Question

What is `int_(0)^(1) (x^2)*e^(-x^2) dx`?

Answer 1

`= -1/(2e) + (sqrt pi)/4 erf (1)`

Explanation:

It's this!!

`int_(0)^(1) (x^2)*e^(-x^2) dx`

`= int_(0)^(1) -1/2x ( - 2 xe^(-x^2)) dx`

`= int_(0)^(1) -1/2x ( e^(-x^2))^prime dx`

Which is Integration by Parts!

`= [-1/2x e^(-x^2)]\_(0)^(1) - int_(0)^(1) (-1/2x)^prime e^(-x^2)dx`

`= -1/(2e) + 1/2 color(blue)( int_(0)^(1) e^(-x^2)dx) qquad triangle`

For that blue bit, I would call on this definition of the error function:

`erf (x) =2/(sqrt (pi )) int _{0}^{x} e^{-t^{2}}\ dt`

We can pattern match this into:

`color(blue)( int_(0)^(1) e^(-x^2)dx) = (sqrt pi)/2 erf (1)`

And so `triangle` becomes

`= -1/(2e) + (sqrt pi)/4 erf (1)`