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Area of a region

  1. Mar 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

    2y=3(x^1/2)
    y=4
    2y+4x=7

    2. Relevant equations



    3. The attempt at a solution

    I decided to solve for y, since that would make it easier to graph. This got me y=(3(x^1/2)/2) and y=7/2-2x. I then graphed these, finding that the span on the x-axis would be from 0 to 2.

    I integrated the problem, getting 3-2*(2^1/2). This seems wrong. What happened? I'm probably overlooking something small...

    (By the way, Latex isn't working, so the x^1/2 things means that that number or variable is square rooted.
     
    Last edited: Mar 21, 2007
  2. jcsd
  3. Mar 22, 2007 #2
    Hang on, what function did you integrate exactly?
     
  4. Mar 22, 2007 #3
    Well, when I graphed it, I noticed that 7/2-2x was above 3(x^1/2)/2. So I integrated (7/2-2x) - (3(x^1/2)/2) with an upper limit of 2 and a lower limit of 0, and that got me 3-2*(2^1/2).

    [tex]\int_{0}^{2} (\frac{7}{2} - 2x) - (\frac{3*\sqrt{x}}{2}) dx[/tex]
     
    Last edited: Mar 22, 2007
  5. Mar 22, 2007 #4
    Okay, I've sort of figured it out. I have to solve for x in the equation, even though I solved for y in the graph. So here it is, revised...

    [tex]\int_{1.5}^{3} \frac{4*y^2}{9} - \frac{-(2*y - 7)}{4} dy[/tex]

    It's still not coming out correct, though (I'm doing an internet-generated problem and it's saying that it's wrong). I'm getting 2.5625, by the way. Am I still doing something wrong?
     
  6. Mar 22, 2007 #5
    Are you sure about " 3-2*(2^1/2)" as an answer, I can see where the 3 comes from, but I don't see how you got the square root of two there, what is the integral of the squareroot of x?
     
  7. Mar 23, 2007 #6

    HallsofIvy

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    Why do you need to solve for x rather than y? That's a crucial part of the problem.
     
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