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Consider the function whose domain is the interval .
(a) Graph .
(b) Approximate the area under the graph of f from to 1 by dividing into five subintervals, each of equal length.
(c) Approximate the area under the graph of f from to 1 by dividing into five subintervals, each of equal length.
(d) Express the area as an
(e) Evaluate the integral using a graphing utility.
(f) What is the actual area?
![Check Mark](/static/check-mark.png)
To solve: The function is defined on the interval ,
a. Graph .
Answer to Problem 32AYU
a.
![Precalculus Enhanced with Graphing Utilities, Chapter 14.5, Problem 32AYU , additional homework tip 1](https://content.bartleby.com/tbms-images/9780134119281/Chapter-14/images/19281-14.5-32sb.jpg)
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ; ; ; ;
a. Graph
![Precalculus Enhanced with Graphing Utilities, Chapter 14.5, Problem 32AYU , additional homework tip 2](https://content.bartleby.com/tbms-images/9780134119281/Chapter-14/images/19281-14.5-32sb.jpg)
![Check Mark](/static/check-mark.png)
To solve: The function is defined on the interval ,
b. Approximate the area under from to 1 into five subintervals of equal length.
Answer to Problem 32AYU
b. 18
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ; ; ; ;
b. Approximate the area under from to 1 into five subintervals of equal length,
The area is approximated as,
![Check Mark](/static/check-mark.png)
To solve: The function is defined on the interval ,
c. Approximate the area under from to 1 into ten subintervals of equal length.
Answer to Problem 32AYU
c. 12
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ; ; ; ;
c. Approximate the area under from to 1 into ten subintervals of equal length,
The area is approximated as,
![Check Mark](/static/check-mark.png)
To solve: The function is defined on the interval ,
d. Express the area as an integral.
Answer to Problem 32AYU
d.
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ; ; ; ;
d. Express the area as an integral,
The area as an integral is .
![Check Mark](/static/check-mark.png)
To solve: The function is defined on the interval ,
e. Evaluate the integral using graphing utility.
Answer to Problem 32AYU
e.
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ; ; ; ;
e. Use a graphing utility to approximate the integral,
That is evaluate the integral,
The value of the integral is , so the area under the graph of from to 1 is .
![Check Mark](/static/check-mark.png)
To solve: The function is defined on the interval ,
f. What is the actual area ?
Answer to Problem 32AYU
f. 12
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ; ; ; ;
f. The actual area under the graph of from to 1 is the area of the semi-circle whose radius is 1. The actual area is,
Therefore,
Chapter 14 Solutions
Precalculus Enhanced with Graphing Utilities
Additional Math Textbook Solutions
Calculus & Its Applications (14th Edition)
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