Question

What is `f(x) = int xe^(x-3)dx` if `f(0)=-2`?

Answer 1

`f(x) = xe^(x-3)-e^(x-3) + e^-3-2`

Explanation:

`f(x)=int xe^(x-3) dx`

Use integration by parts

`int u (dv) = uv - int v (du)`

Selecting `u` and `dv` is the first step we should take.

Let `u=x` and `dv = e^(x-3)dx`

`u=x`
`du = dx`

`dv = e^(x-3)dx`

`int dv = int e^(x-3)dx`

`v = e^(x-3)`

Now our integration becomes

`f(x)=int xe^(x-3) dx = xe^(x-3) - int e^(x-3) dx`

`int xe^(x-3) dx = xe^(x-3)-e^(x-3) + C`

`f(x) = xe^(x-3)-e^(x-3) + C`
Given `f(0) = -2`
`f(0) = 0e^(0-3)-e^(0-3)+C`
`-2=-e^-3+C`
`e^-3-2=C`

`xe^(x-3)-e^(x-3) + e^-3-2`

Answer 2

`f(x) = int_0^x te^{t-3} dt - 2`
`= xe^{x-3} - e^{x-3} + e^{-3} - 2`

Explanation:

Instead of using indefinite integration (refer to the other answer by Karthik), one can also use definite integration for this question. The integration is mostly the same.