Calculus 2 problem. Help please.
y=e^x, y=e^2x, x=ln2, x=0
Find the area enclosed by the curves. Show steps please.

- anonymous

- schrodinger

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- dumbcow

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

- anonymous

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

- anonymous

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

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## More answers

- zepdrix

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

- zepdrix

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

- anonymous

Yes, I got that part, but I did the algebra wrong, looking at it now.
I got the answer now. I have a quick question, however. How do you know which function to use first in the integration process?

- anonymous

And is there anyway to give more than one medal?

- zepdrix

No, just give to cow c: hehe

- zepdrix

You mean which one is being subtracted when you setup the integrand?
Or do you mean which limit you're subtracting when you evaluate the integral (plugging in).

- anonymous

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

- zepdrix

|dw:1359942794387:dw|

- zepdrix

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

- zepdrix

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

- zepdrix

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

- anonymous

Lol, I understood it, so it did its job! Basically, though, we are just sticking to the convention of biggest minus smallest, right? Same with the intervals?

- zepdrix

Yes, same with the limits.
If we did it the other way around, we would end up with the same answer, but negative.
And negative area doesn't make much sense ^^

- anonymous

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

- anonymous

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

- zepdrix

Bah! These ones always confuse me because they didn't give us a second boundary for \(y\).
I think what happens is, our other boundary is just the x-axis \(y=0\).
And in such a case, we're not talking about area between curves.
So in this case we `can` end up with negative area.
\[\huge \int\limits_{x=\pi/4}^{\pi/2} \cos2x-0\; dx\]

- zepdrix

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

- zepdrix

Yah that seems to be accurate c:

- anonymous

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

- zepdrix

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

- anonymous

Ohhhhh, I think I know why.

- anonymous

That part of the curve is under the y axis, but since we can't have a negative area, we abs value the function. I think.. lol

- anonymous

err the x axis*

- zepdrix

Well this is the way I learned in back in calc 2 :D maybe I'm not remembering correctly lol.
If we're looking for area `between curves`, it's always positive because we're thinking of an amount of space between them.
But when we're looking for area under a curve, it can be negative.
You might wanna ask your teacher about that. Because clearly I'm a little rusty XDD

- anonymous

I would, but this hw is due first thing in the morning lol. Thanks for your help, though. It cleared up a lot for me! :)

- zepdrix

ah i see c:

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