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${\lambda ^n}{I_n}$
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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