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Average value

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the average value of f(x,y)=sin(x+y) over the rectangle 0 ≤ x ≤ pi, 0 ≤ y ≤ pi.

    2. Relevant equations
    I need to know if this is the right answer please.


    3. The attempt at a solution
    I got 1/pi as a answer.
     
  2. jcsd
  3. Nov 3, 2009 #2

    Mark44

    Staff: Mentor

    You're not going to make us work this out, are you? Show us how you got it.
     
  4. Nov 3, 2009 #3
    well i dont no how to show the symbols
     
  5. Nov 3, 2009 #4

    Mark44

    Staff: Mentor

    Just click on the integral below, and a window will open with the LaTeX code. All you need to do is fill in where there are question marks.
    For pi, use \pi. A is the area of your region. If you want to switch the order of integration, change the bottom limit from x to y and vice-versa, and switch the dx and dy.
    [tex]\frac{1}{A}\int_{x = ?}^? \int_{y = ?}^? sin(x + y) dy~dx[/tex]
     
  6. Nov 3, 2009 #5
    \frac{1}{A}\int_{x = 0}^\pi \int_{y = 0}^\pi sin(x + y) dy~dx
     
  7. Nov 3, 2009 #6
    yeah i cant figure out how to make it look like yours. It just stays in the formula form
     
  8. Nov 3, 2009 #7

    Mark44

    Staff: Mentor

    It needs to be inside [ tex] [ /tex] tags (without the space I added). Like so:
    [tex]\frac{1}{A}\int_{x = 0}^\pi \int_{y = 0}^\pi sin(x + y) dy~dx[/tex]

    OK, so what do you get for the integral, and what do you get for A?
     
  9. Nov 3, 2009 #8
    For A i get pi^2
     
  10. Nov 3, 2009 #9
    the end integral i get
    1/pi^2[-cosx-x(from 0 to pi)] which i get -1/pi
     
  11. Nov 3, 2009 #10

    jbunniii

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    I tried it and got 0. Can you show us what you got after evaluating the inner integral?

    P.S. This is almost immediate if you use

    [tex]\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)[/tex]

    and break the integral into two parts, integrating the cosine term first in each case (trivial if you use symmetry).
     
    Last edited: Nov 3, 2009
  12. Nov 3, 2009 #11

    Mark44

    Staff: Mentor

    The area is correct, but I'm pretty sure you made an error in calculating the integral. How did you get -cosx -x?
     
  13. Nov 3, 2009 #12
    [math]
    \frac{1}{pi^2}\int_0^pi\int_0^pi sin(x+y)dydx=\frac{1}{pi^2}\int_0^pi sin(x)[-cosy|_0^pi] dx=\frac{1}{pi^2}\int_0^pi (sinx-1)dx=\frac{1}{pi^2}[-cosx-x|_0^pi]=\frac{-pi}{pi^2}=\frac{-1}{pi}
    [/math]
     
  14. Nov 3, 2009 #13
    [tex]
    \frac{1}{pi^2}\int_0^pi\int_0^pi sin(x+y)dydx=\frac{1}{pi^2}\int_0^pi sin(x)[-cosy|_0^pi] dx=\frac{1}{pi^2}\int_0^pi (sinx-1)dx=\frac{1}{pi^2}[-cosx-x|_0^pi]=\frac{-pi}{pi^2}=\frac{-1}{pi}
    [/tex]
     
  15. Nov 3, 2009 #14
    [tex]
    \frac{1}{\pi^2}\int_0^\pi\int_0^\pi sin(x+y)dydx=\frac{1}{\pi^2}\int_0^\pi sin(x)[-cosy|_0^\pi] dx=\frac{1}{\pi^2}\int_0^\pi (sinx-1)dx=\frac{1}{\pi^2}[-cosx-x|_0^\pi]=\frac{-\pi}{\pi^2}=\frac{-1}{\pi}
    [/tex]
     
  16. Nov 3, 2009 #15

    jbunniii

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    I'm not sure what you did in this step. Here is how I did it:

    [tex]\begin{align*}\int_{0}^{\pi} \int_{0}^{\pi} \sin(x+y) dy dx
    &= \int_0^{\pi}\int_{0}^{\pi} \sin x \cos y dy dx + \int_0^{\pi}\int_{0}^{\pi} \cos x \sin y dy dx \\
    &= \int_0^\pi \sin x \left(\int_0^\pi \cos y dy \right) dx + \int_0^\pi \sin y \left(\int_0^\pi \cos x dx \right) dy \\
    &= \int_0^\pi (\sin x) (0) dx + \int_0^\pi (\sin y) (0) dy \\
    &= 0
    \end{align*}[/tex]

    You can also do it without using the trig identity; I highly recommend doing so. (The answer is still 0.)
     
  17. Nov 3, 2009 #16
    i see i was just no where near the right answer.
     
  18. Nov 3, 2009 #17

    jbunniii

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    Wait... are you assuming here that

    [tex]\sin(x+y) = \sin(x) \sin(y)[/tex]?

    It seems that you are, but that's wrong. The inner integral should be

    [tex]\int_0^\pi \sin(x+y) dy = [-\cos(x+y)]|_{y=0}^{y=\pi} = -\cos(x+\pi) + \cos(x) = \cos(x) + \cos(x) = 2\cos(x)[/tex]

    Now do the outer integral:

    [tex]\frac{1}{\pi^2} \int_0^\pi 2 \cos(x) dx[/tex]
     
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