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

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Answer 1 · 2016-11-06T05:16:01

$int_0^1xe^(-x^2)dx=1/2(1-1/e)$

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

As we know if :
$intf(x)=F(x)+C$

Then

$int_a^bf(x)=F(b)-F(a)$

Let us compute the integral $intxe^(-x^2)dx$
Let:
$color(red)(u(x)=e^(-x^2)$

Then

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

$intxe^(-x^2)dx=intcolor(blue)(-1/2du(x))$

$intxe^(-x^2)dx=-1/2intdu(x)$

$intxe^(-x^2)dx=-1/2u(x)+C$

$intxe^(-x^2)dx=-1/2color(red)(e^(-x^2)+C$
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Let us calculate the value of the definite integral:

$int_0^1xe^(-x^2)dx=-1/2(e^(-1^2)-e^(-0^2))$

$int_0^1xe^(-x^2)dx=-1/2(e^(-1)-e^(-0))$

$int_0^1xe^(-x^2)dx=-1/2(e^(-1)-1)$

Therefore,

$int_0^1xe^(-x^2)dx=1/2(1-1/e)$

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