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Math Help - I need help finishing this problem.

  1. #1
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    I need help finishing this problem.

    [\frac{1}{10x^5}-\frac{1}{6x^3}] _{1}^{2}

    The answer is \frac{779}{240}

    I've tried plugging in 2 for x and 1 for x, but I'm not coming up with the answer.



    Thanks
    Jason
    Last edited by Darkhrse99; July 29th 2009 at 01:04 PM.
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  2. #2
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     \int_{1}^{2}[\frac{1/10x^5}{1/6x^3}]<br />

    Should it be this?
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  3. #3
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    Quote Originally Posted by ANDS! View Post
     \int_{1}^{2}[\frac{1/10x^5}{1/6x^3}]<br />

    Should it be this?
    No, it should be what I said, but the intagral shouldn't be there. I didn't know how to place the 2 and 1 after the brackets. the fraction needs to be evaluated from 1 to 2.
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  4. #4
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    Quote Originally Posted by Darkhrse99 View Post
    No, it should be what I said, but the intagral shouldn't be there. I didn't know how to place the 2 and 1 after the brackets. the fraction needs to be evaluated from 1 to 2.
    well, it's not the "answer" you cite (779/240)

    maybe it would help if you posted the original definite integral?
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  5. #5
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    Well

    f(x)]_{1}^{2}

    is

    \int_{1}^{2}f(x)dx

    And agreed. That expression evaluated on those bounds can't be the answer you've got; unless this isn't the full problem.
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  6. #6
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    The original problem is find the arc length of the indicated interval.

    Y=\frac{x^4}{8}+\frac{1}{4x^2}
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  7. #7
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    Whoa. That is markedly different than what you had. What you had simply said "evaluate the integral from 1 to 2". Which is what we're doing when finding arc-length.

    So we have an equation of the form

    \int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx is equal to the arc-legth of  f(x) from a to b.

    So from your equation we need to find the derivative of:

    y=\frac{x^4}{8}+\frac{1}{4x^2}<br />

    Which is equal to:

    \frac{x^3}{2}-\frac{1}{2x^{3}}

    Plugging this into our equation for arc length yields:

    \int_{1}^{2}\sqrt{1+[\frac{x^3}{2}-\frac{1}{2x^{3}}]^{2}}dx

    Can you distribute that squared and go from there?
    Last edited by ANDS!; July 29th 2009 at 02:01 PM. Reason: I suck. . .
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  8. #8
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    Quote Originally Posted by Darkhrse99 View Post
    The original problem is find the arc length of the indicated interval.

    Y=\frac{x^4}{8}+\frac{1}{4x^2}
    y' = \frac{x^3}{2} - \frac{1}{2x^3}

    (y')^2 = \frac{x^6}{4} - \frac{1}{2} + \frac{1}{4x^6}

    1 + (y')^2 = \frac{x^6}{4} + \frac{1}{2} + \frac{1}{4x^6}

    \sqrt{\frac{x^6}{4} + \frac{1}{2} + \frac{1}{4x^6}} = \frac{x^3}{2} + \frac{1}{2x^3}

    \int_1^2 \frac{x^3}{2} + \frac{1}{2x^3} \, dx

    \left[\frac{x^4}{8} - \frac{1}{4x^2}\right]_1^2

    \left(2 - \frac{1}{16}\right) - \left(\frac{1}{8} - \frac{1}{4}\right) = \frac{33}{16}
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  9. #9
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    Got it guys! Sorry for the confusion.
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