Question

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

The Correct Option is C

Let $I=\int\limits^{\infty}_{{0}}x^n\,e^{-ax}=$ $\left[x^{n}\cdot \frac{e^{-ax}}{-a}\right]^{^{\infty}}_{_{_0}}-$ $\int\limits^{\infty}_{{0}}nx^{n-1}\cdot \frac{e^{-ax}}{-a}dx$ $=-\frac{1}{a}$ $\displaystyle \lim_{x \to \infty} $$=\frac{x^{n}}{e^{ax}}+\frac{n}{a}I_{n-1}$ $\therefore I_{n}=\frac{n}{a}I_{n-1}\,\left[\because \displaystyle \lim_{x \to \infty} \frac{x^{n}}{e^{ax}}=0\right]$ $=\frac{n}{a}\cdot \frac{n-1}{a}I_{n-2}=\frac{n\left(n-1\right)\left(n-2\right)}{a^{3}}I_{n-3}$ $.........................................$ $.......................................$ $=\frac{n!}{a^{n}}$$\int\limits^{\infty}_{{0}}$$e^{-ax}dx=\frac{n!}{a^{n+1}}$