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Integral and area

  1. May 19, 2006 #1
    Find the area of the region

    y = x
    x + 2y = 0
    2x + y = 3

    [​IMG]

    So, I decided to integrate with respect to y.

    And got the integral from -1 to 0 of [(3-y)/2 - (-2y)]dy + the integral from 0 to 1 of [(3-y)/2 - y]dy.

    Integrating that, I got

    [3y/2 + 3y^2/4]0-1 + [3y/2 - 3y^2/4]10 = 0

    However, the answer is 3/2.

    The book integrated with respect to x, but I wanted to do it in terms of y.
    Is it possible integrating it with respect to y?
     
  2. jcsd
  3. May 19, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Okay, the lines x+ 2y= 0 and 2x+y= 3 intersect at (3, -3/2) (the lowest point on your graph) while 2x+y= 3 and y= x intersect at (1, 1). At y= 0, of course, the "left edge" changes formula. Integrate (3- y)/2 - (-2y) from -3/2 to 0 and then integrate y- (-2y) from 0 to 1.
     
  4. May 19, 2006 #3
    "Okay, the lines x+ 2y= 0 and 2x+y= 3 intersect at (3, -3/2) (the lowest point on your graph) while 2x+y= 3 and y= x intersect at (1, 1). At y= 0, of course, the "left edge" changes formula."Integrate (3- y)/2 - (-2y) from -3/2 to 0 and then integrate y- (-2y) from 0 to 1."


    Actually, they intersect at (2, -1).
    That's why I was integrating [(3- y)/2 - (-2y)] from -1 to 0
    (at x = 0, the upper function changes from y = x to 2x + y = 3.

    "Integrate (3- y)/2 - (-2y) from -3/2 to 0 and then integrate y- (-2y) from 0 to 1."

    Do you mean integrate [(3-y)/2 - y] from 0 to 1?
    I don't understand where you're getting y - (- 2y) from.
     
  5. May 20, 2006 #4
    Try redoing this step:
    [3y/2 + 3y^2/4] + [3y/2 - 3y^2/4] (substitute the upper and lower limits of integration again).

    You should be able to get the correct answer if you do it carefully.
     
  6. May 20, 2006 #5
    Thanks, I see what I did wrong now.
     
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