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Math Help - Area of Region

  1. #1
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    Area of Region

    Sketch the region enclosed by and . Decide whether to integrate with respect to or . Then find the area of the region.


    I am not looking for the answer, just the name of the topic. I have no idea where to start and was hoping I could use the book to find out how to approach this problem. Thanks. The hw is due in an hour, and i still have 2 more problems to go!
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  2. #2
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    OK so i started by setting them equal to each other. then I solved for x.

    I got (+,- 5/9,6.25) then I subtracted y upper from y lower.

    I am trying to do the problem, but I just cant get it.


    Can anyone help?

    Thanks.
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  3. #3
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    I think im on the right track, I just seem to be missing a fundamental step.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Here is the graph. it is easier to integrate with respect to x. to find the limits of the integration, solve 4x^2 = 5x. obviously one limit will be zero, find the other one. then integrate 5x - 4x^2 since 5x is the higher function

    i'm off to bed, you're on your own
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xcelxp View Post
    OK so i started by setting them equal to each other. then I solved for x.

    I got (+,- 5/9,6.25) then I subtracted y upper from y lower.

    I am trying to do the problem, but I just cant get it.


    Can anyone help?

    Thanks.
    for the limits of integration:

    5x = 4x^2
    => 4x^2 - 5x = 0
    => x(4x - 5) = 0
    => x = 0 or 4x - 5 = 0
    so x = 0 and x = 5/4

    so the limits of integration are 0 up to 5/4

    the area is given by:

    int{0:5/4}(5x - 4x^2)dx
    --
    if there is anything you're unsure about, ask now, i'll be in bed in about a minute, i'm already falling asleep while sitting here

    what is the other question by the way?
    Last edited by ThePerfectHacker; April 19th 2007 at 11:59 AM.
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  6. #6
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    Just wanted to let you know that your explanation really helped

    Ans: ((5/2)(5/4)^2-(4/3)(5/4)^3) - ((5/2)(0)^2-(4/3)(0)^3)

    Ty so much, I gave you your thanks!
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xcelxp View Post
    Just wanted to let you know that your explanation really helped

    Ans: ((5/2)(5/4)^2-(4/3)(5/4)^3) - ((5/2)(0)^2-(4/3)(0)^3)

    Ty so much, I gave you your thanks!
    that seems like the right answer? is it? what about the other question? what explanation, are you talking about the post where i mentioned the fundamental theorem of calculus? you should be sure to look that up by the way, it's VERY important
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  8. #8
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    by the way I figured it out before you wrote that explanation with the numbers, now Im working on:

    Sketch the region enclosed by , , and . Decide whether to integrate with respect to or . Then find the area of the region.
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  9. #9
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    Yup I understand the fundamental theorem now but there is one othe rproblem that is giving me a problem, but the hw is due before 3 am and its 2:55 I doubt ill get everything in on time
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