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\[A = \int\limits_{0}^{\ln(2)} (e^{2x} - e^{x}) dx\]

Right, but what is the actual area. That's where I am having trouble. thanks.

The answer is supposed to be 1/2, but I am not getting that.

Were you able to get through the integration step?
Before plugging in the limits.

Were you able to get this? :)
\[\large \frac{1}{2}e^{2x}-e^x\]

And is there anyway to give more than one medal?

No, just give to cow c: hehe

I guess both. I thought I was okay on the integral, but I am questioning myself on it now.

|dw:1359942794387:dw|

We subtract the `lower` from the `upper` function.

So in that example, the area between the curves would be f(x)-g(x).

Confusing picture? :D
lol so many arrows, maybe that was silly of me :)

I see. So what if you had y=cos 2x, x=pi/ 4, x=pi/2 ? Which interval would be on top?

Is it the one that is biggest relative to the function?

Lemme verify that on wolfram real quick to make sure i'm not talking out my butt :P

Yah that seems to be accurate c:

Hm, when you do it that way, the answer is -1/2. The answer in the back of my book is 1/2.

Really? Hmm wolfram is giving me -1/2 as well... Hmm

Ohhhhh, I think I know why.

err the x axis*

ah i see c: