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Math Help - Integration with a square root

  1. #1
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    Integration with a square root

    <br />
\int_{1}^{4}\sqrt{t^2+4+t^{-2}}dt<br />

    I've never seen an integral of this form, and I need help solving it. Am I expected to use tables?



    The original problem asks for the length of the three dimensional curve given by r(t) = (2t, ln(t), t^2)

    The answer in the back of the book is L = 14 + ln(4)

    I set the arc length forumla up and arrived to this integral. Using the t89, I confirmed my setup results in the correct answer. However, I do not know how to arrive to that answer.
    Last edited by 1005; June 21st 2009 at 12:31 PM.
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  2. #2
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    Quote Originally Posted by 1005 View Post
    <br />
\int_{1}^{4}\sqrt{t^2+4+t^{-2}}dt<br />

    I've never seen an integral of this form, and I need help solving it. Am I expected to use tables?



    The original problem asks for the length of the three dimensional curve given by r(t) = (2t, ln(t), t^2)

    The answer in the back of the book is L = 14 + ln(4)

    I set the arc length forumla up and arrived to this integral. Using the t89, I confirmed my setup results in the correct answer. However, I do not know how to arrive to that answer.
    is this an arc length problem? if so, what is the original function?
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  3. #3
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    Quote Originally Posted by 1005 View Post
    <br />
\int_{1}^{4}\sqrt{t^2+4+t^{-2}}dt<br />

    I've never seen an integral of this form, and I need help solving it. Am I expected to use tables?



    The original problem asks for the length of the three dimensional curve given by r(t) = (2t, ln(t), t^2)

    The answer in the back of the book is L = 14 + ln(4)

    I set the arc length forumla up and arrived to this integral. Using the t89, I confirmed my setup results in the correct answer. However, I do not know how to arrive to that answer.
    Quote Originally Posted by skeeter View Post
    is this an arc length problem? if so, what is the original function?

    sdfsd
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  4. #4
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    integral should be ...

    S = \int_1^4 \sqrt{4 + \frac{1}{t^2} + 4t^2} \, dt<br />

    note that ...

    4t^2 + 4 + \frac{1}{t^2} = \left(2t + \frac{1}{t}\right)^2

    can you finish?
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  5. #5
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    Quote Originally Posted by skeeter View Post
    integral should be ...

    S = \int_1^4 \sqrt{4 + \frac{1}{t^2} + 4t^2} \, dt<br />

    note that ...

    4t^2 + 4 + \frac{1}{t^2} = \left(2t + \frac{1}{t}\right)^2

    can you finish?
    That's the same integral I wrote just with t^-2 instead of 1/t^2(the same thing). Nice factoring. Yes, I can solve that.
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  6. #6
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    Quote Originally Posted by 1005 View Post
    That's the same integral I wrote just with t^-2 instead of 1/t^2(the same thing). Nice factoring. Yes, I can solve that.
    No you left out the 4 in front of t^2 making the integral much more difficult
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