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Chapter 11: Problem 12

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. $$ \int_{-2}^{3}\left[(y+6)-y^{2}\right] d y $$

Short Answer

Expert verified
The shaded region is the area bounded by the curves of the functions \( y+6 \) and \( y^{2} \) from x=-2 to x=3. The area under the line and above the parabola for these x-values is represented by the given definite integral.

Step by step solution

01

Sketch Graphs

Sketch the graph for each function, \( y+6 \) (a straight line) and \( y^{2} \) (a parable). Use an appropriate scale for both x and y to clearly represent the key points of each function.
02

Identify Bounded Area

Identify the area bounded by the two curves. Remember that the limits of the the definite integral (-2 to 3) indicate the x-values for which the area is defined. In this case, the area under the line \( y+6 \) and above the parabola \( y^{2} \) from x=-2 to x=3 is being sought.
03

Shade the Region

Shade the region that is bounded by these two functions from x=-2 to x=3. This shaded area is the one represented by the given definite integral.

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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 \(x\). $$ \frac{d C}{d x}=2.25 \quad x=100 $$

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. $$ f(x)=x^{2}(3-x) \quad[0,3] $$

Use a computer or programmable calculator to approximate the definite integral using the Midpoint Rule and the Trapezoidal Rule for \(n=4\), \(8,12,16\), and 20. $$ \int_{0}^{2} \frac{5}{x^{3}+1} d x $$

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(x)=\sqrt[3]{x}, g(x)=x $$

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 \(x\). $$ \frac{d P}{d x}=\frac{400-x}{150} \quad x=200 $$
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