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

1 Answer
May 5, 2016

f(x)=ln(x)-e^x+e

Explanation:

f(x)= int(1/x-e^x)dx
=int1/xdx-inte^xdx (Linearity)

int1/xdx = ln(x) (Standard integral)

inte^x = e^x (Exponential rule)

Therefore:
f(x) = ln(x)-e^x +C (Where C is the constant of integration)

We are told that f(1)=0 Thus:
ln(1)-e^1+C=0

0 - e +C =0

C=e

Hence: f(x)=ln(x)-e^x+e