If \[\int\limits_{0}^{\infty }{{{e}^{-ax}}dx=\frac{1}{a},}\] then \[\int\limits_{0}^{\infty }{{{x}^{n}}{{e}^{-ax}}dx}\] is

\[\frac{{{(-1)}^{n}}n!}{{{a}^{n+1}}}\]

\[\frac{{{(-1)}^{n}}(n-1)!}{{{a}^{n}}}\]

If \[\int\limits_{0}^{\infty }{{{e}^{-ax}}dx=\frac{1}{a},}\] then \[\i

1.	If \[\int\limits_{0}^{\infty }{{{e}^{-ax}}dx=\frac{1}{a},}\] then \[\int\limits_{0}^{\infty }{{{x}^{n}}{{e}^{-ax}}dx}\] 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» D. None of these

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If \[\int\limits_{0}^{\infty }{{{e}^{-ax}}dx=\frac{1}{a},}\] then \[\int\limits_{0}^{\infty }{{{x}^{n}}{{e}^{-ax}}dx}\] is

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