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Math Help - Integral Problem Help

  1. #1
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    Integral Problem Help

    The problem is y=x^(2/3) and x=1 and x=8. I need to find the exact arc length of the curve over the interval.

    The formula used is the arc length formula: L=integral from a to b of (1+(dy/dx)^2) ^(1/2)

    So far I got this:

    Integral from 1 to 8 of (1+(4/9*x^(-2/3)))^(1/2)dx

    I don't know how to integrate this. Is it a u substitution? I tried u= 1+ 4/9x^(-2/3) and I have an x^(-5/3) left over so that won't work.
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  2. #2
    Ted
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    Quote Originally Posted by Simon777 View Post
    The problem is y=x^(2/3) and x=1 and x=8. I need to find the exact arc length of the curve over the interval.

    The formula used is the arc length formula: L=integral from a to b of (1+(dy/dx)^2) ^(1/2)

    So far I got this:

    Integral from 1 to 8 of (1+(4/9*x^(-2/3)))^(1/2)dx

    I don't know how to integrate this. Is it a u substitution? I tried u= 1+ 4/9x^(-2/3) and I have an x^(-5/3) left over so that won't work.
    For:
    \int_1^8 \sqrt{\frac{4}{9x^{\frac{2}{3}}}+1} dx.

    Make a common denominator to get:

    \int_1^8 \sqrt{ \frac{ 9x^{\frac{2}{3}}+4 }{ 9x^{\frac{2}{3} }} }dx.

    Now, Apply : \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} for b \neq 0.

    Then, think about a substitution.
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  3. #3
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    Quote Originally Posted by Simon777 View Post
    The problem is y=x^(2/3) and x=1 and x=8. I need to find the exact arc length of the curve over the interval.

    The formula used is the arc length formula: L=integral from a to b of (1+(dy/dx)^2) ^(1/2)

    So far I got this:

    Integral from 1 to 8 of (1+(4/9*x^(-2/3)))^(1/2)dx

    I don't know how to integrate this. Is it a u substitution? I tried u= 1+ 4/9x^(-2/3) and I have an x^(-5/3) left over so that won't work.
    f'(x) = \frac{2}{3}x^{-\frac{1}{3}}

    f'(x) = \frac{2}{3x^{\frac{1}{3}}}

    [f'(x)]^2 = \frac{4}{9x^{\frac{2}{3}}}

    1 + [f'(x)]^2 = 1 + \frac{4}{9x^{\frac{2}{3}}} = \frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}<br />

    \sqrt{\frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}} = \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4}



    \int_1^8 \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4} \, dx

    let u = 9x^{\frac{2}{3}}+4

    du = \frac{6}{x^{\frac{1}{3}}} \, dx

    \frac{1}{18} \int_{13}^{40} \sqrt{u} \, du

    can you finish ?
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  4. #4
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    Quote Originally Posted by Ted View Post
    For:
    \int_1^8 \sqrt{\frac{4}{9x^{\frac{2}{3}}}+1} dx.

    Make a common denominator to get:

    \int_1^8 \sqrt{ \frac{ 9x^{\frac{2}{3}}+4 }{ 9x^{\frac{2}{3} }} }dx.

    Now, Apply : \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} for b \neq 0.

    Then, think about a substitution.
    Wow, such a simple solution, I don't know why I didn't think to do that. Thank you so much for your help.
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  5. #5
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    Quote Originally Posted by skeeter View Post
    f'(x) = \frac{2}{3}x^{-\frac{1}{3}}

    f'(x) = \frac{2}{3x^{\frac{1}{3}}}

    [f'(x)]^2 = \frac{4}{9x^{\frac{2}{3}}}

    1 + [f'(x)]^2 = 1 + \frac{4}{9x^{\frac{2}{3}}} = \frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}<br />

    \sqrt{\frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}} = \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4}



    \int_1^8 \frac{1}{3x^{\frac{1}{3}}} \cdot \sqrt{9x^{\frac{2}{3}}+4} \, dx

    let u = 9x^{\frac{2}{3}}+4

    du = \frac{6}{x^{\frac{1}{3}}} \, dx

    \frac{1}{18} \int_{13}^{40} \sqrt{u} \, du

    can you finish ?
    Thank you for working it out so I have something to check my steps with. Yes, I finished and checked my answer with the back of my book and it is correct.
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