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anonymous
 3 years ago
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
 3 years ago
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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[A = \int\limits_{0}^{\ln(2)} (e^{2x}  e^{x}) dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right, but what is the actual area. That's where I am having trouble. thanks.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The answer is supposed to be 1/2, but I am not getting that.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Were you able to get through the integration step? Before plugging in the limits.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Were you able to get this? :) \[\large \frac{1}{2}e^{2x}e^x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0And is there anyway to give more than one medal?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1No, just give to cow c: hehe

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1You 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
 3 years ago
Best ResponseYou've already chosen the best response.0I guess both. I thought I was okay on the integral, but I am questioning myself on it now.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1We subtract the `lower` from the `upper` function.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1So in that example, the area between the curves would be f(x)g(x).

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Confusing picture? :D lol so many arrows, maybe that was silly of me :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lol, 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
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0I see. So what if you had y=cos 2x, x=pi/ 4, x=pi/2 ? Which interval would be on top?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is it the one that is biggest relative to the function?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Bah! 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 xaxis \(y=0\). And in such a case, we're not talking about area between curves. So in this case we `can` end up with negative area. \[\huge \int\limits_{x=\pi/4}^{\pi/2} \cos2x0\; dx\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Lemme verify that on wolfram real quick to make sure i'm not talking out my butt :P

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Yah that seems to be accurate c:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hm, when you do it that way, the answer is 1/2. The answer in the back of my book is 1/2.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Really? Hmm wolfram is giving me 1/2 as well... Hmm

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ohhhhh, I think I know why.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That 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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Well 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
 3 years ago
Best ResponseYou've already chosen the best response.0I would, but this hw is due first thing in the morning lol. Thanks for your help, though. It cleared up a lot for me! :)
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