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

1 Answer
Nov 6, 2016

Answer:

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

Explanation:

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#
#" "#

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)#