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Thefaceless
Group Title
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.
 one year ago
 one year ago
Thefaceless Group Title
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.
 one year ago
 one year ago

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dumbcow Group TitleBest ResponseYou've already chosen the best response.1
\[A = \int\limits_{0}^{\ln(2)} (e^{2x}  e^{x}) dx\]
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
Right, but what is the actual area. That's where I am having trouble. thanks.
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
The answer is supposed to be 1/2, but I am not getting that.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Were you able to get through the integration step? Before plugging in the limits.
 one year ago

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

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
And is there anyway to give more than one medal?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
No, just give to cow c: hehe
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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).
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
I guess both. I thought I was okay on the integral, but I am questioning myself on it now.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1359942794387:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We subtract the `lower` from the `upper` function.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So in that example, the area between the curves would be f(x)g(x).
 one year ago

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

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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 ^^
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
I see. So what if you had y=cos 2x, x=pi/ 4, x=pi/2 ? Which interval would be on top?
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
Is it the one that is biggest relative to the function?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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 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. \[\huge \int\limits_{x=\pi/4}^{\pi/2} \cos2x0\; dx\]
 one year ago

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yah that seems to be accurate c:
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
Hm, when you do it that way, the answer is 1/2. The answer in the back of my book is 1/2.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Really? Hmm wolfram is giving me 1/2 as well... Hmm
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
Ohhhhh, I think I know why.
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
err the x axis*
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Thefaceless Group TitleBest ResponseYou've already chosen the best response.0
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! :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
ah i see c:
 one year ago
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