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

1 Answer
Jul 24, 2017

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.