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Math Help - Integration

  1. #1
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    Integration

    Find the value of p if \int_{1}^{\infty}  \frac{2}{x^2},dx=3 \int_{1}^{p}  \frac{2}{x^2}dx



    many thanks!

    and where can i find the codes for the [tex] function?
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  2. #2
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    Quote Originally Posted by BabyMilo View Post
    Find the value of p if \int_{1}^{\infty}  \frac{2}{x^2},dx=3 \int_{1}^{p}  \frac{2}{x^2}dx



    many thanks!

    and where can i find the codes for the [tex] function?
    what value have you determined for

    \int_{1}^{\infty}  \frac{2}{x^2} \, dx

    ?
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  3. #3
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    Quote Originally Posted by BabyMilo View Post
    Find the value of p if \int_{1}^{\infty} \frac{2}{x^2},dx=3 \int_{1}^{p} \frac{2}{x^2}dx



    many thanks!

    and where can i find the codes for the [tex] function?

    what has you confused about the problem? Evaluate the left hand side. Then evaluate the right hand side, plugging in P as your upper limit, and solve for P
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  4. #4
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    Quote Originally Posted by skeeter View Post
    what value have you determined for

    \int_{1}^{\infty}  \frac{2}{x^2} \, dx

    ?
    2
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  5. #5
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    Quote Originally Posted by artvandalay11 View Post
    what has you confused about the problem? Evaluate the left hand side. Then evaluate the right hand side, plugging in P as your upper limit, and solve for P
    how does the 3 play a role in this equation?

    Just be easy on me as this is my first week of intergration.

    thanks for you help.

    I did:

    \int_{1}^{\infty}  \frac{2}{x^2} dx

     =[\frac{-2}{x}]_{1}^{\infty}

     =0-(-2)

     =2
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  6. #6
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    So do you divide by 3:

     \frac{2}{3} = \int_{1}^{p} \frac{2}{x^2} dx

    = [\frac{-2}{x}]_{1}^{p}

    = \frac{-2}{p}-(-2)=\frac{2}{3}

    = p=\frac{-2}{\frac{2}{3}-2}

      p = 1.5
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  7. #7
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    I thought you had to sub in the 3 in to x
    O.o
    thanks!
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  8. #8
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    lol, I love when you pose questions and then answer them, good job
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