If $\mathop \smallint \limits_0^\infty \,\,{e^{ - ax}}\,\,dx = \frac{1}{a}$, then $\mathop \smallint \limits_0^\infty \,{x^n}{e^{ - ax}}\,\,dx$, is

$\frac{{{{\left( { - 1} \right)}^n}\,\,n!}}{{{a^{n + 1}}}}$

$\frac{{{{\left( { - 1} \right)}^n}\,\,\left( {n - 1} \right)!}}{{{a^n}}}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

\int \limits_0^\infty \,\,{e^{ - ax}}\,\,dx = \frac{1}{a}

, then

\int \limits_0^\infty \,{x^n}{e^{ - ax}}\,\,dx

, is

View answer and explanation

Correct option: C

\frac{{n!}}{{{a^{n + 1}}}}

Option C (

\frac{{n!}}{{{a^{n + 1}}}}

) is correct because it is obtained by differentiating the given integral n times with respect to 'a'.

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