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Chapter 7: Problem 18

Finding the Area of a Region In Exercises \(15 - 28\) sketch the region bounded by the graphs of the equations and find the area of the region. $$y = - x ^ { 2 } + 3 x + 1 , \quad y = - x + 1$$

Short Answer

Expert verified
The area of the region bounded by the graphs of the equations \(y = - x ^ { 2 } + 3 x + 1\) and \(y = - x + 1\) is \(\frac{32}{3}\) square units.

Step by step solution

01

Find the intersection points

First, set the two equations equal to each other to find the x-values where the graphs intersect. \n\n\(-x^2 + 3x + 1 = -x + 1\)\n\nSimplifying this gives us: \n\n\(x^2 - 4x = 0\)\n\nFactoring x out gives: \n\n\(x (x - 4) = 0\)\n\nSetting each factor equal to zero gives the solutions x=0, x=4. These are the points where the graphs intersect.
02

Determine the integral limits and sketch the region

The intersection points x = 0 and x = 4 will be the limits of integration. On a graph, the region bounded by the two curves would appear as a enclosed shape between these two x-values. The region is bounded above by \(y = -x^2 + 3x + 1\) and below by \(y = -x + 1\).
03

Compute the integral

To find the area of the region, we must compute the integral of the absolute difference of the functions over the interval (0, 4).\nThe integral will be: \n\n\(\int_{0}^{4} ((-x^2 + 3x + 1) - (-x + 1)) dx\)\n\nSimplify this equation: \n\n\(\int_{0}^{4}(-x^2 + 3x + 1 + x -1) dx\)\n\nFurther simplify: \n\n\(\int_{0}^{4} (-x^2 + 4x) dx\)\n\nNow, compute the integral: \n\n\([-\frac{1}{3}x^3 + 2x^2]_0^4\)\n\nSubstitute the upper and lower bounds into this equation: \n\n\(-\frac{1}{3}(4)^3 + 2(4)^2 - [ -\frac{1}{3}(0)^3 + 2(0)^2 ] = -\frac{64}{3} + 32 = \frac{32}{3}\)
04

Conclusion

The area of the region bounded by the curves \(y = - x ^ { 2 } + 3 x + 1\) and \(y = - x + 1\) is \(\frac{32}{3}\) square units. So, this is the final answer.

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Key Concepts

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

Area between curves
When working with the concept of finding the area between curves, you're determining the size of the region enclosed by two different function graphs over a certain interval.
To visualize, imagine the space trapped between two curvy lines on a graph.
The region of interest is bounded by these two functions, one usually lying above the other over the given range.To calculate this area, we use the formula:
  • First: Identify the function that forms the top boundary of the region (let's call it the upper function). For example, in the given problem, this function is \(-x^2 + 3x + 1\).
  • Secondly: Identify the lower function, which is the boundary below the region. In the example, this is \(-x + 1\).
  • Subtraction: Calculate the difference. This involves subtracting the value of the lower function from the upper function for each x-value in the interval.
The area is then given by integrating this difference over the observed x-interval.
Definite integrals
Definite integrals are a key tool in calculus used to find the accumulation of quantities, such as areas under curves over a specific interval on the x-axis.
Unlike indefinite integrals, definite integrals result in a numerical value, which often represents an area in many problems.Consider a function \(f(x)\) that is non-negative and continuous over an interval \([a, b]\).
The definite integral of \(f\) from \(a\) to \(b\), denoted \(\int_{a}^{b} f(x) \, dx\), represents the area of the region between the graph of \(f(x)\) and the x-axis over this interval.In the exercise solution:
  • The definite integral was used to find the area between the curves. It was calculated as \(\int_{0}^{4} ((-x^2 + 3x + 1) - (-x + 1)) \, dx\).
  • After simplifying, the integral becomes \(\int_{0}^{4} (-x^2 + 4x) \, dx\), and this is what ultimately calculates the area between the curves.
Intersection points
Intersection points in the context of finding the area between curves are crucial because they determine where the boundaries of the area start and end.
These points occur where the graphs of the two functions cross each other, meaning their y-values are equal for the same x-value.To find these points:
  • Set the two equations equal to each other. For this exercise, solving \(-x^2 + 3x + 1 = -x + 1\) will give us the x-values of intersection points.
  • Simplify and solve: Rearrange and factor the equation, resulting in \(x^2 - 4x = 0\).
    Factorizing gives \(x(x - 4) = 0\), and solving gives the points \(x = 0\) and \(x = 4\).
These intersection points, \(x = 0\) and \(x = 4\), act as the limits for integration when calculating the area between the curves.
Limits of integration
When calculating the area between two curves, the limits of integration play a vital role.
These are simply the x-values that define the range over which you are integrating. In the context of this problem, the limits of integration are determined directly from the intersection points.In integration:
  • The lower limit represents the starting point of the integral on the x-axis.
    In the exercise, the lower limit is \(x = 0\).
  • The upper limit is the endpoint of integration on the x-axis, here being \(x = 4\).
  • These limits encapsulate the entire region over which the area is being computed. So, all calculations for the area take place within these bounds.
Understanding how to determine these limiting values and apply them correctly is critical in solving area problems using integrals.

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Most popular questions from this chapter

Modeling Data The manufacturer of glass for a window in a conversion van needs to approximate the center of mass of the glass. A coordinate system is superimposed on a prototype of the glass (see figure). The measurements (in centimeters) for the right half of the symmetric piece of glass are listed in the table. $$\begin{array}{|c|c|c|c|c|c|}\hline x & {0} & {10} & {20} & {30} & {40} \\\ \hline y & {30} & {29} & {26} & {20} & {0} \\ \hline\end{array}$$ (a) Use the regression capabilities of a graphing utility to find a fourth- degree polynomial model for the glass. (b) Use the integration capabilities of a graphing utility and the model to approximate the center of mass of the glass.

Think About It Consider the equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) (a) Use a graphing utility to graph the equation. (b) Write the definite integral for finding the first-quadrant arc length of the graph in part (a). (c) Compare the interval of integration in part \((\mathrm{b})\) and the domain of the integrand. Is it possible to evaluate the definite integral? Explain. (You will learn how to evaluate this type of integral in Section \(8.8 . )\)

In Exercises 43-46, the integral represents the volume of a solid of revolution. Identify (a) the plane region that is revolved and (b) the axis of revolution. $$2 \pi \int_{0}^{6}(y+2) \sqrt{6-y} d y$$

Approximation In Exercises 31 and \(32,\) approximate the are length of the graph of the function over the interval \([0,4]\) in three ways.(a) Use the Distance Formula to find the distance between the endpoints of the arc. (b) Use the Distance Formula to find the lengths of the four line segments connecting the points on the arc when \(x=0, x=1, x=2, x=3,\) and \(x=4\) .Find the sum of the four lengths. (c) Use the integration capabilities of a graphing utility to approximate the integral yielding the indicated are length. $$f(x)=\left(x^{2}-4\right)^{2}$$

In Exercises 3-6, the area of the top side of a piece of sheet metal is given. The sheet metal is submerged horizontally in 8 feet of water. Find the fluid force on the top side. 25 square feet
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