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Chapter 11: Problem 36

Use a graphing utility to graph the region bounded by the graphs of the functions, and find the area of the region. $$ f(x)=3-2 x-x^{2}, g(x)=0 $$

Short Answer

Expert verified
The area of the region bounded by the graphs of the functions is 6 square units.

Step by step solution

01

Graph the Functions

The first step involves graphing both functions to visualize them and find the intersection points. By graphing the function \(f(x) = 3 - 2x - x^2\) and \(g(x) = 0\), it is evident that they intersect at x = -1 and x = 3.
02

Define the Interval

The region bound by the two curves is noticeable on the interval from x = -1 to x = 3. We can see that for any point in this interval, the value of the function \(f(x)\) is always greater than the value of the function \(g(x)=0\).
03

Calculate Area

We can calculate the area under the curve \(f(x)\) from x = -1 to x = 3 by using the formula for the definite integral: \[A = \int_{a}^{b} f(x) - g(x) \,dx = \int_{-1}^{3} (3 - 2x - x^2 - 0) \,dx\]. Hence, \[A = \left. [3x - x^2 - \frac{x^3}{3}]\right|_{-1}^{3} = 12 - 9 - 9 = -6\]. Since area cannot be negative, we take the absolute value to get the answer 6 square units.

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Most popular questions from this chapter

Use the Midpoint Rule with \(n=4\) to approximate the area of the region bounded by the graph of \(f\) and the \(x\) -axis over the interval. Compare your result with the exact area. Sketch the region. $$ f(x)=3 x^{2}+1 \quad[-1,3] $$

Use the Midpoint Rule with \(n=4\) to approximate the area of the region. Compare your result with the exact area obtained with a definite integral. $$ f(x)=1-x^{2}, \quad[-1,1] $$

Find the area of the region. $$ \begin{aligned} &f(x)=3\left(x^{3}-x\right) \\ &g(x)=0 \end{aligned} $$

The integrand of the definite integral is a difference of two functions. Sketch the graph of each function and shade the region whose area is represented by the integral. $$ \int_{-2}^{3}\left[(y+6)-y^{2}\right] d y $$

Find the consumer and producer surpluses. $$ p_{1}(x)=50-0.5 x \quad p_{2}(x)=0.125 x $$
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