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Finding the Area of a Region In Exercises \(31-36,\) (a) use a graphing utility to graph the region bounded by the graphs of the equations, (b) find the area of the region analytically, and (c) use the integration capabilities of the graphing utility to verify your results. $$ f(x)=\frac{1}{1+x^{2}}, \quad g(x)=\frac{1}{2} x^{2} $$

Short Answer

Expert verified
We find the area between the curves analytically by integrating the difference of the two functions over their intersection interval, and then verify the result using the graphing utility's integration capabilities.

Step by step solution

01

Graph the Functions

Use a graphing utility to graph the functions \(f(x)=\frac{1}{1+x^{2}}\) and \(g(x)=\frac{1}{2} x^{2}\). Identify the area between the curves where \(f(x)\) is greater than \(g(x)\).
02

Find the Points of Intersection

Solve \(f(x) = g(x)\) to find the points where the two functions intersect. We will set \(\frac{1}{1+x^{2}} = \frac{1}{2} x^{2}\) and solve for x. The solutions will be the bounds of the integral for calculating the area between the curves.
03

Integrate to Find the Area

Next, integrate the difference between the two functions from the lower intersection point to the higher intersection point, that is \(\int_{a}^{b} (f(x) - g(x)) dx\), where a and b are the lower and upper bounds found in step 2.
04

Use Graphing Utility's Integration Capability to Verify the Result

Lastly, use the graphing utility to find the area under the curve between the two points found in step 2. This should verify the analytical result obtained in step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Graphing Utility
In the digital age, integration by graphing utility is a powerful tool for visualizing and calculating the area between curves. A graphing utility, such as a graphing calculator or software like Desmos and GeoGebra, allows students to enter the functions and quickly see where they intersect and how they are positioned relative to each other.

Utilizing this technology simplifies two parts of the process: identifying the region of interest and verifying the results of an analytical integration. In our exercise, after graphing the given functions, students can adjust the viewing window to focus on the areas where one function exceeds the other, marking the region whose area we seek to find. This visual representation bolsters comprehension and serves as a precursor to more rigorous analytical methods.
Area of Region Calculus
In calculus, finding the area of a region bounded by two curves is a fundamental concept. The area calculation hinges on understanding that it represents the integral of the difference between the upper and the lower function over the interval bounded by their intersection points.

For the curves represented by the functions in our exercise, the area is the integral of \( f(x) - g(x) \) over the bounded region. This is reflective of summing slices or differential elements of area, a core idea behind integral calculus. Start by sketching the area to gain a clear picture, then proceed with the calculation by integrating the vertical strips of area between the functions along the x-axis.
Intersection Points of Functions
Identifying the intersection points of functions is crucial for determining the limits of integration when calculating the area between curves. These points are found where the functions have equal values; mathematically, this means solving the equation \( f(x) = g(x) \).

In the case of our exercise, we need to solve the equation \( \frac{1}{1+x^{2}} = \frac{1}{2} x^{2} \) to find the x-values that serve as the lower and upper bounds for the definite integral. Accurately determining these points ensures the integration is performed over the correct interval, encapsulating the entire area of interest without extraneous sections.
Definite Integrals
The concept of definite integrals is a central theme in calculus, especially for calculating the area under a curve or between curves. A definite integral is essentially the net area under a curve from one point to another, which, in many cases, represents a physical quantity such as distance or the area bounded by curves.

In our exercise, the definite integral \( \int_{a}^{b} (f(x) - g(x)) \, dx \) specifies the exact area between the functions \( f(x) \) and \( g(x) \) from point \( a \) to \( b \) (the intersection points). Comprehending the use of definite integrals is key to moving from the abstract concept of integration to real-world applications, as it encapsulizes a specific, measurable outcome: the area between curves.

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Most popular questions from this chapter

Approximating Arc Length or Surface Area In Exercises \(63-66,\) set up the definite integral for finding the indicated arc length or surface area. Then use the integration capabilities of a graphing utility to approximate the are length or surface area. (You will learn how to evaluate this type of integral in Section \(8.8 . )\) Using a Loop Consider the graph of $$y^{2}=\frac{1}{12} x(4-x)^{2}$$ shown in the figure. Find the area of the surface formed when the loop of this graph is revolved about the \(x\) -axis.

Verifying a Formula (a) Given a circular sector with radius \(L\) and central angle \(\theta\) (see figure), show that the area of the sector is given by $$S=\frac{1}{2} L^{2} \theta .$$ (b) By joining the straight-line edges of the sector in part (a), a right circular cone is formed (see figure) and the lateral surface area of the cone is the same as the area of the sector. Show that the area is \(S=\pi r L,\) where \(r\) is the radius of the base of the cone. (Hint: The arc length of the sector equals the circumference of the base of the cone.) (c) Use the result of part (b) to verify that the formula for thelateral surface area of the frustum of a cone with slant height \(L\) and radii \(r_{1}\) and \(r_{2}\) (see figure) is \(S=\pi\left(r_{1}+r_{2}\right) L .\) (Note: This formula was used to develop the integral for finding the surface area of a surface of revolution.)

Modeling Data The circumference \(C\) (in inches) of a vase is measured at three-inch intervals starting at its base. The measurements are shown in the table, where \(y\) is the vertical distance in inches from the base. $$ \begin{array}{|c|c|c|c|c|c|c|}\hline y & {0} & {3} & {6} & {9} & {12} & {15} & {18} \\ \hline C & {50} & {65.5} & {70} & {66} & {58} & {51} & {48} \\\ \hline\end{array} $$ (a) Use the data to approximate the volume of the vase by summing the volumes of approximating disks. (b) Use the data to approximate the outside surface area (excluding the base) of the vase by summing the outside surface areas of approximating frustums of right circular cones. (c) Use the regression capabilities of a graphing utility to find a cubic model for the points \((y, r),\) where \(r=C /(2 \pi) .\) Use the graphing utility to plot the points and graph the model. (d) Use the model in part (c) and the integration capabilities of a graphing utility to approximate the volume and outside surface area of the vase. Compare the results with your answers in parts ( a ) and (b).

Hooke's Law In Exercises \(5-10\) , use Hooke's Law to determine the variable force in the spring problem. Eighteen foot-pounds of work is required to stretch a spring 4 inches from its natural length. Find the work required to stretch the spring an additional 3 inches.

Pumping Gasoline In Exercises 23 and \(24,\) find the work done in pumping gasoline that weighs 42 pounds per cubic foot. (Hint: Evaluate one integral by a geometric formula and the other by observing that the integrand is an odd function.) A cylindrical gasoline tank 3 feet in diameter and 4 feet long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 feet above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.

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