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Chapter 11: Problem 28

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(y)=y(2-y), g(y)=-y $$

Short Answer

Expert verified
The area of the region bounded by the curves \(y(2-y)\) and \(-y\) from y=0 to 2, is found by evaluating the integral \( \int_{0}^{2} [(y(2-y))-(-y)] dy \).

Step by step solution

01

Identify the points of intersection

The points of intersection are obtained by setting the two functions equal to each other and solving for y. So set \(y(2-y) = -y\). This will yield two solutions, which are y=0 and y=2. These are the values of y that the region of interest is between.
02

Sketch the region

Sketch the graphs of both functions \(f(y)=y(2-y)\) and \(g(y)=-y\) between y=0 and y=2. This will give a visual representation of the region whose area needs to be found. \(f(y)\) will give a parabolic curve opening downwards with a maximum value at y=1, while \(g(y)\) will give a straight line with negative slope.
03

Compute the area using integration

The area of region between two curves from a to b is found using the definite integral \( \int_{a}^{b} [f(y)-g(y)] dy \). Here, \(f(y)\) is the upper curve and \(g(y)\) is the lower curve. Since \(f(y)>g(y)\), substitute \( y(2-y) \) for \(f(y)\) and \(-y\) for \(g(y)\) resulting in the integral \( \int_{0}^{2} [(y(2-y))-(-y)] dy \). Evaluate this integral to get the area of the region.

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Most popular questions from this chapter

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ \begin{aligned} &y=x^{2}-4 x+3, y=3+4 x-x^{2} \\ &y=4-x^{2} \cdot y=x^{2} \end{aligned} $$

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. $$ f(x)=\frac{1}{x}, \quad[1,5] $$

Find the amount of an annuity with income function \(c(t)\), interest rate \(r\), and term \(T\). $$ c(t)=\$ 2000, \quad r=3 \%, \quad T=15 \text { years } $$

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. $$ f(y)=2 y, \quad[0,2] $$

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