Question

The value of the integral $I=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }exp(-\frac{x^2}{8})dx$ is

• A. $1$
• B. $\pi$
• C. $2$
• D. $2\pi$

Answer 1

The correct option is A $1$

$I=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }{e}^{-\frac{x^2}{8}}dx$
Let $x^2=8t$
$x=2\sqrt{2}\sqrt{t}$
$dx=\frac{\sqrt{2}}{\sqrt{t}}dt$
$I=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }{e}^{-t}\frac{\sqrt{2}}{\sqrt{t}}dt$
$=\frac{1}{\sqrt{\pi }}{\int }_0^{\infty }{e}^{-t}{t}^{-1/2}dt$
$=\frac{1}{\sqrt{\pi }}{\int }_0^{\infty }{e}^{-t}.t^{\frac{1}{2}-1}dt$


$I=\frac{1}{\sqrt{\pi }}\sqrt{\pi }$
$I=1$