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

1 Answer

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.