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Chapter 11: Problem 16

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ y=x^{3}-2 x+1, y=-2 x, x=1 $$

Short Answer

Expert verified
The area is computed by the previously mentioned integration formula with the found limits of integration.

Step by step solution

01

Graphing the function

First, the functions \(y=x^{3}-2x+1\), \(y=-2x\), and \(x=1\) must be graphed. This can be done using graph paper or a graphing software program. After drawing the graph, it's time to find the region bounded by the three curves.
02

Detect the bounded region

Once the graph of the function is drawn, the next step is to identify the bounded region. This is the region enclosed by all three functions. As observed from the graph, the bounded region will be between the graphs of \(y=x^{3}-2x+1\), \(y=-2x\), and the vertical line \(x=1\). Notice that the intersection points of the curves are the limits of integration.
03

Find the limits of integration

The intersection points of the curves are essentially the limits for the definite integral to calculate the area. Given that the 2 functions intersect when \(x^{3}-2x+1=-2x\), solving this equation reveals the two points which will be used as the limits of integral. The intersection point with the line \(x=1\) also need to be identified.
04

Compute the area

Now that the limits of integration are known, the area can be computed. The area of a region bounded by two curves \(f(x)\) and \(g(x)\) is given by the formula \(Area=\int_{a}^{b} |f(x)-g(x)| dx\). Here, \(f(x)\) and \(g(x)\) refer to \(y=x^{3}-2x+1\) and \(y=-2x\) respectievly, and 'a' and 'b' are the points of intersection.

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Most popular questions from this chapter

Find the change in cost \(C\), revenue \(R\), or profit \(P\), for the given marginal. In each case, assume that the number of units \(x\) increases by 3 from the specified value of \(x\). $$ \frac{d C}{d x}=\frac{20,000}{x^{2}} \quad x=10 $$

Find the amount of an annuity with income function \(c(t)\), interest rate \(r\), and term \(T\). $$ c(t)=\$ 250, \quad r=8 \%, \quad T=6 \text { years } $$

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. $$ \int_{0}^{4}\left[(x+1)-\frac{1}{2} x\right] d x $$

Use the Trapezoidal Rule with \(n=4\) to approximate the definite integral. $$ \int_{0}^{2} \frac{1}{x+1} d x $$

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. $$ f(x)=2 x^{2} $$
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