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Math Help - An Integral

  1. #1
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    An Integral

    It's been a while since I did integrals, and i'm having trouble with this problem.

    √( (2t)^2 + (t^-2) + (4) ) And the integral goes from 1 to e

    Is there an easier way I can simplify the stuff under the square root, in order to make this less messy? Or do I have to go straight to substitution? And if so, whats the best way to start? Thanks for your time.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by zellster87 View Post
    It's been a while since I did integrals, and i'm having trouble with this problem.

    √( (2t)^2 + (t^-2) + (4) ) And the integral goes from 1 to e

    Is there an easier way I can simplify the stuff under the square root, in order to make this less messy? Or do I have to go straight to substitution? And if so, whats the best way to start? Thanks for your time.
    \int(2t)^2+t^{-2}+4dt

    simplifying to

    \int{4t^2+t^{-2}+4dt}=\frac{4}{3}t^3-t^{-1}+4t+C
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zellster87 View Post
    It's been a while since I did integrals, and i'm having trouble with this problem.

    √( (2t)^2 + (t^-2) + (4) ) And the integral goes from 1 to e

    Is there an easier way I can simplify the stuff under the square root, in order to make this less messy? Or do I have to go straight to substitution? And if so, whats the best way to start? Thanks for your time.
    Start by noting that
    4t^2 + \frac{1}{t^2} + 4
    is a perfect square.

    Hint: It is of the form
    \left ( at + \frac{b}{t} \right )^2

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int(2t)^2+t^{-2}+4dt

    simplifying to

    \int{4t^2+t^{-2}+4dt}=\frac{4}{3}t^3-t^{-1}+4t+C
    You left off the square root.

    -Dan
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  5. #5
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    Not big deal, since \sqrt{4t^{2}+\frac{1}{t^{2}}+4}=\frac{\sqrt{4t^{4}  +4t^{2}+1}}{t}=\frac{2t^{2}+1}{t}.

    Mathstud28 misread the problem.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by topsquark View Post
    You left off the square root.

    -Dan
    where is there a...oh...sorry...its hard to see those little things
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  7. #7
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    first of all, the 1st term inside the root is it (2t)^2 or just 2t^2, if there is a bracket round 2t then you have to make it to 4t^2

    edit: wait, my mistake I'm just being stupid lol
    Last edited by Ihatemccormick; June 2nd 2008 at 05:10 PM. Reason: error
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  8. #8
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    Quote Originally Posted by topsquark View Post
    Start by noting that
    4t^2 + \frac{1}{t^2} + 4
    is a perfect square.

    Hint: It is of the form
    \left ( at + \frac{b}{t} \right )^2

    -Dan
    Thanks for the quick replies guy, I really appreciate it.

    Now when I take the integral of (2t^2 + 1) / t from 1 to e

    I simplified it to ∫ 2t^2 / t + 1/t dt -----> ∫2t + t^-1 dt

    But I have a problem with the "t^-1" term when I take the anti-derivative of it. It comes out as t^0 / 0 which cant be right. Did i simplify this wrong?

    Thanks
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zellster87 View Post
    Thanks for the quick replies guy, I really appreciate it.

    Now when I take the integral of (2t^2 + 1) / t from 1 to e

    I simplified it to ∫ 2t^2 / t + 1/t dt -----> ∫2t + t^-1 dt

    But I have a problem with the "t^-1" term when I take the anti-derivative of it. It comes out as t^0 / 0 which cant be right. Did i simplify this wrong?

    Thanks
    \int \frac{dt}{t} = ln|t| + C

    -Dan
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  10. #10
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    Woah, I really am rusty. Thanks for you help guys.
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