Question

What is `f(x) = int (x^2-2x)(e^x-3x) dx` if `f(1 ) = 2`?

Answer 1

`f(x)=(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+3/4-e`

Explanation:

`f(x)=int(x^2-2x)*e^x*dx-int(3x^3-6x^2)*dx`

= `[(x^2-2x)-(2x-2)+2]*e^x-(3/4*x^4-2x^3)+C`

= `(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+C`

After imposing `f(1)=2` condition,

`(1-4+4)*e-(3/4-2)+C=2`

`C=3/4-e`

Thus, `f(x)=(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+3/4-e`

1) I integrated the right side of the equation.

2) I found C for `f(1)=2` condition.