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Bonrozzy

  • 2 years 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
    • 2 years ago
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    So you have talked about improper integrals?

  2. Bonrozzy
    • 2 years ago
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    yes

  3. myininaya
    • 2 years 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
    • 2 years ago
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    we want that one area to be half of that other area

  5. myininaya
    • 2 years ago
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    did you get this far?

  6. Bonrozzy
    • 2 years ago
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    i see

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

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

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

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

  11. myininaya
    • 2 years 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
    • 2 years ago
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    \[-1/e^k = 1/e\]

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

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

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

  16. Bonrozzy
    • 2 years ago
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    the antiderivative of e^-x is -e^x right?

  17. Bonrozzy
    • 2 years ago
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    -E^-x*

  18. myininaya
    • 2 years 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
    • 2 years ago
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    yep you are right

  20. Bonrozzy
    • 2 years 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
    • 2 years ago
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    i got it

  22. myininaya
    • 2 years ago
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    i wrote the equation just a little wrong

  23. Bonrozzy
    • 2 years ago
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    oh

  24. myininaya
    • 2 years 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
    • 2 years ago
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    you will get the answer you got using your calculator except it will be exact

  26. myininaya
    • 2 years ago
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    i'm idiot sorry

  27. Bonrozzy
    • 2 years 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
    • 2 years ago
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    Did you get it?

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

  30. Bonrozzy
    • 2 years ago
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    thats the function right?

  31. myininaya
    • 2 years ago
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    yep good so far

  32. Bonrozzy
    • 2 years ago
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    Then just solve for k?

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

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

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

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

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

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

  39. Bonrozzy
    • 2 years ago
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    true haha

  40. myininaya
    • 2 years 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
    • 2 years ago
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    Yes I noticed that haha. I can't thank you enough.

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

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

  44. myininaya
    • 2 years 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
    • 2 years ago
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    anyways fun problem we are going over improper integrals today in class :)

  46. Bonrozzy
    • 2 years ago
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    cool.

  47. Bonrozzy
    • 2 years 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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