Question 1039145
The roots are correct.
int (x-2)(1-e^x)dx
let u=x-2
du=dx
dv=(1-e^x)dx
v=x-e^x
the integral is uv-int v*du
that is (x-2)(x-e^x)- int (x-e^x)dx; the integral is int x dx-int (e^x dx), so that is x^2/2-e^x
putting it together, it is x^2-xe^x-2x+2e^x-x^2/2+e^x, evaluated at 2 and 0.
at 2, it is 4-2e^2-4+2e^2-2+e^2. The fours cancel and the 2e^2 cancel to leave -2+e^2.
at 0, it is 0-0-0+2-0+1=3
subtracting -2+e^2-3=e^2-5 or 2.389
{{{graph(300,200,-5,5,-5,5,(x-2)(1-e^x))}}}