How do you find the definite integral of #xe^(x^2 + 2) # in the interval #[0, 2]#?

Question

1297 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-07T08:11:56

#= e^2/2 ( e^(4) - 1)#

#int_0^2 xe^(x^2 + 2) dx = int_0^2 xe^(x^2)e^2 dx#

#= e^2 int_0^2 xe^(x^2) dx#

Now noting that #d/dx(e^(x^2)) = 2 x e^(x^2#

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

and by the FTC

#= e^2/2 [ e^(x^2) ]_0^2#

#= e^2/2 ( e^(4) - e^0)#

#= e^2/2 ( e^(4) - 1)#