# How do you find the integral int_0^1x*e^(-x^2)dx ?

Sep 22, 2014

We begin by making a substitution.

Let $u = - {x}^{2}$

$\int x {e}^{u} \mathrm{dx}$

$\mathrm{du} = - 2 x \mathrm{dx}$

$\frac{\mathrm{du}}{- 2} = \frac{- 2 x \mathrm{dx}}{- 2}$

$\frac{\mathrm{du}}{- 2} = x \mathrm{dx}$

$\frac{- 1}{2} \cdot \mathrm{du} = x \mathrm{dx}$

$\int {e}^{u} x \mathrm{dx}$, Notice that xdx can be replaced by $\frac{- 1}{2} \cdot \mathrm{du}$

$\int {e}^{u} \frac{- 1}{2} \mathrm{du}$, Move the constant to the front of the integral

$\frac{- 1}{2} \int {e}^{u} \mathrm{du}$

After integration look back to the original substitution to find the value for $u$. In this case that value is $- {x}^{2}$.

I switched back to $- {x}^{2}$ so that I could use the original boundaries.

$\frac{- 1}{2} {\left[{e}^{- {x}^{2}}\right]}_{0}^{1} = \frac{- 1}{2} \left[{e}^{- 1} - {e}^{0}\right] = \frac{- 1}{2} \left[\frac{1}{e} - 1\right] = \frac{- 1}{2 e} + \frac{1}{2} =$

$= 0.31606$