Exercise 9.7 - 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 $9.7$
Question 1.
Evaluate the following
(i) $\int_0^{\infty} x^5 e^{-3 x} d x$
Solution:
We know that
$
\begin{aligned}
& \int_0^{\infty} x^n e^{-a x} d x=\frac{n !}{a^{n+1}} \\
\therefore & \int_0^{\infty} x^5 e^{-3 x} d x=\frac{5 !}{3^{5+1}}=\frac{5 !}{3^6} \quad(\text { Here } n=5, a=3)
\end{aligned}
$

(ii) $
\int_0^{\frac{\pi}{2}} \frac{e^{-\tan x}}{\cos ^6 x} d x
$
Solution:

$
\begin{aligned}
& =\int_0^{\frac{\pi}{2}} e^{-\tan x} \sec ^6 x d x \\
& =\int_0^{\frac{\pi}{2}} e^{-\tan x} \sec ^4 x \sec ^2 x d x \\
\text { Put } t & =\tan x
\end{aligned}
$

Differentiate with respect to ' $x$ '
$
d t=\sec ^2 x d x
$
Substitute (2) and (3) in (1), we get,
$
\begin{aligned}
(1) \Rightarrow \quad & =\int_0^{\infty} e^{-t}\left(\sec ^2 x\right)^2 d t \quad\left(\because \sec ^2 x=1+\tan ^2 x=1+t^2\right) \\
& =\int_0^{\infty} e^{-t}\left(1+t^2\right)^2 d t \\
& =\int_0^{\infty} e^{-t}\left(1+t^4+2 t^2\right) d t \\
& =\int_0^{\infty} e^{-t} d t+\int_0^{\infty} e^{-t} t^4 d t+2 \int_0^{\infty} e^{-t} t^2 d t
\end{aligned}
$
By using Gamma Integral,
$
\begin{aligned}
& =\left(-e^{-t}\right)_0^{\infty}+\frac{4 !}{(1)^{4+1}}+2 \frac{(2 !)}{(1)^{2+1}} \\
& =-e^{-\infty}+e^{o+4 !+2(2 !)}
\end{aligned}
$
$
=0+1+24+4=29
$

Question 2.
If $\int_0^{\infty} e^{-\alpha x^2} x^3 d x=32, \alpha>0$, find $\alpha$
Solution:

$
\begin{aligned}
\int_0^{\infty} e^{-\alpha x^2} x^3 d x & =32 \\
\int_0^{\infty} e^{-\alpha x^2} x^2 \cdot x d x & =32 \\
\text { Put } t & =x^2
\end{aligned}
$
Differentiate with respect to ' $x$ '
$
\begin{aligned}
d t & =2 x d x \\
\frac{d t}{2} & =x d x
\end{aligned}
$
Substitute (2) and (3) in (1), we get,
(1) $\Rightarrow$
$
\begin{aligned}
& \int_0^{\infty} e^{-\alpha t} t \cdot \frac{d t}{2}=32 \\
& \frac{1}{2} \int_0^{\infty} t e^{-\alpha t} d t=32
\end{aligned}
$
By using Gamma integral $(n=1 ; a=\alpha)$

$
\begin{aligned}
\frac{1}{2}\left[\frac{1}{\alpha^2}\right] & =32 \\
\frac{1}{\alpha^2} & =32 \times 2 \\
\alpha^2 & =\frac{1}{64} \\
\alpha & =\pm \frac{1}{8}\left[\alpha=-\frac{1}{8}(\text { not possible })\right] \\
\alpha & =\frac{1}{8} .
\end{aligned}
$