Question

What is `f(x) = int e^(2x)-e^x+x dx` if `f(4 ) = 0`?

Answer 1

`f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-8`
`f(x)=e^(2x)/2-e^x+x^2/2-1444`

Explanation:

Let's split this into:
`inte^(2x)dx-inte^xdx+intxdx`

The two rules of integrating for this are:
`inte^(ax+b)dx=e^(ax+b)/a+C`
`intx^ndx=x^(n+1)/(n+1)+C`

This gives us:
`f(x)=e^(2x)/2-e^x+x^2/2+C`

Though we need `C`, however we are told that `f(4)=0`

`0=e^(2(4))/2-e^4+4^2/2+C`

`0=e^8/2-e^4+8+C`

`C=e^4-e^8/2-8~~-1444`

`f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-8`
`f(x)=e^(2x)/2-e^x+x^2/2-1444`