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Integral calculus: plane areas in rectangular coordinates

by delapcsoncruz
Tags: areas, calculus, coordinates, integral, plane, rectangular
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delapcsoncruz
#1
Feb28-12, 07:21 AM
P: 20
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?
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#2
Feb28-12, 09:53 AM
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P: 3,307
looks good to me
SammyS
#3
Feb28-12, 09:58 AM
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Quote Quote by delapcsoncruz View Post
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?
Yes.

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


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