Question

If ${\int }_0^{\infty }{e}^{-ax}dx=\frac{1}{a}$, then ${\int }_0^{\infty }{x}^n{e}^{-ax}dx$ is

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

Answer 1

The correct option is C $\frac{n!}{a^{n+1}}$

Let $I_n={\int }_0^{\infty }{x}^n{e}^{-ax}dx$

$={[x^n.\frac{e^{-ax}}{-a}]}_0^{\infty }-{\int }_0^{\infty }n{x}^{n-1}.\frac{e^{-ax}}{-a}dx$

$=-\frac{1}{a}\underset{x\to \infty }{lim}\frac{x^n}{e^{ax}}+\frac{n}{a}{I}_{n-1}$

$\therefore I_n=\frac{n}{a}{I}_{n-1}[\because \underset{x\to \infty }{lim}\frac{x^n}{e^{ax}}=0]$

$=\frac{n}{a}.\frac{n-1}{a}{I}_{n-2}$

$=\frac{n(n-1)(n-2)}{a^3}{I}_{n-3}$

$......$

$.......$

$=\frac{n!}{a^n}{\int }_0^{\infty }{e}^{-ax}dx=\frac{n!}{a^{n+1}}$