Integral calculus: plane areas in rectangular coordinates

  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. lanedance

    lanedance 3,307
    Homework Helper

    looks good to me
     
  4. SammyS

    SammyS 8,209
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes.

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