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Math Help - help evaluating a double integral

  1. #1
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    help evaluating a double integral

    heres the integral

    x is between 0 and 2
    y is between 0 and 5x

    I need to find the exact value of the double integral 5x^2e^(xy)

    I'm not really sure how to do this so any help would be greatly appreciated. Thanks in advance.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ryu991 View Post
    heres the integral

    x is between 0 and 2
    y is between 0 and 5x

    I need to find the exact value of the double integral 5x^2e^(xy)

    I'm not really sure how to do this so any help would be greatly appreciated. Thanks in advance.
    \int_0^2\int_0^{5x}5x^2e^{xy}\,dy\,dx.

    Where are you stuck in the calculation? If you showed us any attempt you made at the problem, it would give us a better idea of how to best help you.
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  3. #3
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    <br />
\int_0^2\int_0^{5x}5x^2e^{xy}\,dy\,dx<br />

    <br />
=\int_0^2 5x^2 [\int_0^{5x}e^{xy}\,dy]\,dx<br />

    <br />
=\int_0^2 5x^2 [\frac{e^{xy}}{x}]|_0^{5x}\,dx<br />

    <br /> <br />
=\int_0^2 5x [e^{xy}]|_0^{5x}\,dx<br />

    <br />
=\int_0^2 5x (e^{5x^2}-1)\,dx<br />

    <br />
=\int_0^2 5x e^{5x^2}\,dx - \int_0^2 5x \,dx<br />

    <br />
=\frac{1}{2}\int_0^2  e^{5x^2}\,d{(5x^2)} - \int_0^2 5x \,dx<br />

    Get the idea?
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  4. #4
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    Thank you, what I didn't realize was to factor out the 5x^2, I was staring at the whole integral and I had no idea how to approach it I was thinming integration by parts or something but it just did not seem right.

    Quote Originally Posted by qmech View Post
    [tex]


    <br />
=\frac{1}{2}\int_0^2  e^{5x^2}\,d{(5x^2)} - \int_0^2 5x \,dx<br />

    Get the idea?
    just a few questions about the end?

    first where does the 1/2 come from?

    second how would I integrate that expression e^(5x^2)d(5x^2)? The other half of it I know, i just have a lot of trouble integrating expressions with e.

    thanks again for helping me with this.
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  5. #5
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     1/2 \int_0^2 \! e^{5x^2}\, d(5x^2) is another way to write u-substitution where u = 5x^2.
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  6. #6
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    Quote Originally Posted by lvleph View Post
     1/2 \int_0^2 \! e^{5x^2}\, d(5x^2) is another way to write u-substitution where u = 5x^2.
    so after the u substitution I would have to evaluate

    (1/2)e^(5x^2) - (5/2)x^2 from 0 -2? making my answer be

    [(1/2)e^20 -10] - [(1/2)e^0 - 0]

    or whatever that evaluates to? or did I make a mistake along the way?
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  7. #7
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    As long as you meant \left[\frac{1}{2}e^{5x^2} - \frac{5}{2}x^2\right|^2_0 then yes.
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  8. #8
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    Quote Originally Posted by lvleph View Post
    As long as you meant (1/2)e^{5x^2} - {5/2}x^2\left|^2_0\right. then yes.
    yeah thats what I meant(I saw what the syntax error was when I quoted you) thanks for all of your help.
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