Question

What is `f(x) = int xe^(2-x) -2 x^2 dx` if `f(0 ) = 1`?

Answer 1

`f(x)=-xe^(2-x)-e^(2-x)-2/3x^3+1+e^2`

Explanation:

`f(x)=int(xe^(2-x)-2x^2)dx`

and `f(0)=1`

`f(x)=int(xe^(2-x)-2x^x2)dx=color(red)(int(xe^(2-x))dx)-int2x^2dx`

the second integral is integrated using the power rule; the first integral is done by integrating by parts#

`I_1=int2x^2dx=2/3x^3`

`I_2=int(xe^(2-x))dx`

#IBP formula

`I_2=intu(dv)/(dx)dx=uv-intv(du)/(dx)dx`

`u=x=>du=dx`

`(dv)/(dx)=e^(2-x)=>v=-e^(2-x)`

`:. I_2=-xe^(2-x)-int(-e^(2-x))dx`

`I_2=-xe^(2-x)-e^(2-x)`

`:. f(x)=I_2-I_1`

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

`f(0)=1`

`=>1=0-e^2-0+c`

`:. c=1+e^2`

`f(x)=-xe^(2-x)-e^(2-x)-2/3x^3+1+e^2`

which can be simplified as required