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Thread: integration problem

  1. #1
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    integration problem

    1. $\displaystyle \int_{\bold{R}} xy \ dx \ dy, \ \ \ R = \text{triangle with vertices at} \ (1,0), (2,2) \ \text{and} \ (1,2) $.

    Is this equaled to $\displaystyle \int\limits_{1}^{2} \int\limits_{2x-2}^{2} xy \ dy \ dx = -\frac{7}{12}$?


    2. $\displaystyle \int_{\bold{R}} e^{xy} \ dx \ dy, \ \ \ R = \{(x,y) \ | \ 0 \leq x \leq 1 + (\log y)/y \}, 2 \leq y \leq 3 \} $.

    Here I am assuming that $\displaystyle \log y = \ln y $.

    So $\displaystyle 0 \leq x \leq 1 + (\log 3)/3 $ and $\displaystyle 2 \leq y \leq 3 $.

    So is the integral: $\displaystyle \int\limits_{0}^{1 + (\log 3)/3} \int\limits_{2}^{3} e^{xy} \ dy \ dx $?
    Last edited by heathrowjohnny; Feb 23rd 2008 at 03:53 PM.
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  2. #2
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    The lower limit in the first one should be 2x-2.

    Are you finding area?. Then it shouldn't be negative.
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  3. #3
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    Quote Originally Posted by heathrowjohnny View Post
    1. $\displaystyle \int_{\bold{R}} xy \ dx \ dy, \ \ \ R = \text{triangle with vertices at} \ (1,0), (2,2) \ \text{and} \ (1,2) $.

    Is this equaled to $\displaystyle \int\limits_{1}^{2} \int\limits_{2x-2}^{2} xy \ dy \ dx = -\frac{7}{12}$?
    You set up the double integral correctly, but its value is $\displaystyle \frac{11}6.$

    Quote Originally Posted by heathrowjohnny View Post
    2. $\displaystyle \int_{\bold{R}} e^{xy} \ dx \ dy, \ \ \ R = \{(x,y) \ | \ 0 \leq x \leq 1 + (\log y)/y \}, 2 \leq y \leq 3 \} $.
    As it's written, it's just

    $\displaystyle \int_2^3 {\int_0^{1 + \frac{{\ln y}}
    {y}} {e^{xy} \,dx} \,dy} .$
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