Problem 13 - Calculus Early Transcendentals

Question

Sketch the region and find its area. The region bounded by \(y=\frac{2}{1+x^{2}}\) and \(y=1\)

Step-by-Step Solution

Answer

Answer: The area between the two functions is 0.

Step 1: Identify the bounded region

First, we need to identify the region bounded by the two given functions: \(y=\frac{2}{1+x^2}\) and \(y=1\). By observation, we can see that the region will be formed where the first function lies below the second function.

Step 2: Determine the intersection points

To find the intersection points, we need to set both functions equal to each other and solve for \(x\): \[\frac{2}{1+x^2} = 1\] To solve for \(x\), we'll first manipulate the equation to get rid of the fraction: \[2 = 1 + x^2\] Now, solve for \(x\): \[x^2 = 1\] \[x = \pm 1\] Thus, the intersection points occur at \(x= -1\) and \(x=1\).

Step 3: Set up an integral to calculate the area

Since the first function lies below the second function, we can set up an integral from \(x=-1\) to \(x=1\) to calculate the area between the two functions. The integral will be the difference between the two functions: \[\text{Area} = \int_{-1}^1 (1 - \frac{2}{1+x^2})dx\]

Step 4: Evaluate the integral

Now, we evaluate the integral: \[\text{Area} = \int_{-1}^1 (1 - \frac{2}{1+x^2})dx = \int_{-1}^1 \frac{1+x^2-2}{1+x^2} dx = \int_{-1}^1 \frac{x^2-1}{1+x^2} dx\] This integral appears symmetrical, and since the function \(\frac{x^2-1}{1+x^2}\) is an odd function, the integral of an odd function over a symmetrical interval is 0. Thus: \[\text{Area} = 0\] In this case, the area between the two functions is 0.

Key Concepts

Integral calculus

Integral calculus is a fundamental branch of mathematics that deals with the concept of finding areas under curves, among other things. When working with area between curves, integral calculus is utilized to find the region's measure, especially when descriptions are given by functions. To calculate the area, set up an integral that reflects the difference between the upper and lower functions over the region of interest. In this exercise, since the region is defined by two functions, finding this area involves setting up an integral for the difference between the bounding functions over a specified range. Here's a step-by-step approach:

• Identify the upper curve and the lower curve over the region you are interested in.
• The difference between the two functions will be the integrand, as we are interested in the vertical distance between them.
• Integrate this difference within the limits set by the intersection points of the functions.

This systematic application of integral calculus gives us the area between the curves, which in many cases represents a distinct geometric space bounded by the intersecting curves.

Symmetrical functions

Symmetry in functions can greatly simplify integral calculations and problem-solving. A function is considered symmetrical if it exhibits the same characteristics about a certain axis or point. Common types include even and odd function symmetries:

• Even Functions: These are symmetrical about the y-axis, meaning that if you fold the function along the y-axis, both sides would match. Mathematically, a function is even if: \[f(x) = f(-x)\]
• Odd Functions: These are symmetric about the origin such that rotating 180 degrees results in the same function. A function is odd if: \[f(-x) = -f(x)\]

Symmetry plays a crucial role because for odd functions integrated over an interval that is symmetric about the y-axis, the result is zero. This property helps simplify calculations by eliminating need for detailed integration processes. In the given exercise, the integrand emerges as an expression of an odd function constrained within equally sized and opposite limits, which simplifies the solution considerably.

Intersection points

Intersection points are locations where two or more curves meet or cross each other. Finding these points is critical in problems involving areas between curves as they dictate the limits of integrals. To find intersection points, set the equations of the curves equal to each other and solve for the variable in question. This will give you the \(x\)-values where the functions intersect. In this exercise, setting \(y = \frac{2}{1+x^2}\) equal to \(y =1\) provided the points at which the curves intersect. Solving \[\frac{2}{1+x^2} = 1\] shows that the intersection occurs at \(x = 1\) and \(x = -1\). Once the intersection points are found:

• They serve as the bounds of the integral when calculating the area between curves.
• These points ensure the accurate measurement of the total region enclosed by the curves over that particular interval.

Finding and using intersection points correctly ensures that the integral captures the entire desired area without left-out spaces or unwanted extras.