# Thread: Integration of 1 over e

1. ## Integration of 1 over e

Hi,

I am trying to integrate $\displaystyle \int \frac{1}{e^x} dx$

I have tried using substitution of $\displaystyle u = x$ or $\displaystyle u=e^x$ both does not help.

On a side note, is it correct that $\displaystyle \frac{1}{e^x} = e^{-x}$?

2. ## Re: Integration of 1 over e

Hi M.R!

Originally Posted by M.R
Hi,

I am trying to integrate $\displaystyle \int \frac{1}{e^x} dx$

I have tried using substitution of $\displaystyle u = x$ or $\displaystyle u=e^x$ both does not help.

On a side note, is it correct that $\displaystyle \frac{1}{e^x} = e^{-x}$?
Yes, it is correct that $\displaystyle \frac{1}{e^x} = e^{-x}$.

Do you know what the derivative of $\displaystyle e^{x}$ is?

3. ## Re: Integration of 1 over e

Yes. Derivative of $\displaystyle e^x = e^x$ and derivative of $\displaystyle e^{-x} = -e^{-x}$

So it follows that $\displaystyle \int e^{-x} dx = -e^{-x}$ ?

4. ## Re: Integration of 1 over e

Yes, although you should add an integration constant to an indeterminate integral.

That is:

$\displaystyle \int e^{-x} dx = -e^{-x} + C$

5. ## Re: Integration of 1 over e

Correct, but don't forget the $\displaystyle +C$ , you can check it by finding $\displaystyle \frac{d}{dx}\left(-e^{-x}\right)$

Thank you