Question

The value of ${\int }_0^{\infty }x.e^{-x^2}d{x}_{=}$

• A. $1$
• B. $-1/2$
• C. $1/2$
• D. $0$

Answer 1

The correct option is C $1/2$

${\int }_0^{\infty }x{e}^{-x^2}dx=\frac{1}{2}{\int }_0^{\infty }{e}^{-x^2}d{x}^2$
$=\frac{1}{2}{\int }_0^{\infty }{e}^{-x^2}d{x}^2=\frac{1}{2}[-\frac{1}{e^{x^2}}+c]{\int }_0^{\infty }$
$=\frac{1}{2}\left[\right(0+e)-(-1+e\left)\right]$
$=\frac{1}{2}$
${\int }_0^{\infty }x{e}^{-x^2}dx=\frac{1}{2}$