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Math Help - Double integrals problem

  1. #1
    Senior Member chella182's Avatar
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    Double integrals problem

    Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
    \int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx  dy
    ...and I'm told to solve it by changing the order of integration. I get:
    \int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx
    I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate 2x^5y\cos{(xy^2)} with respect to y, taking x as a constant.

    Can anyone help?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by chella182 View Post
    Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
    \int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx  dy
    ...and I'm told to solve it by changing the order of integration. I get:
    \int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx
    I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate 2x^5y\cos{(xy^2)} with respect to y, taking x as a constant.


    Can anyone help?
    I didn't check your change of order of inegration but use this sub to evaluate the integral

    u=xy^2 \implies du=2xydy

    and we get

    \int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx

    \int_{0}^{1}\int x^4\cos(u)dudx
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by chella182 View Post
    Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
    \int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx  dy
    ...and I'm told to solve it by changing the order of integration. I get:
    \int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx
    I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate 2x^5y\cos{(xy^2)} with respect to y, taking x as a constant.

    Can anyone help?
    Your reordering of the integration is correct. Proceeding from there, let u=y^2. Thus du = 2y\,dy

    You now have \int_0^1\int_0^{x^2}x^5\cos(xu)\,du = \frac{1}{x}\cdot x^5\sin(xu) = \left[x^4\sin(xy^2)\right]_0^{x^2} = x^4\sin(x^5)

    So we have \int_0^1 x^4\sin(x^5)\,dx

    Let u=x^5 and du=5x^4. Now you have:

    \frac{1}{5}\int_0^1\sin(u)\,du = -\frac{1}{5}\cos(u) = \left[-\frac{1}{5}\cos(x^5)\right]_0^1 = \boxed{\frac{1}{5}-\frac{1}{5}\cos(1)}
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  4. #4
    Senior Member chella182's Avatar
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    I totally don't understand the second line of your working, sorry where does the 2y disappear to? And the x^4 in the second integral? And I don't get where the \frac{1}{x} has come from
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  5. #5
    Senior Member Pinkk's Avatar
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    2y is the derivative of y^{2}. All he did was a u-substitution.
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  6. #6
    Senior Member chella182's Avatar
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    No, I know that, where does it disappear to though?
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  7. #7
    Senior Member Pinkk's Avatar
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    Into the du. It never "disappears", it's just part of his new variables u and du. And the \frac{1}{x} is part of the integration of cos(xy).

    The antiderivative of cos(xy) where x is treated as a variable is \frac{sin(xy)}{x}.
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  8. #8
    Senior Member chella182's Avatar
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    So because the 2y appears in the du it's not in the problem anymore? :S I'm well confused.

    I get where the \frac{1}{x} comes from now though, cheers.
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  9. #9
    Senior Member Pinkk's Avatar
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    It's just u-substitution.

    Just replace the y^{2} with a u and 2ydy with a du. Don't let the fact that it's a double integral throw you off. It's just a straightforward u-substitution.
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  10. #10
    Senior Member chella182's Avatar
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    I know I just... don't get it :S urgh. I'll go see my lecturer tomorrow if I get chance.
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  11. #11
    Senior Member Pinkk's Avatar
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    Let's look at a single integral:

    \int 2ycos(y^{2})\,\,dy

    Let u=y^{2} and du=2ydy.

    Now just substitute:

    \int cos(u)\,\,du

    Now just integrate:

    \int cos(u)\,\,du = sin(u)

    Now just resubstitute for u=y^{2}:

    sin(y^{2}).


    Take the derivative to clarify:

    \frac{d}{dy}sin(y^{2})=cos(y^{2})\cdot \frac{d}{dy}y^{2} = 2ycos(y^{2})



    All you're doing is substituting one type of variable for another. The actual function isn't changed, just the way it's being represented.
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  12. #12
    Senior Member chella182's Avatar
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    Ohhh right, I get it. All you needed to do to explain was say dy=\frac{du}{2y} so the 2y's cancel
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  13. #13
    Senior Member Pinkk's Avatar
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    It hasn't disappeared. du=2ydy

    I'm just representing 2ydy as du so I can have an easier integral to evaluate.


    And yes, you can think of it that way if you want. Whatever clicks with you.
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  14. #14
    Senior Member chella182's Avatar
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    It has cancelled out therefore it's disappeared in my mind. I do understand the whole thing, all I didn't get was why that disappeared 'cause no one was telling me where it'd gone. Although this does seem rather more complicated than examples we've done in lectures :S
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