What is f(x) = int e^(2x)-e^x+x dxf(x)=e2xex+xdx if f(4 ) = 0 f(4)=0?

1 Answer
May 20, 2018

f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-8f(x)=e2x2ex+x22+e4e828
f(x)=e^(2x)/2-e^x+x^2/2-1444f(x)=e2x2ex+x221444

Explanation:

Let's split this into:
inte^(2x)dx-inte^xdx+intxdxe2xdxexdx+xdx

The two rules of integrating for this are:
inte^(ax+b)dx=e^(ax+b)/a+Ceax+bdx=eax+ba+C
intx^ndx=x^(n+1)/(n+1)+Cxndx=xn+1n+1+C

This gives us:
f(x)=e^(2x)/2-e^x+x^2/2+Cf(x)=e2x2ex+x22+C

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

0=e^(2(4))/2-e^4+4^2/2+C0=e2(4)2e4+422+C

0=e^8/2-e^4+8+C0=e82e4+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