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Bonrozzy Group Title

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.

  • 6 months ago
  • 6 months ago

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  1. myininaya Group Title
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    So you have talked about improper integrals?

    • 6 months ago
  2. Bonrozzy Group Title
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    yes

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

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

    • 6 months ago
  5. myininaya Group Title
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    did you get this far?

    • 6 months ago
  6. Bonrozzy Group Title
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    i see

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

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

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

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

    • 6 months ago
  11. myininaya Group Title
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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

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

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

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

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

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

    • 6 months ago
  17. Bonrozzy Group Title
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    -E^-x*

    • 6 months ago
  18. myininaya Group Title
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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

    • 6 months ago
  19. myininaya Group Title
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    yep you are right

    • 6 months ago
  20. Bonrozzy Group Title
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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

    • 6 months ago
  21. myininaya Group Title
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    i got it

    • 6 months ago
  22. myininaya Group Title
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    i wrote the equation just a little wrong

    • 6 months ago
  23. Bonrozzy Group Title
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    oh

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

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

    • 6 months ago
  26. myininaya Group Title
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    i'm idiot sorry

    • 6 months ago
  27. Bonrozzy Group Title
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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!

    • 6 months ago
  28. myininaya Group Title
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    Did you get it?

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

    • 6 months ago
  30. Bonrozzy Group Title
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    thats the function right?

    • 6 months ago
  31. myininaya Group Title
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    yep good so far

    • 6 months ago
  32. Bonrozzy Group Title
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    Then just solve for k?

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

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

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

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

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

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

    • 6 months ago
  39. Bonrozzy Group Title
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    true haha

    • 6 months ago
  40. myininaya Group Title
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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

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

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

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

    • 6 months ago
  44. myininaya Group Title
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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

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

    • 6 months ago
  46. Bonrozzy Group Title
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    cool.

    • 6 months ago
  47. Bonrozzy Group Title
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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

    • 6 months ago
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