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To solve: The function is defined on the interval ,
a. Graph , indicating the area under from 0 to .
![Check Mark](/static/check-mark.png)
Answer to Problem 22AYU
a.
![Precalculus, Chapter 14.5, Problem 22AYU , additional homework tip 1](https://content.bartleby.com/tbms-images/9780135189405/Chapter-14/images/19281-14.5-22sb.jpg)
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ;
a. Graph
![Precalculus, Chapter 14.5, Problem 22AYU , additional homework tip 2](https://content.bartleby.com/tbms-images/9780135189405/Chapter-14/images/19281-14.5-22sb.jpg)
To solve: The function is defined on the interval ,
b. Approximate the area by Partition into four subintervals of equal length and choose u as the left endpoint of each subinterval.
![Check Mark](/static/check-mark.png)
Answer to Problem 22AYU
b. The area is approximated as .
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ;
b. Partition into four subintervals of equal length and choose u as the left endpoint of each subinterval.
The area is approximated as
To solve: The function is defined on the interval ,
c. Approximate the area by Partition into eight subintervals of equal length and choose u as the left endpoint of each subinterval.
![Check Mark](/static/check-mark.png)
Answer to Problem 22AYU
c. The area is approximated as .
Explanation of Solution
Given:
The function is defined on the interval .
Calculation:
; ; ; ; ; ; ;
c. Partition into eight subintervals of equal length and choose u as the left endpoint of each subinterval.
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 22AYU
d. The area as an integral is .
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. Use a graphing utility to approximate the integral.
![Check Mark](/static/check-mark.png)
Answer to Problem 22AYU
e. The area under the graph of from 0 to is 1.
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 1.
So the area under the graph of from 0 to is 1.
Chapter 14 Solutions
Precalculus
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