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

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Answer 1 · 2018-05-20T20:17:05

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

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 $f(4)=0$

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

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

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

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

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