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Integration of a triangle

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫∫D sin(x) / x dA, where A is the triangle with vertices (0, 0),(0, 1),(1, 1).

    2. Relevant equations


    3. The attempt at a solution
    I've set up the integral, with x≤y≤x^2, 0≤x≤1 for the upper and lower bands of each integral. dxdy
    I'm not sure how to go about it though.
    I can re-write the integral to be sin(x)*x-1 and do integration by parts but I get stuck in a loop. Is there some method of integration I should be using instead?
     
  2. jcsd
  3. Apr 22, 2016 #2

    LCKurtz

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    What does ##y=x^2## have to do with this problem? Try setting it up as a dydx integral with corrected limits.
     
  4. Apr 22, 2016 #3
    As said above, you need to use the correct limits. It is not good enough to choose limits which agree at the boundaries (which yours doesn't even do), you need to have the correct domain throughout (e.g. for a triangle use linear functions).
     
  5. Apr 22, 2016 #4
    ok, so for dydx
    x≤y≤1
    0≤x≤y
     
  6. Apr 22, 2016 #5
    sin(x)/x cannot be integrated.
    If I set up the integral with the correct regions, which I think they are in the post above, I still won't be able to integrate it, no?
     
  7. Apr 22, 2016 #6

    Samy_A

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    The integration limits for the outer integral (over x), can't contain a y.
    Your regions are not correct.
    Even with correct regions, I'm not sure that this integral can be easily computed.
    Are you sure about the three points? Maybe it should be (1,0) and not (0,1).
     
  8. Apr 22, 2016 #7
    But the rightmost value of x=y and the leftmost value of x=0
    The highest value of y=1 and the lowest value of y=x
    x≤y≤1
    0≤x≤y

    Is this not saying the same thing as:
    0≤y≤1
    0≤x≤1

    ?
    I wrote in x and y into my first region because I thought it had to be more general. i.e. if we changed the points to be say, (0,0), (0,5) (5,5) we could say the region is:

    x≤y≤5
    0≤x≤y

    or do we just put the actual values in the domain? So it would be:

    0≤y≤5
    0≤x≤5
     
  9. Apr 22, 2016 #8
    I think I got the limit of integration for y, just by rewriting it. I can now compute the first integral.
    0 ≤ y ≤ x
     
  10. Apr 22, 2016 #9
    I think I got the limit of integration just by rewriting it. I can now compute the first integral.
    0 ≤ y ≤ x
    0 ≤ x ≤ 1

    ∫sin(x)/x dy
    =sin(x)y/x
    | evaluated at 0 ≤ y ≤ x
    =sin(x)
    ∫sin(x) dx
    = -cos(x)
    | evaluated at 0 ≤ x ≤ 1
    =1-cos(1)

    I'm still having trouble setting up limits of integration. We had 3 points of the triangle, and (1,1) was the highest point for y, so how come I can't say 0 ≤ y ≤ x ? I guess I could say that, but it wouldn't get me very far computing the integral. Is that really all there is to it? Looking at the problem more and trying to find equivalent statements?
     
  11. Apr 22, 2016 #10

    Samy_A

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    Don't guess the integration limits. Draw your triangle, and try to deduce them.
    This is your triangle (if the points are indeed as you posted in post #1):
    triangle.jpg

    Since you first integrate over y, fix x, and determine what values of y are relevant. That will give you the limits for y, and these limits can depend on x.
    Then determine the limits for x. These can't depend on y.
     
  12. Apr 22, 2016 #11

    Samy_A

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    These limits correspond to the triangle defined by the points (0,0), (1,0), (1,1). Not the points you stated in post #1.
     
  13. Apr 22, 2016 #12
    Damn! The points are as stated in post #1
     
  14. Apr 22, 2016 #13
    Argh! Just checked my email -- teacher said to correct the point to (1,0). ARGH!

    Thanks again for your help, Samy_A!
     
  15. Apr 22, 2016 #14

    Samy_A

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    That makes more sense. Glad you sorted this out.
     
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