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

1 Answer
Apr 13, 2017

f(x) = 1/3x^3 - 2e^x(x - 1) - 4

Explanation:

Separate the integral.

f(x) = int x^2dx - int 2xe^xdx

The first integral is 1/3x^3.

f(x) = 1/3x^3 - int 2xe^xdx

We will use integration by parts for the last integral. We let u = 2x and dv = e^x. Then du = 2dx and v= e^x.

int 2xe^xdx = 2x(e^x) - int 2e^x

int2xe^xdx = 2xe^x - 2inte^x

int2xe^x = 2xe^x - 2e^x

int2xe^x = 2e^x(x - 1)

We put the integral back together to find

f(x) = 1/3x^3 - 2e^x(x - 1) + C

We must now find the value of C. We know that when x = 0, y= -2, therefore:

-2 = 1/3(0)^3 - 2e^0(0 - 1) + C

-2 = 0 + 2 + C

C = -4

This means that f(x) = 1/3x^3 - 2e^x(x - 1) - 4.

Hopefully this helps!