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Math Help - Another Integral Question

  1. #1
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    Another Integral Question

    <br />
\int \frac{{{2x}^2+2}}{x^2}dx<br />

    There's an upper limit of 4 and lower limit of 1.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Lolcats View Post
    <br />
\int \frac{{{2x}^2+2}}{x^2}dx<br />

    There's an upper limit of 4 and lower limit of 1.

    Note that \frac{2x^2+2}{x^2}=2+2x^{-2}

    So, can you find \int_1^4(2+2x^{-2})dx ?
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  3. #3
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    Quote Originally Posted by Lolcats View Post
    <br />
\int \frac{{{2x}^2+2}}{x^2}<br />

    There's an upper limit of 4 and lower limit of 1.
    Hello .
    \frac{{{2x}^2+2}}{x^2}=\frac{2x^2}{x^2}+\frac{2}{x  ^2}
    =2+2x^{-2}
    Now,
    \int_{1}^{4} \frac{{{2x}^2+2}}{x^2} dx
    =2\int_{1}^{4}dx+2\int_{1}^{4}x^{-2}dx.

    Can you complete it ?
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  4. #4
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    Jeez these are easy problems I'm screwing up. Thanks !
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  5. #5
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    Sorry to be a bother, the answer should be 4 shouldn't it?
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  6. #6
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    Quote Originally Posted by Lolcats View Post
    Sorry to be a bother, the answer should be 4 shouldn't it?
    No.
    See this:
    integrate [ 2 + 2 x^(-2) ] from x=1 to x=4 - Wolfram|Alpha

    Please show us your work, to see your mistakes.
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Lolcats View Post
    Sorry to be a bother, the answer should be 4 shouldn't it?
    no

    15/2
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  8. #8
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    K well i'd integrate and get 2x-(2/x), then insert the upper limit so:
    2(4) + 2/4 = 4 then minus that from 2(1) + 2/1 = 0

    Sorry i'm replying slow I have a seperated shoulder atm and can only type with 1 hand.
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  9. #9
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    Quote Originally Posted by Lolcats View Post
    K well i'd integrate and get 2x-(2/x), then insert the upper limit so:
    2(4) + 2/4 = 4 then minus that from 2(1) + 2/1 = 0

    Sorry i'm replying slow I have a seperated shoulder atm and can only type with 1 hand.

    See the red.
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  10. #10
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    Wops.. thats embarrasing. Thanks again.
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