If I_n = _0^ e^ - x x^ n - 1 dx, then _0^ e^ - x x^ n - 1 dx = - Vidyadip

MCQ

If ${I_n} = \int_0^\infty {{e^{ - x}}{x^{n - 1}}dx,} $ then $\int_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx = } $

A
$\lambda {I_n}$
B
$\frac{1}{\lambda }{I_n}$
✓
$\frac{{{I_n}}}{{{\lambda ^n}}}$
D
${\lambda ^n}{I_n}$

✓

Answer

Correct option: C.

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

c
(c) 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}}}} $.

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