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

• May 29th 2011, 04:43 AM
Hydralisk
Slight mistake with Integral, I can't see why though.
I am having a small problem with this question,
$\displaystyle \int\frac{1}{2}(e^x-e^{-x})dx$
My workings are:
Breaking it down into two terms gives:
$\displaystyle \int(e^{x})dx=\frac{1}{1}e^{1x}+c=e^{x}+c$ and $\displaystyle \int(-e^{-x})dx=\frac{1}{-1}e^{-1x}+c=-e^{-x}+c$
Combining it all together then gives:
$\displaystyle \int\frac{1}{2}(e^x-e^{-x})dx=\frac{1}{2}(e^{x}-e^{-x})+c$
My mistake is inside the brackets, I am getting $\displaystyle -e^{-x}$ when the solution gives $\displaystyle +e^{-x}$
Can anybody see where I am going wrong?
Many thanks.
• May 29th 2011, 04:47 AM
Quacky
Quote:

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

Here is the mistake.

When you integrate $\displaystyle -e^{-x}$, you get $\displaystyle \frac{-1}{-1}e^{-x}+c$ Can you see why?
• May 29th 2011, 04:56 AM
Hydralisk
I'm really sorry but I can't see why, I don't understand why the numerator is negative.
• May 29th 2011, 05:02 AM
Hydralisk
I think I've got it, it's because the $\displaystyle e$ is negative, ie $\displaystyle -e=-1e$
So $\displaystyle \frac{1}{-1}(-1e)=\frac{-1}{-1 }e$
Is this correct?
• May 29th 2011, 05:04 AM
Quacky
Quote:

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

Yes, and obviously this gives $\displaystyle +e^{-x}$

You had integrated $\displaystyle e^{-x}$ in the initial post, whereas you were trying to integrate $\displaystyle -e^{-x}$
• May 29th 2011, 05:06 AM
Hydralisk
That's clear now, thank you!