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

1 Answer
Mar 22, 2018

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