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Bonrozzy

  • 11 months ago

Find the value of k that the line x=k divides the area of the first quadrant region of y=e^-x and the x-axis x >= 1 into two equal parts.

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  1. myininaya
    • 11 months ago
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    So you have talked about improper integrals?

  2. Bonrozzy
    • 11 months ago
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    yes

  3. myininaya
    • 11 months ago
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    so we want this: \[1/2 \int\limits_{1}^{k}e^{-x} dx=\lim_{a \rightarrow \infty}\int\limits_{1}^{a}e^{-x} dx\]

  4. myininaya
    • 11 months ago
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    we want that one area to be half of that other area

  5. myininaya
    • 11 months ago
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    did you get this far?

  6. Bonrozzy
    • 11 months ago
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    i see

  7. Bonrozzy
    • 11 months ago
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    I hadn't thought about constructing it that way. It makes sense.

  8. myininaya
    • 11 months ago
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    so we can try to evaluate both integrals first then we just solve the equation for k

  9. Bonrozzy
    • 11 months ago
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    That's true haha it seems so simple now thank you :)

  10. Bonrozzy
    • 11 months ago
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    I get that k=1, but i know that's not the answer.

  11. myininaya
    • 11 months ago
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    what is the answer? also yep if k is one, then that one area is 0 and we know 0 is not half the other area

  12. Bonrozzy
    • 11 months ago
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    \[-1/e^k = 1/e\]

  13. Bonrozzy
    • 11 months ago
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    sorry it wouldnt be one, but this is the end result i believe

  14. myininaya
    • 11 months ago
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    that negative is weird killing me because e^(to a positive number) can't be negative

  15. myininaya
    • 11 months ago
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    i was agreeing it isn't 1 because it would give us 0 for that one area

  16. Bonrozzy
    • 11 months ago
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    the antiderivative of e^-x is -e^x right?

  17. Bonrozzy
    • 11 months ago
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    -E^-x*

  18. myininaya
    • 11 months ago
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    maybe since we can't solve that equation for k there is no way possible to divide the area with a vertical line so we have two equal areas on both sides

  19. myininaya
    • 11 months ago
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    yep you are right

  20. Bonrozzy
    • 11 months ago
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    Okay, well here's some help. plugging the e^-x in my calculator and then taking the integral of that I find the area, i divided it by 2 and then checked the x y table to find the x that gives half the area. It is about 1.693

  21. myininaya
    • 11 months ago
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    i got it

  22. myininaya
    • 11 months ago
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    i wrote the equation just a little wrong

  23. Bonrozzy
    • 11 months ago
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    oh

  24. myininaya
    • 11 months ago
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    \[ \int\limits\limits_{1}^{k}e^{-x} dx=1/2 \lim_{a \rightarrow \infty}\int\limits\limits_{1}^{a}e^{-x} dx\]

  25. myininaya
    • 11 months ago
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    you will get the answer you got using your calculator except it will be exact

  26. myininaya
    • 11 months ago
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    i'm idiot sorry

  27. Bonrozzy
    • 11 months ago
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    oh such an easily overseen mistake! Ill check this out right now hahhaaha it's perfectly fine :D thanks for taking the time to figure it out!

  28. myininaya
    • 11 months ago
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    Did you get it?

  29. Bonrozzy
    • 11 months ago
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    \[-1/e^k+1/e=1/2e\]

  30. Bonrozzy
    • 11 months ago
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    thats the function right?

  31. myininaya
    • 11 months ago
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    yep good so far

  32. Bonrozzy
    • 11 months ago
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    Then just solve for k?

  33. myininaya
    • 11 months ago
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    yep! :) I left it as \[e^{-k}+e^{-1}=e^{-1}/2 \]

  34. myininaya
    • 11 months ago
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    oops with a negative in front of the e^(-k)

  35. myininaya
    • 11 months ago
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    \[-e^{-k}+e^{-1}=1/2e^{-1} => -e^{-k}=-1/2 e^{-1} \]

  36. myininaya
    • 11 months ago
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    get rid of those negatives then take natural log of both sides

  37. Bonrozzy
    • 11 months ago
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    Yes I got it! Thank you so much :) it's gonna be: \[-\ln (1/2e)\]

  38. myininaya
    • 11 months ago
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    or you could write it as ln(2e)

  39. Bonrozzy
    • 11 months ago
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    true haha

  40. myininaya
    • 11 months ago
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    but yep if you put that in your calculator you will see that its approximation is the same as the approximation you got using just the calculator to find what was it 1.6 something

  41. Bonrozzy
    • 11 months ago
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    Yes I noticed that haha. I can't thank you enough.

  42. myininaya
    • 11 months ago
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    I'm going to need to learn that cool calculator trick for myself

  43. Bonrozzy
    • 11 months ago
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    I'm rather inept at using the calculator, but that's one trick I know ;)

  44. myininaya
    • 11 months ago
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    i know the basics and i know there are really cool tricks in stuff i could learn just never got around to it

  45. myininaya
    • 11 months ago
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    anyways fun problem we are going over improper integrals today in class :)

  46. Bonrozzy
    • 11 months ago
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    cool.

  47. Bonrozzy
    • 11 months ago
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    Well, where i am it's just about dinner time. Thanks again it really saved a lot of hair pulling. See ya

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