If {I_n} = int_0^infty {{e^{ - x}}{x^{n - 1}} | vedclass

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(C) Let the given integral be $J = \int_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx}$.Substitute $\lambda x = t$,which implies $x = \frac{t}{\lambda}$ a

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

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(C) Let the given integral be $J = \int_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx}$.Substitute $\lambda x = t$,which implies $x = \frac{t}{\lambda}$ and $dx = \frac{dt}{\lambda}$.As $x 	o 0$,$t 	o 0$,and as $x 	o \infty$,$t 	o \infty$.Substituting these into the integral:$J = \int_0^\infty {{e^{ - t}}{(\frac{t}{\lambda})^{n - 1}} \frac{dt}{\lambda}}$$J = \int_0^\infty {{e^{ - t}} \frac{{{t^{n - 1}}}}{{{\lambda ^{n - 1}}}} \frac{dt}{\lambda}}$$J = \frac{1}{{{\lambda ^n}}} \int_0^\infty {{e^{ - t}}{t^{n - 1}}dt}$Since ${I_n} = \int_0^\infty {{e^{ - x}}{x^{n - 1}}dx}$,the integral $\int_0^\infty {{e^{ - t}}{t^{n - 1}}dt}$ is also equal to ${I_n}$.Therefore,$J = \frac{{{I_n}}}{{{\lambda ^n}}}$.

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

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