Integral calculus: plane areas in rectangular coordinates

[delapcsoncruz]

Homework Statement

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

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?

 

[lanedance]

looks good to me

 

[SammyS]

Homework Statement

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

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.