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

1 Answer
May 20, 2018

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