Question

If $I_n={\int }_0^{\infty }{e}^{-x}{x}^{n-1}dx$, then ${\int }_0^{\infty }{e}^{-\lambda x}{x}^{n-1}dx$ is equal to?

• A. $\lambda I_n$
• B. $\frac{1}{\lambda }{I}_n$
• C. $\frac{I_n}{{\lambda }^n}$
• D. ${\lambda }^n{I}_n$

Answer 1

The correct option is C $\frac{I_n}{{\lambda }^n}$

$I_n={\int }_0^{\infty }{e}^{-x}{x}^{n-1}dx$

then $I_1={\int }_0^{\infty }{e}^{-x}{x}^{n-1}dx$

$I_1={\int }_0^{\infty }{e}^{-x}\frac{\left(\right(\lambda x{)}^{n-1})}{\lambda }dx$

$I_1=\frac{1}{{\lambda }^n}{\int }_0^{\infty }{e}^{-t}{t}^{n-1}dt$

$I_1=\frac{I_n}{{\lambda }^n}.$

$So,{\int }_0^{\infty }{e}^{-t}{x}^{n-1}dt=\frac{I_n}{{\lambda }^n}.$