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To solve: The function is defined on the interval ,
a. Graph .
![Check Mark](/static/check-mark.png)
Answer to Problem 32AYU
a.
![Precalculus, Chapter 14.5, Problem 32AYU , additional homework tip 1](https://content.bartleby.com/tbms-images/9780135189405/Chapter-14/images/19281-14.5-32sb.jpg)
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ; ; ; ;
a. Graph
![Precalculus, Chapter 14.5, Problem 32AYU , additional homework tip 2](https://content.bartleby.com/tbms-images/9780135189405/Chapter-14/images/19281-14.5-32sb.jpg)
To solve: The function is defined on the interval ,
b. Approximate the area under from to 1 into five subintervals of equal length.
![Check Mark](/static/check-mark.png)
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,
To solve: The function is defined on the interval ,
c. Approximate the area under from to 1 into ten subintervals of equal length.
![Check Mark](/static/check-mark.png)
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,
To solve: The function is defined on the interval ,
d. Express the area as an integral.
![Check Mark](/static/check-mark.png)
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 .
To solve: The function is defined on the interval ,
e. Evaluate the integral using graphing utility.
![Check Mark](/static/check-mark.png)
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 .
To solve: The function is defined on the interval ,
f. What is the actual area ?
![Check Mark](/static/check-mark.png)
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
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