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Math
Mathematical Applications for the Management, Life, and Social Sciences11th Edition

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Chapter 13.3, Problem 13E
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Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Solutions

Chapter
Section
Problem 1CP
Problem 2CP
Problem 3CP
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
Problem 49E
Problem 50E
BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

In Problems 13-26, equations are given whose graphs enclose a region. In each problem, find the area of the region.

f ( x ) = x 2 + 2 ; g ( x ) = − x 2 ; x = 0 ; x = 2

To determine

To calculate: The area of the region bounded by equation f(x)=x2+2 and g(x)=−x2 from x=0 to x=2.

Explanation

Given information:

The provided equation are:

f(x)=x2+2, g(x)=−x2 from x=0 to x=2.

Formula used:

The area of the region bounded by equation y=f(x) and y=g(x) from x=a to x=b is given by

A=∫ab[f(x)−g(x)dx]

Here, f(x) and g(x) are continuous functions on [a,b] and if f(x)≥g(x) on [a,b].

Calculation:

Consider the equation,

f(x)=x2+2

Calculate the values f(x) at different values of x,

Substitute x=0 in the function f(x)=x2+2.

y=(02+2)=2

Substitute x=1 in the function f(x)=x2+2.

y=(12+2)=3

Substitute x=2 in the function f(x)=x2+2.

y=(22+2)=6

Substitute x=−1 in the function f(x)=x2+2.

y=((−1)2+2)=3

The following table shows the different coordinate (x,y) for equation f(x)=x2+2.

x f(x)=x2+2 Coordinates (x,y)
0 2 (0,2)
1 3 (1,3)
2 6 (2,6)
−1 3 (−1,3)

Now, consider,

g(x)=−x2

Substitute x=0 in the function g(x)=−x2.

y=−(0)2=0

Substitute x=1 in the function g(x)=−x2.

y=−(1)2=−1

Substitute x=2 in the function g(x)=−x2.

y=−(2)2=−4

Substitute x=−1 in the function g(x)=−x2

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