Question

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

Answer 1

`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.!