Question

${\int }_0^{\infty }{x}^6{e}^{-\frac{x}{2}}dx=$

• A. $\frac{6!}{2^7}$
• B. $\frac{6!}{2^6}$
• C. $2^{6}6!$
• D. $2^{7}6!$

Answer 1

Answer: (D)

$2^{7}6!$

Let $\frac{x}{2}=t$.

Therefore, $dx=2dt$

Hence

$I={\int }_0^{\infty }{x}^6{e}^{-\frac{x}{2}}.dx$

$={\int }_0^{\infty }(2t{)}^6{e}^{-t}.2dt$

$=2^7{\int }_0^{\infty }{t}^6{e}^{-t}dt$ ....$\left(i\right)$

Now the integral term is gamma function, where gamma function is defined as

$\tau \left(z\right)={\int }_0^{\infty }{x}^{z-1}{e}^{-x}dx$.

Also, $\tau \left(z\right)=(z-1)!$.

Comparing with the integral in equation $\left(i\right)$,

$z-1=6$ or $z=7$.

Therefore,

$2^7{\int }_0^{\infty }{t}^6{e}^{-t}dt$

$=2^7.\tau \left(7\right)$

$=2^7(7-1)!$

$=2^7.6!$