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Chapter 7: Problem 17

Finding the Area of a Region In Exercises \(17-30,\) sketch the region bounded by the graphs of the equations and find the area of the region. $$ y=x^{2}-1, \quad y=-x+2, \quad x=0, \quad x=1 $$

Short Answer

Expert verified
The area of the region bounded by the given equations is 1 unit square.

Step by step solution

01

Sketch the Graphs

The equations given are \(y=x^{2}-1\), \(y=-x+2\), \(x=0\), and \(x=1\). The first two equations represent a parabola and a line, respectively, while \(x=0\) and \(x=1\) both represent vertical lines. By sketching these on the same graph, we can visualize the region we are interested in.
02

Find the Points of Intersection

To find where the lines intersect, we set the y-values of the equations equal to each other. Instead of calculating it symbolically, we can infer from the graph that the points of intersection are (0,1) where the line and the parabola meet, (0,-1) and (1,1) where the parabola intersects the vertical lines.
03

Calculate the Area

By finding the integrals of the equations from the lower limit (the x-value of the leftmost intersection point) to the upper limit (the x-value of the rightmost intersection point) we get areas under the curves. The absolute difference between these areas will give the area of the region. Calculate the integral of \(y=x^{2}-1\) from \(x=0\) to \(x=1\), and likewise for the equation \(y=-x+2\). Subtract the smaller result from the larger to get the required area.

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Key Concepts

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

Area Between Curves
The concept of finding the area between curves is fundamental in calculus, often used to determine the size of a region enclosed by different functions. In our exercise, we are dealing with a parabola, straight line, and two vertical lines creating a closed area. Determining the area of such a region involves calculating how much space lies between these curves. To do this:
  • Identify the boundaries, such as vertical lines or points where the curves intersect.
  • Draw the graphs to extract visual understanding. This helps to see which function lies on top over a certain interval.
  • Integrate the top function minus the bottom function within the interval determined by these boundaries.
By integrating the difference of the functions, you obtain the net area between them. A clear understanding of where the functions cross each other (the points of intersection) is crucial for setting the correct limits of integration.
Definite Integrals
Definite integrals are a way to find the area under a curve, from one point to another on the x-axis. In the context of our example, definite integrals help in calculating the area between the curves as a number鈥攁n actual value, not just a general formula. Here's how it works:
  • Set up an integral for each curve over the desired interval. This interval is bounded by the x-values from the intersection points or given limits.
  • By using definite integrals of each function, you effectively calculate the entire area beneath each curve from the start to end of the interval.
  • Finally, the difference between these areas gives the area between the curves over that specific interval.
Definite integrals are essential for precisely computing the size of regions formed by intersecting functions, as they not only account for areas above x-axis but also handle areas beneath it when needed.
Functions Intersection
Intersection points between functions are the points where their graphs touch or cross each other. These points are crucial when calculating the area between curves, as they define the limits of integration. Here's the process to determine these points:
  • Equate the two functions and solve for x. This gives the x-coordinates where the graphs intersect.
  • Plug these x-values back into the original equations to find the corresponding y-values, forming the coordinate pairs of intersection.
  • Use these points to set the limits for integration when calculating areas.
In our example, identifying intersections without hefty calculations was possible by sketching, yet in trickier problems, precise algebraic solutions ensure accuracy. Correctly identifying these points ensures you're capturing the exact region intended for measuring with integration.

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Most popular questions from this chapter

Revenue In Exercises 75 and \(76,\) two models \(R_{1}\) and \(R_{2}\) are given for revenue (in billions of dollars) for a large corporation. Both models are estimates of revenues from 2015 through 2020 , with \(t=15\) corresponding to \(2015 .\) Which model projects the greater revenue? How much more total revenue does that model project over the six-year period? $$ \begin{array}{l}{R_{1}=7.21+0.26 t+0.02 t^{2}} \\ {R_{2}=7.21+0.1 t+0.01 t^{2}}\end{array} $$

Finding Arc Length In Exercises \(17-26,\) (a) sketch the graph of the function, highlighting the part indicated by the given interval, (b) find a definite integral that represents the arc length of the curve over the indicated interval and observe that the integral cannot be evaluated with the techniques studied so far, and (c) use the integration capabilities of a graphing utility to approximate the are length. $$ y=x^{2}+x-2, \quad-2 \leq x \leq 1 $$

Finding the Area of a Surface of Revolution In Exercises \(37-42,\) set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=2 \sqrt{x} $$

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 \(8.8 . )\) Bulb Design An ornamental light bulb is designed by revolving the graph of $$y=\frac{1}{3} x^{1 / 2}-x^{3 / 2}, \quad 0 \leq x \leq \frac{1}{3}$$ about the \(x\) -axis, where \(x\) and \(y\) are measured in feet (see figure). Find the surface area of the bulb and use the result to approximate the amount of glass needed to make the bulb. (Assume that the glass is 0.015 inch thick.)

Constant Force In Exercises \(1-4,\) determine the work done by the constant force. A 1200 -pound steel beam is lifted 40 feet.
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