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Chapter 5: Problem 23

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ f(y)=y^{2}+1, g(y)=0, \quad y=-1, \quad y=2 $$

Short Answer

Expert verified
The area of the region bounded by the two curves from y=-1 to y=2 is 7 square units

Step by step solution

01

Sketch the Functions

Begin with plotting the functions \(f(y)=y^{2}+1\) and \(g(y)=0\). The function \(f(y)=y^{2}+1\) is a parabola opening upwards with vertex at (0, 1), and \(g(y)=0\) is a horizontal line along the x-axis.
02

Identify the Bounded Region

Bounded Region is the area enclosed between \(f(y)\) and \(g(y)\) from \(y=-1\) to \(y=2\). This is the region which needs to be calculated.
03

Setting up the Integral

Since the area lies between two functions over an interval of y, we may use the formula for the area between curves. In this case, it's the difference between the formula of the curves over the boundary points. So, it's the integral from -1 to 2 of [(y^2 + 1) - 0], \(A = \int_{-1}^{2} [(y^2 + 1) - 0] dy \)
04

Evaluate the Integral

Calculating the integral, A = [(y^{3}/3) + y] evaluated from -1 to 2. So, A = [(8/3) + 2 - ((-1/3) + (-1))] = 8/3 + 2 + 1/3 + 1 = 4 + 3 = 7 square units

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