Exercise 2.10 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.10$
Question 1.
Evaluate the following.
(i) $\Gamma(4)$
Solution:
$
\Gamma(4)=\Gamma(3+1)=3 !=6
$
(ii) $\Gamma\left(\frac{9}{2}\right)$
Solution:
$
\begin{aligned}
\Gamma\left(\frac{9}{2}\right) & =\left(\frac{9}{2}-1\right) \Gamma\left(\frac{9}{2}-1\right) \\
& =\frac{7}{2} \Gamma \frac{7}{2} \\
& =\frac{7}{2} \frac{5}{2} \Gamma \frac{5}{2}=\frac{7}{2} \frac{5}{2} \frac{3}{2} \Gamma \frac{3}{2} \\
& =\frac{7}{2} \frac{5}{2} \frac{3}{2} \frac{1}{2} \Gamma\left(\frac{1}{2}\right) \\
& =\frac{105 \sqrt{\pi}}{16}
\end{aligned}
$
(iii) $\int_0^{\infty} e^{-m x} x^6 d x$
Solution:
$
\int_0^{\infty} e^{-m x} x^6 d x=\int_0^{\infty} x^n e^{-a x} d x=\frac{n !}{a^{n+1}}
$
Where $n=6, a=m$
So the integral becomes $\frac{6 !}{m^7}$

(iv) $\int_0^{\infty} e^{-4 x} x^4 d x$
Solution:
$
\int_0^{\infty} e^{-4 x} x^4 d x=\int_0^{\infty} x^n e^{-a x} d x=\frac{n !}{a^{n+1}}
$
Where $n=4, a=4$
So the integral becomes $\frac{4 !}{4^5}=\frac{4 \times 3 \times 2}{4 \times 4 \times 4 \times 4 \times 4}=\frac{3}{128}$
(v) $\int_0^{\infty} e^{-\frac{x}{2}} x^5 d x$
Solution:
$
\int_0^{\infty} e^{-\frac{x}{2}} x^5 d x=\frac{5 !}{\left(\frac{1}{2}\right)^6}=\left(2^6\right) 5 !
$
Question 2.
If, $f(x)=\left\{\begin{array}{l}x^2 e^{-2 x}, x \geq 0 \\ 0, \text { otherwise }\end{array}\right.$, then evaluate $\int_0^{\infty} f(x) d x$
Solution:
$
\int_0^{\infty} x^2 e^{-2 x} d x=\int_0^{\infty} x^n e^{-a x} d x=\frac{n !}{a^{n+1}}
$
Where $n=2, a=2$
So $\int_0^{\infty} f(x) d x=\frac{2 !}{2^3}=\frac{2}{8}=\frac{1}{4}$