Question

The value of the definite integral, $I={\int }_0^{\sqrt{10}}\frac{x}{e^{x^2}}dx$ is equal to

Answer 1

$I={\int }_0^{\sqrt{10}}\frac{x}{e^{x^2}}dx{x}^2=t$

$\Rightarrow 2x.dx=dt\Rightarrow xdx=\frac{dt}{2}$

$I={\int }_0^{\sqrt{10}}\frac{dt}{2\times e^t}=\frac{1}{2}{\int }_0^{\sqrt{10}}{e}^{-t}dt$

$I=\frac{-1}{2}[e^{-t}{]}_0^{\sqrt{10}}=\frac{-1}{2}[e^{-x^2}{]}_0^{\sqrt{10}}$

$I=\frac{-1}{2}[e^{-10}-e^0]=\frac{-1}{2}\left[\right[\frac{1}{e^2}{]}^5-1]$

$=0.499$