What is $f(x) = int e^(2x)-e^x+x^2 dx$ if $f(1) = 0$ ?

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Answer 1 · 2017-07-10T03:10:42

Zero.

First integrate:

$f(x)=inte^(2x)-e^x+x^2dx$

$=1/2e^(2x)-e^x+1/3x^3+C$

So, for $f(1)=0$ , we have:

$0=1/2e^2-e+1/3+C$

Then you can solve for C.

$C=-e^2/2+e-1/3$

So,

$f(x)=inte^(2x)-e^x+x^2dx=1/2e^2-e+1/3+(-e^2/2+e-1/3)$

$=0$

What is f(x) = int e^(2x)-e^x+x^2 dxf(x) = int e^(2x)-e^x+x^2 dx if f(1) = 0 f(1) = 0 ?