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Integral calculus: plane areas in rectangular coordinates

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the area between y= 1/(x2+1) and the x-axis, from x=0 to x=1


    3. The attempt at a solution

    so when x=0, y=1
    and when x=1, y=1/2

    next i plot the points, so the intersection of the given equation is (0,1) and (1,1/2)
    Yh= Y-higher= 1/(x2+1)
    Yl= Y-lower= 0
    the strip is vertical, so the length (L) = (Yh-Yl) and the width (W) is dx


    dA=LW
    dA=(Yh-Yl)dx
    dA=(1/(x2+1))dx
    A=∫from 0-1 dx/(x2+1)
    A=Arctan x from 0-1
    A=Arctan 1 -Arctan 0
    A=pi/4 sq.units


    was my solution correct?
     
  2. jcsd
  3. Feb 28, 2012 #2

    lanedance

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    looks good to me
     
  4. Feb 28, 2012 #3

    SammyS

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    Yes.

    You went through a lot of steps to get the answer.
     
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