# How do you integrate int xe^(-x^2)dx from [0,1]?

Nov 6, 2016

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = \frac{1}{2} \left(1 - \frac{1}{e}\right)$

#### Explanation:

To determine the definite integral ,we compute the integral itself first then work on the boundaries.

As we know if :
$\int f \left(x\right) = F \left(x\right) + C$

Then

${\int}_{a}^{b} f \left(x\right) = F \left(b\right) - F \left(a\right)$

Let us compute the integral $\int x {e}^{- {x}^{2}} \mathrm{dx}$
Let:
color(red)(u(x)=e^(-x^2)

Then

color(blue)(du(x)=-2xe^(-x^2)dx
rArrcolor(blue)(xe^(-x^2)=-1/2du(x)

$\int x {e}^{- {x}^{2}} \mathrm{dx} = \int \textcolor{b l u e}{- \frac{1}{2} \mathrm{du} \left(x\right)}$

$\int x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} \int \mathrm{du} \left(x\right)$

$\int x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} u \left(x\right) + C$

intxe^(-x^2)dx=-1/2color(red)(e^(-x^2)+C
$\text{ }$

Let us calculate the value of the definite integral:

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} \left({e}^{- {1}^{2}} - {e}^{- {0}^{2}}\right)$

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} \left({e}^{- 1} - {e}^{- 0}\right)$

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = - \frac{1}{2} \left({e}^{- 1} - 1\right)$

Therefore,

${\int}_{0}^{1} x {e}^{- {x}^{2}} \mathrm{dx} = \frac{1}{2} \left(1 - \frac{1}{e}\right)$