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Math Help - double integral

  1. #1
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    double integral

    evaluate double integral...

    integral from 0 to 1 integral from 2x to 2 of e^y^2 dy dx.

    i don't know how to show that other than words...

    the part that is messing me up is evaluating the e^y^2. i think you have to maybe change the bounds and change it to dx dy but i have no idea...can any one help me reset up the integral????
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  2. #2
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    Quote Originally Posted by chris25 View Post
    evaluate double integral...

    integral from 0 to 1 integral from 2x to 2 of e^y^2 dy dx.

    i don't know how to show that other than words...

    the part that is messing me up is evaluating the e^y^2. i think you have to maybe change the bounds and change it to dx dy but i have no idea...can any one help me reset up the integral????
    2x \le y \le 2 , 0 \le x \le 1 \Leftrightarrow  0 \le 2x \le y \le 2 \Leftrightarrow 0 \le x \le \frac{y}2 \le 1


    Thus:
    \int_0^1 \int_{2x}^2 e^{y^2} \, dy\,  dx =  \int_0^2 \int_{0}^{\frac{y}{2}} e^{y^2} \, dx\,  dy
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  3. #3
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    so the new integral is what??? it says syntax error
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  4. #4
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    Read again Isomorphism's post, he just played with inequalities.

    First two determine the region where the double integral is taken, then after some algebraic manipulations he got the new region where the double integral is taken in reverse order.
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  5. #5
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    Quote Originally Posted by chris25 View Post
    so the new integral is what??? it says syntax error
    You should draw a sketch graph that shows the region of integration.

    Getting the new integral limits for reversing the order of integration is then much easier to see and it will then be very clear to you where Isomorphism's integral limits have come from.
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  6. #6
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    double integral answer?

    Is the answer for the integral 2e^4 or did I do something wrong?
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  7. #7
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    Quote Originally Posted by jme44 View Post
    Is the answer for the integral 2e^4 or did I do something wrong?
    Well I get

    \int_0^2 \int_{0}^{\frac{y}{2}} e^{y^2} \, dx\,  dy = \int_0^2 \frac{y}{2} e^{y^2}\,  dy = \frac14 e^{y^2} \bigg{|}_0^2 = \frac{e^4  - 1}{4}
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