# Thread: Slight mistake with Integral, I can't see why though.

1. ## Slight mistake with Integral, I can't see why though.

I am having a small problem with this question,
$\int\frac{1}{2}(e^x-e^{-x})dx$
My workings are:
Breaking it down into two terms gives:
$\int(e^{x})dx=\frac{1}{1}e^{1x}+c=e^{x}+c$ and $\int(-e^{-x})dx=\frac{1}{-1}e^{-1x}+c=-e^{-x}+c$
Combining it all together then gives:
$\int\frac{1}{2}(e^x-e^{-x})dx=\frac{1}{2}(e^{x}-e^{-x})+c$
My mistake is inside the brackets, I am getting $-e^{-x}$ when the solution gives $+e^{-x}$
Can anybody see where I am going wrong?
Many thanks.

2. Originally Posted by Hydralisk
I am having a small problem with this question,
$\int(-e^{-x})dx=\frac{1}{-1}e^{-1x}+c=-e^{-x}+c$
Here is the mistake.

When you integrate $-e^{-x}$, you get $\frac{-1}{-1}e^{-x}+c$ Can you see why?

3. I'm really sorry but I can't see why, I don't understand why the numerator is negative.

4. I think I've got it, it's because the $e$ is negative, ie $-e=-1e$
So $\frac{1}{-1}(-1e)=\frac{-1}{-1 }e$
Is this correct?

5. Originally Posted by Hydralisk
I think I've got it, it's because the $e$ is negative, ie $-e=-1e$
So $\frac{1}{-1}(-1e)=\frac{-1}{-1 }e$
Is this correct?
Yes, and obviously this gives $+e^{-x}$

You had integrated $e^{-x}$ in the initial post, whereas you were trying to integrate $-e^{-x}$

6. That's clear now, thank you!