Problem 31 - Essential Calculus

Question

(a) use a graphing utility to graph the region bounded by the graphs of the equations, \((b)\) find the area of the region, and (c) use the integration capabilities of the graphing utility to verify your results. $$ f(x)=1 /\left(1+x^{2}\right), \quad g(x)=\frac{1}{2} x^{2} $$

Step-by-Step Solution

Answer

To find the area between the curves \(f(x) = \frac{1}{1+x^2}\) and \(g(x) = \frac{1}{2} x^2\), the integral \(\int_{-1}^{1}[\frac{1}{1+x^2}-\frac{1}{2} x^2] dx\) must be evaluated. The result can be confirmed by integrating the functions within the same bounds of -1 and 1, using a graphing utility's integration feature.

Step 1: Plotting the graphs

Using a graphing utility, plot the functions \(f(x) = \frac{1}{1 + x^2}\) and \(g(x) = \frac{1}{2} x^2\). Observe where these functions intersect each other, as this will be important for the integration step.

Step 2: Find the area between the curves

The area between the two curves is given by the definite integral \(A = \int [(f(x) - g(x)] dx\) over the interval where the curves intersect. The points of intersection need to be identified by setting the two equations equal to each other and solving.

Step 3: Set the equations equal to each other

\(f(x) = g(x)\) gives \(\frac{1}{1 + x^2} = \frac{1}{2} x^2\). By cross multiplying and simplifying, one arrives at \(2 = x^4+x^2\) which is rearranged as \(x^4 + x^2 - 2 = 0\). This quadratic in \(x^2\) can be factored into \((x^2-1)(x^2+2)=0\), yielding \(x = 1, -1\) as solutions.

Step 4: Evaluate the definite integral

Evaluate \(A = \int_{-1}^{1} [(f(x) - g(x)] dx = \int_{-1}^{1}[\frac{1}{1+x^2}-\frac{1}{2} x^2] dx\). Solving this definite integral will provide the area of the region between the curves.

Step 5: Verify the results using a graphing utility

Use the integral feature of the graphing utility to evaluate the definite integral figured out in theprevious step. If done correctly, the calculated area should match up with the area obtained from graphic tool.

Key Concepts

Graphing Utilities in Calculus

When dealing with calculus problems involving curves, graphing utilities become indispensable. These tools allow you to visualize functions easily and identify crucial points such as intersections, peaks, and troughs. In the given problem, graphing the functions

• \(f(x) = \frac{1}{1 + x^2}\)
• \(g(x) = \frac{1}{2} x^2\)

helps us immediately see where these curves intersect. This visual insight is particularly vital when determining the points over which you will integrate to find areas.
By inputting these functions into your graphing calculator or software, you can adjust views to zoom in on the relevant intersection points. Seeing the graphs helps confirm your algebraic work and provides a clearer understanding of each function's behavior.

Area Between Curves

Finding the area between two curves involves integrating the difference between their equations over a certain interval. You first determine where the curves intersect, which defines the boundaries of integration.
For the functions

• \(f(x) = \frac{1}{1+x^2}\) and
• \(g(x) = \frac{1}{2} x^2\),

the points of intersection are calculated by setting the two functions equal, leading to solutions of \(x = 1\) and \(x = -1\).
These points become the limits of integration. The area can be represented as \[A = \int_{-1}^{1} \left[f(x) - g(x)\right] \, dx = \int_{-1}^{1} \left[\frac{1}{1+x^2} - \frac{1}{2}x^2\right] \, dx.\]
By solving this definite integral, you compute the exact area enclosed by the graphs.

Polynomial Equations Solving

Solving polynomial equations is essential in finding intersection points between curves. In this exercise, equating \[\frac{1}{1 + x^2} = \frac{1}{2} x^2\]leads to the polynomial equation \[x^4 + x^2 - 2 = 0.\]
This equation can be seen as a quadratic in terms of \(x^2\). Factoring it gives \[(x^2 - 1)(x^2 + 2) = 0.\]
Solving these factors, we find \(x^2 = 1\) or \(x^2 = -2\). Since \(x^2 = -2\) does not yield real solutions, we focus on the real solutions \(x = 1\) and \(x = -1\).
Recognizing this allows us to determine the interval, from \(-1\) to \(1\), necessary for integration.