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Thefaceless

  • 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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  1. dumbcow
    • 3 years ago
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    \[A = \int\limits_{0}^{\ln(2)} (e^{2x} - e^{x}) dx\]

  2. Thefaceless
    • 3 years ago
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    Right, but what is the actual area. That's where I am having trouble. thanks.

  3. Thefaceless
    • 3 years ago
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    The answer is supposed to be 1/2, but I am not getting that.

  4. zepdrix
    • 3 years ago
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    Were you able to get through the integration step? Before plugging in the limits.

  5. zepdrix
    • 3 years ago
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    Were you able to get this? :) \[\large \frac{1}{2}e^{2x}-e^x\]

  6. Thefaceless
    • 3 years ago
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    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?

  7. Thefaceless
    • 3 years ago
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    And is there anyway to give more than one medal?

  8. zepdrix
    • 3 years ago
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    No, just give to cow c: hehe

  9. zepdrix
    • 3 years ago
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    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).

  10. Thefaceless
    • 3 years ago
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    I guess both. I thought I was okay on the integral, but I am questioning myself on it now.

  11. zepdrix
    • 3 years ago
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    |dw:1359942794387:dw|

  12. zepdrix
    • 3 years ago
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    We subtract the `lower` from the `upper` function.

  13. zepdrix
    • 3 years ago
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    So in that example, the area between the curves would be f(x)-g(x).

  14. zepdrix
    • 3 years ago
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    Confusing picture? :D lol so many arrows, maybe that was silly of me :)

  15. Thefaceless
    • 3 years ago
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    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?

  16. zepdrix
    • 3 years ago
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    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 ^^

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

  18. Thefaceless
    • 3 years ago
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    Is it the one that is biggest relative to the function?

  19. zepdrix
    • 3 years ago
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    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. \[\huge \int\limits_{x=\pi/4}^{\pi/2} \cos2x-0\; dx\]

  20. zepdrix
    • 3 years ago
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    Lemme verify that on wolfram real quick to make sure i'm not talking out my butt :P

  21. zepdrix
    • 3 years ago
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    Yah that seems to be accurate c:

  22. Thefaceless
    • 3 years ago
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    Hm, when you do it that way, the answer is -1/2. The answer in the back of my book is 1/2.

  23. zepdrix
    • 3 years ago
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    Really? Hmm wolfram is giving me -1/2 as well... Hmm

  24. Thefaceless
    • 3 years ago
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    Ohhhhh, I think I know why.

  25. Thefaceless
    • 3 years ago
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    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

  26. Thefaceless
    • 3 years ago
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    err the x axis*

  27. zepdrix
    • 3 years ago
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    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

  28. Thefaceless
    • 3 years ago
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    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! :)

  29. zepdrix
    • 3 years ago
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    ah i see c:

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