What is f(x) = int xe^(x+2)-x dx if f(-1) = 2 ?

1 Answer
Aug 22, 2016

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

Explanation:

f(x)=int(xe^(x+2)-x)dx=intxe^(x+2)dx-intxdx=I-x^2/2, where,

I=intxe^(x+2)dx.

To evaluate I, we use the Rule of Integration by Parts :

intuvdx=uintvdx-int[(du)/dxintvdx]dx.

We take, u=x rArr (du)/dx=1, and,

v=e^(x+2) rArr intvdx=e^(x+2). Hence,

I=xe^(x+2)-inte^(x+2)dx=xe^(x+2)-e^(x+2)

:. f(x)=xe^(x+2)-e^(x+2)-x^2/2+C...........(1).

To determine C, we use the cond. : f(-1)=2 in (1).

:. -e-e-1/2+C=2 rArr C=5/2+2e. Therefore, (1) gives,

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

Enjoy Maths.!