Integral Calculus Question 238
Question: If $ \int\limits_0^{\infty }{{e^{-ax}}dx=\frac{1}{a},} $ then $ \int\limits_0^{\infty }{x^{n}{e^{-ax}}dx} $ is
Options:
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
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ I_{n}=\int\limits_0^{\infty }{x^{n}{e^{-ax}}=[ x^{n}.\frac{{e^{-ax}}}{-a} ]0^{\infty }-\int\limits_0^{\infty }{n{x^{n-1}}.\frac{{e^{-ax}}}{-a}dx}} $ $ =-\frac{1}{a}\underset{x\to \infty }{\mathop{\lim }},\frac{x^{n}}{e^{ax}}+\frac{n}{a}{I{n-1}}\therefore I_{n}=\frac{n}{a}{I_{n-1}} $ $ =\frac{n}{a}\cdot \frac{n-1}{a}{I_{n-2}}=\frac{n(n-1)(n-2)}{a^{3}}{I_{n-3}} $ $ =\frac{n!}{a^{n}}\int\limits_0^{\infty }{{e^{-ax}}dx=\frac{n!}{{a^{n+1}}}} $
BETA
