anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
dumbcow
  • dumbcow
\[A = \int\limits_{0}^{\ln(2)} (e^{2x} - e^{x}) dx\]
anonymous
  • anonymous
Right, but what is the actual area. That's where I am having trouble. thanks.
anonymous
  • anonymous
The answer is supposed to be 1/2, but I am not getting that.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zepdrix
  • zepdrix
Were you able to get through the integration step? Before plugging in the limits.
zepdrix
  • zepdrix
Were you able to get this? :) \[\large \frac{1}{2}e^{2x}-e^x\]
anonymous
  • 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
  • anonymous
And is there anyway to give more than one medal?
zepdrix
  • zepdrix
No, just give to cow c: hehe
zepdrix
  • 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
  • anonymous
I guess both. I thought I was okay on the integral, but I am questioning myself on it now.
zepdrix
  • zepdrix
|dw:1359942794387:dw|
zepdrix
  • zepdrix
We subtract the `lower` from the `upper` function.
zepdrix
  • zepdrix
So in that example, the area between the curves would be f(x)-g(x).
zepdrix
  • zepdrix
Confusing picture? :D lol so many arrows, maybe that was silly of me :)
anonymous
  • 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
  • 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
  • anonymous
I see. So what if you had y=cos 2x, x=pi/ 4, x=pi/2 ? Which interval would be on top?
anonymous
  • anonymous
Is it the one that is biggest relative to the function?
zepdrix
  • 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
  • zepdrix
Lemme verify that on wolfram real quick to make sure i'm not talking out my butt :P
zepdrix
  • zepdrix
Yah that seems to be accurate c:
anonymous
  • 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
  • zepdrix
Really? Hmm wolfram is giving me -1/2 as well... Hmm
anonymous
  • anonymous
Ohhhhh, I think I know why.
anonymous
  • 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
  • anonymous
err the x axis*
zepdrix
  • 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
  • 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
  • zepdrix
ah i see c:

Looking for something else?

Not the answer you are looking for? Search for more explanations.