Question

What is `f(x) = int xe^x-5x dx` if `f(-1) = 5`?

Answer 1

`f(x)=xe^x-e^x-5/2x^2+2e^(-1)+15/2`

Explanation:

The given

`int (xe^x-5x) dx` and `f(-1)=5`

Integration by parts for the `int (xe^x) dx`

Let `u=x` and `dv=e^x*dx` and `v=e^x` and `du=dx`

Using the integration by parts formula

`int u*dv=uv-int v*du`

`int x*e^x*dx=xe^x-int e^x*dx`

`int x*e^x*dx=xe^x- e^x`

Therefore

`f(x)=int (xe^x-5x) dx=xe^x- e^x-5/2x^2+C`

`f(x)=xe^x- e^x-5/2x^2+C`

At `f(-1)=5`

`f(-1)=5=-1*e^(-1)- e^(-1)-5/2(-1)^2+C`

`5=-1*e^(-1)- e^(-1)-5/2(-1)^2+C`

`5=-2*e^(-1)-5/2+C`

`C=15/2+2e^(-1)`

finally

`f(x)=xe^x- e^x-5/2x^2+15/2+2e^(-1)`

God bless....I hope the explanation is useful.