What is f(x) = int e^(5x-1)+x dxf(x)=∫e5x−1+xdx if f(0) = -4 f(0)=−4?

1 Answer
Gió
Sep 9, 2017

I tried this:

Explanation:

Given:
f(x)=int(e^(5x-1)+x)dxf(x)=∫(e5x−1+x)dx
let us solve the integral:
f(x)=int(e^(5x-1))dx+intxdxf(x)=∫(e5x−1)dx+∫xdx
f(x)=(e^(5x-1))/5+x^2/2+cf(x)=e5x−15+x22+c
so now we need to evaluate the value of the constant cc. We know that at x=0x=0 we have: f(0)=-4f(0)=−4 so using our function:
f(0)=-4=(e^(5*color(red)(0)-1))/5+color(red)(0)^2/2+cf(0)=−4=e5⋅0−15+022+c
-4=e^-1/5+c−4=e−15+c
so that:
c=-e^-1/5-4c=−e−15−4
and our function finally becomes:
f(x)=(e^(5x-1))/5+x^2/2-e^-1/5-4f(x)=e5x−15+x22−e−15−4

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