)_0^1=\frac{8}{3}
$
Question 9.
The value of $\int_0^1 x(1-x)^{99} d x$ is
(a) $\frac{1}{11000}$
(b) $\frac{1}{10100}$
(c) $\frac{1}{10010}$
(d) $\frac{1}{10001}$
Solution:

(b) $\frac{1}{10100}$
$
\begin{aligned}
& \text { Hint: } \quad\left[\int_0^1 x(1-x)^n d x=\frac{1}{(n+1)(n+2)}\right] \\
& =\int_0^1 x(1-x)^{99} d x=\frac{1}{(100)(101)}=\frac{1}{10100} \text {. } \\
&
\end{aligned}
$
Question 10.
The value of $\int_0^\pi \frac{d x}{1+5^{\cos x}}$ is
(a) $\frac{\pi}{2}$
(b) $\pi$
(c) $\frac{3 \pi}{2}$
(d) $2 \pi$
Solution:
(a) $\frac{\pi}{2}$
$
\begin{aligned}
& \text { Hint: } \quad \text { I }=\int_0^\pi \frac{d x}{1+5^{\cos x}} \\
& \int_0^a f(x) d x=\int_0^a f(a-x) d x \\
& 2 \mathrm{I}=\int_0^\pi d x=[x]_0^\pi \\
& I=\int_0^\pi \frac{5^{\cos x}}{5^{\cos x}+1} d x \\
&
\end{aligned}
$
Question 11.
If $\frac{\Gamma(n+2)}{\Gamma(n)}=90$ then $n$ is.
(a) 10
(b) 5
(c) 8
(d) 9
Solution:
(d) 9

Hint:
$
\begin{aligned}
\frac{\sqrt{n+2}}{\sqrt{n}} & =90 \\
\frac{(n+1) \sqrt{n+1}}{\sqrt{n}} & =90 \Rightarrow \quad \frac{(n+1) n \sqrt{n}}{\sqrt{n}}=90 \\
n^2+n-90 & =0 \Rightarrow(n-9)(n+10)=0 \\
n & =9 .
\end{aligned}
$
Question 12.
The value of $\int_0^{\frac{\pi}{6}} \cos ^3 3 x d x$ is.
(a) $\frac{2}{3}$
(b) $\frac{2}{9}$
(c) $\frac{1}{9}$
(d) $\frac{1}{3}$
Solution:
(b) $\frac{2}{9}$
Hint:
$
\begin{aligned}
& \int_0^{\frac{\pi}{6}} \cos ^3 3 x d x \\
& \text { Put } t=3 x \\
& \quad=\frac{1}{3} \int_0^{\pi / 2} \cos ^3 t d t=\frac{1}{3}\left(\frac{2}{3}\right)=\frac{2}{9} .
\end{aligned}
$

Question 13.
The value of $\int_0^\pi \sin ^4 x d x$ is.
(a) $\frac{3 \pi}{10}$
(b) $\frac{3 \pi}{8}$
(c) $\frac{3 \pi}{4}$
(d) $\frac{3 \pi}{2}$
Solution:
$\frac{3 \pi}{8}$
Hint:
$
=2 \int_0^{\frac{\pi}{2}} \sin ^4(\pi-x) d x \quad \int_0^{2 a} f(x) d x=2 \int_0^a f(2 a-x) d x
$

$
=2 \int_0^{\frac{\pi}{2}} \sin ^4 x d x=2\left(\frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right)=\frac{3 \pi}{8} .
$
Question 14.
The value of $\int_0^{\infty} e^{-3 x} x^2 d x$ is
(a) $\frac{7}{27}$
(b) $\frac{5}{27}$
(c) $\frac{4}{27}$
(d) $\frac{2}{27}$
Solution:
(d) $\frac{2}{27}$
Hint:
$
\begin{array}{r}
\cdot\left(\therefore \int_0^{\infty} x^n e^{-a x} d x=\frac{n !}{a^{n+1}}\right) \\
\int_0^{\infty} e^{-3 x} x^2 d x=\frac{2 !}{(3)^{2+1}}=\frac{2}{27} .
\end{array}
$

Question 15.
If $\int_0^a \frac{1}{4+x^2} d x=\frac{\pi}{8}$ then $a$ is.
(a) 4
(b) 1
(c) 3
(d) 2
Solution:
(d) 2
Hint:
$
\begin{aligned}
\int_0^a \frac{1}{4+x^2} d x & =\frac{\pi}{8} \\
{\left[\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right]_0^a } & =\frac{\pi}{8} \\
\tan ^{-1}\left(\frac{a}{2}\right) & =\frac{\pi}{4} \Rightarrow \frac{a}{2}=\tan \left(\frac{\pi}{4}\right)=1 \\
a & =2
\end{aligned}
$
Question 16.
The volume of solid of revolution of the region bounded by $y^2=x(a-x)$ about $x$-axis is
(a) $\pi a^3$
(b) $\frac{\pi a^3}{4}$
(c) $\frac{\pi a^3}{5}$
(d) $\frac{\pi a^3}{6}$
Solution:
(d) $\frac{\pi a^3}{6}$
Hint:
$
\begin{aligned}
\mathrm{V} & =\pi \int_0^a y^2 d x \quad \text { Put } y=0 ; \text { we get } x=0 \text { and } x=a \\
& =\pi \int_0^a\left(a x-x^2\right) d x=\pi\left[\frac{a x^2}{2}-\frac{x^3}{3}\right]_0^a \\
& =\pi\left[\frac{a^3}{2}-\frac{a^3}{3}\right]=\pi\left[\frac{a^3}{6}\right]=\frac{\pi a^3}{6} .
\end{aligned}
$

Question 17.
If $f(x)=\int_1^x \frac{e^{\sin u}}{u} d u, x>1$ and $\int_1^3 \frac{e^{\sin x^2}}{x} d x=\frac{1}{2}[f(a)-f(1)]$, then one of the possible value of $a$ is
(a) 3
(b) 6
(c) 9
(d) 5
Solution:
(c) 9
Hint:

$\begin{aligned}
& \int_1^3 \frac{e^{\sin x^2}}{x} d x \\
& \text { Put } t=x^2 \\
& \frac{d t}{2 x}=d x \Rightarrow \frac{d t}{2 \sqrt{t}}=d x \\
& \Rightarrow \int_1^9 \frac{e^{\sin t}}{\sqrt{t}} \cdot \frac{d t}{2 \sqrt{t}}=\frac{1}{2}[f(a)-f(1)] \\
& \Rightarrow \frac{1}{2} \int_1^9 \frac{e^{\sin t}}{t} d t \\
& \Rightarrow \frac{1}{2} \int_1^9 \frac{d}{d x}[f(x)]=\frac{1}{2}[f(a)-f(1)] \\
& f(a)-f(1)=f(a)-f(1) \\
& a=9 .
\end{aligned}$
Question 18.
The value of $\int_0^1\left(\sin ^{-1} x\right)^2 d x$ is

(a) $\frac{\pi^2}{4}-1$
(b) $\frac{\pi^2}{4}+2$
(c) $\frac{\pi^2}{4}+1$
(d) $\frac{\pi^2}{4}-2$
Solution:
(d) $\frac{\pi^2}{4}-2$
Hint:
Put $x=\sin t$
$
d x=\cos t d t
$
$
=\int_0^{\frac{\pi}{2}} t^2 \cos t d t=\left[t^2(\sin t)-2 t(-\cos t)+2(-\sin t)\right]_0^{\frac{\pi}{2}}
$

Question 19.
The value of $\int_0^a\left(\sqrt{a^2-x^2}\right)^3 d x$ is.
(a) $\frac{\pi a^3}{16}$
(b) $\frac{3 \pi a^4}{16}$
(c) $\frac{3 \pi a^2}{8}$
(d) $\frac{3 \pi a^4}{8}$
Solution:
(b) $\frac{3 \pi a^4}{16}$
Hint:
Put $x=a \sin \theta$
$
d x=a \cos \theta d \theta
$
$
\begin{aligned}
\left(\sqrt{a^2-x^2}\right)^3 & =a^3 \cos ^3 \theta \\
& =\int_0^{\pi / 2} a^3 \cos ^3 \theta(a \cos \theta) d \theta=a^4 \int_0^{\pi / 2} \cos ^4 \theta d \theta \\
& =a^4\left[\frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right]=\frac{3 \pi a^4}{16} .
\end{aligned}
$

Question 20.
If $\int_0^x f(t) d t=x+\int_x^1 t f(t) d t$, then the value of $f(1)$ is.
(a) $\frac{1}{2}$
(b) 2
(c) 1
(d) $\frac{3}{4}$
Solution:
(a) $\frac{1}{2}$
Hint:
$
\begin{aligned}
\frac{d}{d x}\left[\int_0^x f(t) d t\right] & =\frac{d}{d x}\left[x+\int_x^1 t f(t) d t\right] \\
f(x) & =1+\frac{d}{d x}(1) \cdot 1 f(1)-\frac{d}{d x}(x) \cdot x f(x) \\
f(x) & =1+0-x f(x) \\
f(x)(1+x) & =1 \\
f(x) & =\frac{1}{1+x} \Rightarrow f(1)=\frac{1}{2}
\end{aligned}
$