What is f(x) = int xe^(2-x) + 3x^2 dx if f(0 ) = 1 ?

1 Answer
Jul 7, 2016

-xe^(2-x)-e^(2-x)+x^3+1+e^2

Explanation:

Begin by using the sum rule for integrals and splitting these into two separate integrals:
intxe^(2-x)dx+int3x^2dx

The first of these mini-integrals is solved using integration by parts:
Let u=x->(du)/dx=1->du=dx
dv=e^(2-x)dx->intdv=inte^(2-x)dx->v=-e^(2-x)

Now using the integration by parts formula intudv=uv-intvdu, we have:
intxe^(2-x)dx=(x)(-e^(2-x))-int(-e^(2-x))dx
=-xe^(2-x)+inte^(2-x)dx
=-xe^(2-x)-e^(2-x)

The second of these is a case of the reverse power rule, which states:
intx^ndx=(x^(n+1))/(n+1)

So int3x^2dx=3((x^(2+1))/(2+1))=3(x^3/3)=x^3

Therefore, intxe^(2-x)+3x^2dx=-xe^(2-x)-e^(2-x)+x^3+C (remember to add the constant of integration!)

We are given the initial condition f(0)=1, so:
1=-(0)e^(2-(0))-e^(2-(0))+(0)^3+C
1=-e^2+C
C=1+e^2

Making this final substitution, we obtain our final solution:
intxe^(2-x)+3x^2dx=-xe^(2-x)-e^(2-x)+x^3+1+e^2