Question

If $\underset{0}{\overset{\infty }{\int }}{e}^{-ax}dx=\frac{1}{a},$ then $\underset{0}{\overset{\infty }{\int }}{x}^n{e}^{-ax}dx$ is

• A. $\frac{{\left(-1\right)}^{n}n!}{a^{n+1}}$
• B. $\frac{{\left(-1\right)}^n\left(n-1\right)!}{a^n}$
• C. $\frac{n!}{a^{n+1}}$
• D. None of these

Answer 1

Answer: (C)

Let $I=\underset{0}{\overset{\infty }{\int }}{x}^n{e}^{-ax}=$ ${\left[x^n\cdot \frac{e^{-ax}}{-a}\right]}_{{}_{{}_0}}^{{}^{\infty }}-$ $\underset{0}{\overset{\infty }{\int }}n{x}^{n-1}\cdot \frac{e^{-ax}}{-a}dx$
$=-\frac{1}{a}$ $\underset{x\to \infty }{\lim }$$=\frac{x^n}{e^{ax}}+\frac{n}{a}{I}_{n-1}$
$\therefore I_n=\frac{n}{a}{I}_{n-1}\left[\because \underset{x\to \infty }{\lim }\frac{x^n}{e^{ax}}=0\right]$
$=\frac{n}{a}\cdot \frac{n-1}{a}{I}_{n-2}=\frac{n\left(n-1\right)\left(n-2\right)}{a^3}{I}_{n-3}$
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$=\frac{n!}{a^n}$$\underset{0}{\overset{\infty }{\int }}$$e^{-ax}dx=\frac{n!}{a^{n+1}}$