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Sketch the graphs of the functions \(f\) and \(g\) and find the area of the region enclosed by these graphs and the vertical lines \(x=a\) and \(x=b\). $$f(x)=x+2, g(x)=x^{2}-4 ; a=-1, b=2$$

Short Answer

Expert verified
The area enclosed by the graphs of \(f(x)=x+2\) and \(g(x)=x^2-4\), and the vertical lines \(x=-1\) and \(x=2\), is \(\frac{5}{3}\) square units.

Step by step solution

01

Graph the Functions

First, we need to graph the functions f(x) and g(x). The function f(x) is a linear function, and g(x) is a quadratic function. To graph the functions, create a table of x and y values and plot the points on the graph. For f(x)=x+2: x | y=f(x) -1 | 1 0 | 2 +1 | 3 +2 | 4 For g(x)=x^2-4: x | y=g(x) -1 | 3 0 | -4 +1 | -3 +2 | 0 Plot the points on the xy-plane, and the graphs of f(x) and g(x) will look like: ![graphs](https://i.imgur.com/Wn4Lpl2.png)
02

Find the Points of Intersection

To find the points where the graphs of f(x) and g(x) intersect, we need to solve the equation: $$f(x) = g(x)$$ $$x+2 = x^2-4 $$ $$x^2-x-6 = 0 $$ Factoring the equation, we get, $$(x-3)(x+2) = 0$$ The solutions for x are x = 3 and x = -2. However, we are only interested in the intersection points within the limits x = -1 and x = 2. Since x = -2 is outside the given limits, we can ignore it. Thus, our point of intersection is at (3, 5).
03

Find the Area Enclosed by the Graphs and Vertical Lines

We want to find the area enclosed by the graphs of f(x) and g(x), and the vertical lines x = -1 and x = 2. To do this, we integrate the difference of the two functions with respect to x, from the lower limit a = -1 to the upper limit b = 2: $$A = \int_{-1}^{2} [f(x) - g(x)] dx$$ Substitute f(x) and g(x) into the formula: $$A = \int_{-1}^{2} [(x+2) - (x^2-4)] dx$$ Simplify the expression: $$A = \int_{-1}^{2} (1-x-x^2) dx$$ Now apply the power rule for integration to integrate the expression: $$A = \int_{-1}^{2}(1-x-x^2) dx = x-\dfrac{1}{2}x^2-\dfrac{1}{3}x^3 \Big|_{-1}^{2} $$ Evaluate the integral at the limits of integration: $$A = [2 - \frac{1}{2}(2)^2 - \frac{1}{3}(2)^3] - [-1 - \frac{1}{2}(-1)^2 - \frac{1}{3}(-1)^3]$$ $$A = [2 - 4 - \frac{8}{3}] - [-1 - \frac{1}{2} + \frac{1}{3}]$$ $$A = \frac{5}{3}$$
04

Conclusion

The area of the region enclosed by the graphs of the functions f(x) = x + 2 and g(x) = x^2 - 4, and the vertical lines x = -1 and x = 2, is \(\frac{5}{3}\) square units.

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus that involves finding the area under a curve. In the context of finding the area between curves, integration allows us to calculate the 'accumulated' area bounded by the curves on a certain interval.

Specifically, when the region between two curves is mentioned, we usually mean integration of the absolute difference between the two functions over the interval given. This is carried out by taking a series of infinitesimally thin vertical slices between the curves and summing their areas from one end of the interval to the other. Integration is useful not only for simple shapes but also for complex regions that cannot be easily computed using basic geometric formulas.
Linear Function
A linear function is one of the simplest forms of functions we have in mathematics. It is an algebraic equation in which each term is either a constant or the product of a constant and a single variable.

Linear functions are graphically represented by straight lines and mathematically described by the formula: \( f(x) = mx + b \), where \( m \) represents the slope, and \( b \) represents the y-intercept. To graph a linear function, you can simply plot a couple of points that satisfy the equation and draw a line through them. The steepness of the line reflects the value of the slope, and where it crosses the y-axis indicates the y-intercept.
Quadratic Function
On the other hand, a quadratic function represents a parabola on the Cartesian plane. Its general form is \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is non-zero.

The graph of a quadratic function is a curve called a parabola that either opens upwards or downwards, depending on the sign of \( a \). The highest or lowest point of the parabola is known as the vertex, and it is a point of symmetry for the graph. Quadratic functions are very important in physics and other sciences, as they often describe the motion of objects under constant acceleration.
Definite Integral
The definite integral of a function over an interval provides the net area between the function and the x-axis on that interval. The net area can be thought of as 'signed area', where the parts of the graph below the x-axis contribute negatively to the total.

When we speak of the definite integral, such as the one expressed by \( \int_{a}^{b} f(x)dx \), we are saying that we want to calculate the total area from \( x=a \) to \( x=b \) under the curve of \( f(x) \). In the context of our problem, where we have two functions, we take the definite integral of the difference between two functions to find the area enclosed by them, as demonstrated in the provided step-by-step solution. This is a powerful tool for understanding and calculating real-world quantities such as distances, areas, and volumes.

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Most popular questions from this chapter

The manager of TeleStar Cable Service estimates that the total number of subscribers to the service in a certain city \(t\) yr from now will be $$ N(t)=-\frac{40,000}{\sqrt{1+0.2 t}}+50,000 $$ Find the average number of cable television subscribers over the next \(5 \mathrm{yr}\) if this prediction holds true.

Sinclair wishes to supplement his retirement income by $$\$ 300 /$$ month for the next 10 yr. He plans to obtain a reverse annuity mortgage (RAM) on his home to meet this need. Estimate the amount of the mortgage he will require if the prevailing interest rate is \(8 \% /\) year compounded continuously.

Sketch the graphs of the functions \(f\) and \(g\) and find the area of the region enclosed by these graphs and the vertical lines \(x=a\) and \(x=b\). $$f(x)=x, g(x)=e^{2 x} ; a=1, b=3$$

The White House wants to cut the gasoline usage from 140 billion gallons per year in 2007 to 128 billion gallons per year in 2017 . But estimates by the Department of Energy's Energy Information Agency suggest that won't happen. In fact, the agency's projection of gasoline usage from the beginning of 2007 until the beginning of 2017 is given by $$ A(t)=0.014 t^{2}+1.93 t+140 \quad(0 \leq t \leq 10) $$ where \(A(t)\) is measured in billions of gallons/year and \(t\) is in years, with \(t=0\) corresponding to 2007 . a. According to the agency's projection, what will be gasoline consumption at the beginning of \(2017 ?\) b. What will be the average consumption/year over the period from the beginning of 2007 until the beginning of 2017 ?

Sketch the graph and find the area of the region bounded below by the graph of each function and above by the \(x\) -axis from \(x=a\) to \(x=b\). $$f(x)=x^{2}-5 x+4 ; a=1, b=3$$

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