Can I have some help finding the original function of
f'(x)=(e^-x)x^2(x-2)
I'm really confused about how to solve this!
f'(x) = x^3 e^-x - 2x^2 e^-x
you do these using integration by parts.. That is just the product rule in reverse
d(uv) = u dv + v du
so, u dv = d(uv) - v du
∫ u dv = uv - ∫ v du
So let u = x^3, dv = e^-x dx
du = 3x^2 dx, v = -e^-x
Then
∫ x^3 e^-x dx = -x^3 e^-x + 3∫x^2 e^-x dx
Now the power of x has dropped by 1, and you do it again. You wind up with
∫ x^3 e^-x dx = -e^-x (x^3+3x^2+6x+6) + C