Question

What is `f(x) = int xe^(2-x) -2 x^2 +2 dx` if `f(0 ) = 2`?

Answer 1

`I=int(xe^(2-x)-2x^2+2)dx`

`=intxe^(2-x)dx-2intx^2dx+2int1dx`

`=intxe^(2-x)dx-2/3x^3+2x`

Let `X=2-x`

`x=2-X`

`dx=-1dX`

So:

`I=-int(2-X)e^Xdx-2/3x^3+2x`

Using integration by parts:
`f(X)=(2-X)`, `f'(X)=-1`, `g'(X)=e^X`, `g(X)=e^X`

So:

`I=-(2-X)e^X+int-e^XdX-2/3x^3+2x`

`I=(X-2)e^X-e^X-2/3x^3+2x`

`I=-xe^(2-x)-e^(2-x)-2/3x^3+2x+c`, `c in RR`

And knowing that `f(0)=2`,

`c-e^2=2`

So:

`I=-(x+1)e^(2-x)-2/3x^3+2x+2+e^2`