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

1 Answer
Gió
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#

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