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Math Help - Integration of a square root

  1. #1
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    Integration of a square root

    While working on finding the arc length of the parametrized curve g(t)=(logt,2t,t^2) where t is in [1,e]. i got to an integral that i couldn't manage to integrate. Here is the work so far leading up to the problematic integral so that it may be checked for errors.

    arc length =\int_{a}^{b} |g'(t)| dt by definition

    =\int_{1}^{e} |( 1/t,2,2t)| dt by taking the partials of g

    =\int_{1}^{e} \sqrt{(1/t)^2+2^2+(2t)^2} dt by taking the length

    =\int_{1}^{e} \sqrt{1/t^2+4+4t^2} dt by simplification

    = ...? by integration

    I'm thinking it has something to do with simplifying the inside of the sqrt some more, or maybe a trig substitution (ugh!)

    By the way, the answer is e^2

    (Appologies if this is posted in the wrong thread as it is coming from an analysis course, but it really is only the integral that i am needing help with, so i see it fit that i place this thread here. Thank you)
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  2. #2
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    Quote Originally Posted by DonnMega View Post
    While working on finding the arc length of the parametrized curve g(t)=(logt,2t,t^2) where t is in [1,e]. i got to an integral that i couldn't manage to integrate. Here is the work so far leading up to the problematic integral so that it may be checked for errors.

    arc length =\int_{a}^{b} |g'(t)| dt by definition

    =\int_{1}^{e} |( 1/t,2,2t)| dt by taking the partials of g

    =\int_{1}^{e} \sqrt{(1/t)^2+2^2+(2t)^2} dt by taking the length

    =\int_{1}^{e} \sqrt{1/t^2+4+4t^2} dt by simplification

    = ...? by integration

    I'm thinking it has something to do with simplifying the inside of the sqrt some more, or maybe a trig substitution (ugh!)

    By the way, the answer is e^2
    does this help?

    \frac{1}{t^2} + 4 + 4t^2 = \left(\frac{1}{t} + 2t\right)^2
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  3. #3
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    That does....a lot. As they always say, calculus is easy, it's the algebra that'll get ya......
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