If ${I_n} = \mathop \smallint \limits_0^\infty \,\,{e^{ - x}}\,{x^{n - 1}}\,\,dx$, then $\mathop \smallint \limits_0^\infty \,\,{e^{ - \lambda x}}\,\,{x^{n - 1}}\,\,dx$ is equal to

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

{I_n} = \int \limits_0^\infty \,\,{e^{ - x}}\,{x^{n - 1}}\,\,dx

, then

\int \limits_0^\infty \,\,{e^{ - \lambda x}}\,\,{x^{n - 1}}\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{{{I_n}}}{{{\lambda ^n}}}

Option C (

\frac{{{I_n}}}{{{\lambda ^n}}}

) is correct because it is the result of a correct u-substitution and simplification.

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