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Math Help - Arc length problem

  1. #1
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    Arc length problem

    L= definite Integral a - b [sqrt(1+(dy/dx)^2)]dx

    y - 2 = (x+2)^3/2 points (-2,2) to (2,10)
    y=(x+2)^3/2 +2
    y' = 3/2 (x+2)^1/2
    1 + (y')^2 = 9x/4 +9/2 + 1 = (9/4)x +11/2

    I am getting stuck when it comes to setting up the integral, i forgot the law is it correct to distribute the integral when adding ?

    The answer I get is:
    Def. Int. (-2 --> 2) of [sqrt((9/4)x +11/2)]dx =
    [(3/2)x^3/2 +2.345x + C]-2 --> 2

    but I really don't like that last term 2.345x, i think i did it wrong. If you can point me in the right direction it would be greatly appreciated.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    Quote Originally Posted by bgonzal8 View Post
    L= definite Integral a - b [sqrt(1+(dy/dx)^2)]dx

    y - 2 = (x+2)^3/2 points (-2,2) to (2,10)
    y=(x+2)^3/2 +2
    y' = 3/2 (x+2)^1/2
    1 + (y')^2 = 9x/4 +9/2 + 1 = (9/4)x +11/2

    I am getting stuck when it comes to setting up the integral, i forgot the law is it correct to distribute the integral when adding ? This is true, but you can't forget about the radical. It is Not OK to distribute the integral through that.

    The answer I get is:
    Def. Int. (-2 --> 2) of [sqrt((9/4)x +11/2)]dx =
    [(3/2)x^3/2 +2.345x + C]-2 --> 2

    but I really don't like that last term 2.345x, i think i did it wrong. If you can point me in the right direction it would be greatly appreciated.
    Here's the set up...

    s=\int_{-2}^2\sqrt{1+\frac{9}{4}(x+2)}dx

    =\int_{-2}^2\sqrt{\frac{9}{4}x+\frac{11}{2}}dx

    Now let u=the radicand, then

    du=\frac{9}{4}dx\Rightarrow{\frac{4}{9}{du}}=dx

    So we now have

    s=\frac{4}{9}\int_{1}^{10}\sqrt{u}du.

    Note the new limits of integration.
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  3. #3
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    Thank you.
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