Results 1 to 6 of 6

Math Help - Double Integral Question

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    8

    Double Integral Question

    I am trying to evaluate the following double integral, by changing the order of integration first.

    cosh(x^2).dx.dy for 3y < or equal to: x < or equal to 3
    and 0 < or equal to: y < or equal to 1

    I changed the order of integration to get:

    cosh(x^2).dy.dx for x/3 < or equal to: y < or equal to 1
    and 0 < or equal to: x < or equal to 3

    From here I used the fact that cosh(x^2) = 1/2e^(x^2) + 1/2e^(-x^2)

    So I have:

    1/2e^(x^2) + 1/2e^(-x^2).dy.dx
    for x/3 < or equal to: y < or equal to 1
    and 0 < or equal to: x < or equal to 3

    From here I integrated and ended up with an answer of 0.

    I'm not sure if the approach I used was correct. can i use cosh(x^2) = 1/2e^(x^2) + 1/2e^(-x^2) for the double integration.

    Any advice on whether I went about this the right way would be greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,396
    Thanks
    1847
    So you had \int_{y=0}^3\int_{x= 3y}^3 cosh(x^2)dx dy and have changed it to
    \int_{x= 0}^1\int_{y= x/3}^1 cosh(x^2) dy dx.

    That conversion is valid.

    However, when you integrate with respect to y, you will have
    \int_{x=0}^1 \left[ y cosh(x^2)\right]_{y= x/3}^1 y cosh(x^2) dx
    = \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx.

    You certainly can make the substitution u= x^2 in the second integral because then du= 2x dx so \frac{1}{2}du= xdx. But I see no good way to integrate cosh(x^2)= \frac{e^{x^2}+ e^{-x^2}}{2}.

    How did you do that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    8
    Thank you soo much for the reply and help. I basically just split it up into [e*(x^2)]/2 + [e*-(x^2)]/2, initially to try and integrate.

    After using the way you suggested and evaluating it I got 0 again.

    Not sure if I correctly went about the rest of the integration to get 0 though.

    But from integrating cosh(x^2).dx for x between 0 and 3 i got:

    First part: [sinh(x^2) / 2x] for x=0 and x=3, which led me to sinh(9)/6 - 0.

    Second part: From integating (x/3)cosh(x^2).dx for x between 0 and 3 i got (1/6)sinh(9) - (1/6)sinh(0) = (1/6)sinh(9) - 0 = (1/6)sinh(9).

    Hence subtracting the second integral from the first gives me 0 again.

    Does that look right? Thanks again!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by HallsofIvy View Post
    So you had \int_{y=0}^3\int_{x= 3y}^3 cosh(x^2)dx dy and have changed it to
    \int_{x= 0}^1\int_{y= x/3}^1 cosh(x^2) dy dx.

    That conversion is valid.

    However, when you integrate with respect to y, you will have
    \int_{x=0}^1 \left[ y cosh(x^2)\right]_{y= x/3}^1 y cosh(x^2) dx
    = \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx.

    You certainly can make the substitution u= x^2 in the second integral because then du= 2x dx so \frac{1}{2}du= xdx. But I see no good way to integrate cosh(x^2)= \frac{e^{x^2}+ e^{-x^2}}{2}.

    How did you do that?
    I think it might be sufficient to label  \int \frac{e^{x^2}+ e^{-x^2}}{2} as  \frac{erf(-x) + erf(x)}{2} where erf(x) is defined as  erf(x) = \int e^{-x^2} dx Note: http://mathworld.wolfram.com/Erf.html


    Thank you soo much for the reply and help. I basically just split it up into [e*(x^2)]/2 + [e*-(x^2)]/2, initially to try and integrate.

    After using the way you suggested and evaluating it I got 0 again.

    Not sure if I correctly went about the rest of the integration to get 0 though.

    But from integrating cosh(x^2).dx for x between 0 and 3 i got:

    First part: [sinh(x^2) / 2x] for x=0 and x=3, which led me to sinh(9)/6 - 0.

    Second part: From integating (x/3)cosh(x^2).dx for x between 0 and 3 i got (1/6)sinh(9) - (1/6)sinh(0) = (1/6)sinh(9) - 0 = (1/6)sinh(9).

    Hence subtracting the second integral from the first gives me 0 again.

    Does that look right? Thanks again!
    I don't think is is correct. You're saying that

     \int cosh(x^2) = \frac{sinh(x^2)}{2x}

    But the derivative of  \frac{sinh(x^2)}{2x} is  \frac{ cosh(x^2)x^{-1} - sinh(x^2)x^{-2} }{2}

    In fact the parts of  cosh(x^2) is a well known error function in statistics, but does not have an elementary anti-derivative. I think it would follow that  cosh(x^2) does not either.
    Last edited by AllanCuz; May 19th 2010 at 06:39 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2010
    Posts
    8
    But the derivative of is ......

    I see how you did this, so yep my anti-derivative must have been wrong. So you are saying that cosh(x^2) doesn't have a specific anti-derivative at all?
    I wasnt expecting that!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,396
    Thanks
    1847
    Well, you knew that \int e^{x^2}dx and \int e^{-x^2}dx, individually, had no elementary anti-derivatives, didn't you?

    It is not that they, and cosh(x^2) and sinh(x^2) "have no specific anti-derivatives". Of course, the have anti-derivatives but they cannot be written in terms of elementary functions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. double integral question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 28th 2010, 08:04 PM
  2. Double Integral and Triple Integral question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 3rd 2010, 01:47 PM
  3. double integral question 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 23rd 2009, 04:10 PM
  4. double integral question 3
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 23rd 2009, 07:54 AM
  5. a double integral question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 1st 2007, 02:43 PM

Search Tags


/mathhelpforum @mathhelpforum