Question

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

Answer 1

`-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`