What is f(x) = int e^(5x-1)+x dx if f(0) = -4 ?

1 Answer
Sep 9, 2017

I tried this:

Explanation:

Given:
f(x)=int(e^(5x-1)+x)dx
let us solve the integral:
f(x)=int(e^(5x-1))dx+intxdx
f(x)=(e^(5x-1))/5+x^2/2+c
so now we need to evaluate the value of the constant c. We know that at x=0 we have: f(0)=-4 so using our function:
f(0)=-4=(e^(5*color(red)(0)-1))/5+color(red)(0)^2/2+c
-4=e^-1/5+c
so that:
c=-e^-1/5-4
and our function finally becomes:
f(x)=(e^(5x-1))/5+x^2/2-e^-1/5-4