Integral calculus: plane areas in rectangular coordinates

To find the area between the given function and the x-axis, you simply need to integrate the function from 0 to 1. In summary, the solution involves finding the indefinite integral of 1/(x^2+1), evaluating it at 0 and 1, and finding the difference between the two values. This gives the area as pi/4 square units.
  • #1
delapcsoncruz
20
0

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?
 
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  • #2
looks good to me
 
  • #3
delapcsoncruz said:

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.
 

1. What is integral calculus?

Integral calculus is a branch of mathematics that deals with finding the area under a curve or the accumulation of infinitesimal quantities. It is used to solve problems involving continuous change, such as finding the total distance traveled by an object with varying velocity.

2. What are plane areas in rectangular coordinates?

Plane areas in rectangular coordinates refer to the measurement of two-dimensional shapes, such as squares, rectangles, and triangles, on a coordinate plane. This is done by finding the length and width of the shape and multiplying them together to determine the area.

3. How is integral calculus used to find plane areas in rectangular coordinates?

Integral calculus is used to find plane areas in rectangular coordinates by finding the integral of a function that represents the boundary of the shape. This integral represents the area under the curve and can be evaluated to determine the exact area of the shape.

4. What are some real-life applications of integral calculus in finding plane areas in rectangular coordinates?

Some real-life applications of integral calculus in finding plane areas in rectangular coordinates include calculating the area of a field or garden, determining the amount of material needed for a construction project, and finding the volume of a container or storage tank.

5. What are the limitations of using integral calculus to find plane areas in rectangular coordinates?

One limitation of using integral calculus to find plane areas in rectangular coordinates is that it can only be used for shapes with straight edges. It also requires a known function that can represent the boundary of the shape, which may not always be available. Additionally, the process can be time-consuming for complex shapes.

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