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-   -   Very difficult math equation (https://www.askmehelpdesk.com/showthread.php?t=392591)

  • Sep 1, 2009, 03:27 PM
    survivorboi
    1 Attachment(s)
    very difficult math equation
    Prove it!
  • Sep 2, 2009, 07:21 AM
    ebaines
    Quote:
    Originally Posted by survivorboi View Post
    Prove it!
    Where did you get this equation from? I think I can show graphically that the sum of the two integrals is precisely 9, so I think the inequality (less than sign) on the left is in error.
  • Sep 2, 2009, 08:32 AM
    galactus
    Running them through my TI, I see the integrals add to 9.00004286557

    Which is between 9 and 9.001

    Try using

    Suppose f is integrable on [a,b] and
  • Sep 2, 2009, 01:17 PM
    ebaines
    1 Attachment(s)

    OK, here's a method - I'm going to do a bit of "hand waving" here but I trust it'll be reasonably clear...

    First, note that the two integrals are functions that are inverses of each other. That is, if



    then



    Hence we know that the area between the lower curve and the x axis is equal to the area between the y axis and upper curve y. These areas are shown as A and A' in the figure. This also means that the areas between each curve and the diagonal are equal -- I have marked these areas as B and B' in the figure. Similarly, the two areas marked C and C' are identical. Finally, there is a little tail out at the end of the upper curve, labelled D.

    Now, the integral for the lower curve from x = 1 to x = 3 is equal to the area A. This area is also equal to the area under the diagonal minus B minus C. Since the area under the diagonal is simply a triangle, we know its area is 1/2*3*3 = 4.5.

    The integral for the upper curve is equal to the area under the diagonal, plus the areas B' plus C' + D.

    Add the two integral together and you have (4.5 - B - C)+(4.5 + B'+ C' +D). Or in other words, the sum of the integrals is simply 9 +D.

    The area D is slightly smaller than a triangle, and with a little work you can show that the length of the legs of the triangle are approximately 0.0092 and 0.0093, so the area D is 1/2* 0.0092 * 0.0093 = .000043. Hence the total area is greater than 9, and less than 9.000043. QED.
  • Sep 2, 2009, 01:59 PM
    jcaron2
    Quote:
    Originally Posted by ebaines View Post
    The area D is slightly smaller than a triangle, and with a little work you can show that the length of the legs of the triangle are approximately 0.0092 and 0.0093, so the area D is 1/2* 0.0092 * 0.0093 = .000043. Hence the total area is greater than 9, and less than 9.000043. QED.
    Brilliant! And so simple now that you point it out. Great job!
  • Sep 3, 2009, 04:16 AM
    galactus
    I wish they would adjust the 'rate this answer'. Every time I try to give feedback, it tells me I can't. I am just going to quit altogether. :(
    I tried ebaines. Nice proof.
  • Sep 13, 2009, 07:51 AM
    survivorboi
    Quote:
    Originally Posted by ebaines View Post
    Where did you get this equation from? I think I can show graphically that the sum of the two integrals is precisely 9, so I think the inequality (less than sign) on the left is in error.
    It was one of the Math olympiad questions in 1990's something/
  • Sep 13, 2009, 07:53 AM
    survivorboi
    Quote:
    Originally Posted by ebaines View Post
    Where did you get this equation from? I think I can show graphically that the sum of the two integrals is precisely 9, so I think the inequality (less than sign) on the left is in error.
    It's not an equation is it?
  • Sep 13, 2009, 08:14 AM
    Unknown008

    How could I miss this thread! I think its more appropriate name would be inequality.
  • Jan 25, 2011, 03:37 AM
    marijaISKRA
    A little les than 9 ;D

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