Definite Integration Question 440
Question: If $ I _{n}=\int_0^{\infty }{{e^{-x}}{x^{n-1}}dx,} $ then $ \int_0^{\infty }{{e^{-\lambda x}}{x^{n-1}}dx=} $
Options:
A) $ \lambda I _{n} $
B) $ \frac{1}{\lambda }I _{n} $
C) $ \frac{I _{n}}{{{\lambda }^{n}}} $
D) $ {{\lambda }^{n}}I _{n} $
Show Answer
Answer:
Correct Answer: C
Solution:
Putting $ \lambda x=t,\lambda dx=dt $
we get , $ \int_0^{\infty }{{e^{-\lambda x}}{x^{n-1}}dx} $
$ =\frac{1}{{{\lambda }^{n}}}\int_0^{\infty }{{e^{-t}}{t^{n-1}}}dt $
$ =\frac{1}{{{\lambda }^{n}}}\int_0^{\infty }{{e^{-x}}{x^{n-1}}dx=\frac{I _{n}}{{{\lambda }^{n}}}} $ .
BETA
