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

Question

1207 views around the world

You can reuse this answer
Creative Commons License

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#