Question

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

Answer 1

`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!