Answer a. Explanation Given: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 2 ( 0 ) + 8 = 8 ; ƒ ( 0.5 ) = − 2 ( 0.5 ) + 8 = 7 ; ƒ ( 1 ) = − 2 ( 1 ) + 8 = 6 ; ƒ ( 1.5 ) = − 2 ( 1.5 ) + 8 = 5 ƒ ( 2 ) = − 2 ( 2 ) + 8 = 4 ; ƒ ( 2.5 ) = − 2 ( 2.5 ) + 8 = 3 ; ƒ ( 3 ) = − 2 ( 3 ) + 8 = 2 a. Graph ƒ ( x ) = − 2 x + 8 To determine To solve: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] , b. Partition [ 0 , 3 ] into three subintervals of equal length and choose u as the left endpoint of each subinterval. Answer b. 18 Explanation Given: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 2 ( 0 ) + 8 = 8 ; ƒ ( 0.5 ) = − 2 ( 0.5 ) + 8 = 7 ; ƒ ( 1 ) = − 2 ( 1 ) + 8 = 6 ; ƒ ( 1.5 ) = − 2 ( 1.5 ) + 8 = 5 ƒ ( 2 ) = − 2 ( 2 ) + 8 = 4 ; ƒ ( 2.5 ) = − 2 ( 2.5 ) + 8 = 3 ; ƒ ( 3 ) = − 2 ( 3 ) + 8 = 2 b. Partition [ 0 , 3 ] into three subintervals of equal length 1 and choose u as the left endpoint of each subinterval. The area A is approximated as A = f ( 0 ) 1 + f ( 1 ) 1 + f ( 2 ) 1 = 8 ( 1 ) + 6 ( 1 ) + 4 ( 1 ) = 8 + 6 + 4 = 18 To determine To solve: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] , c. Partition [ 0 , 3 ] into three subintervals of equal length and choose u as the right endpoint of each subinterval. Answer c. 12 Explanation Given: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 2 ( 0 ) + 8 = 8 ; ƒ ( 0.5 ) = − 2 ( 0.5 ) + 8 = 7 ; ƒ ( 1 ) = − 2 ( 1 ) + 8 = 6 ; ƒ ( 1.5 ) = − 2 ( 1.5 ) + 8 = 5 ƒ ( 2 ) = − 2 ( 2 ) + 8 = 4 ; ƒ ( 2.5 ) = − 2 ( 2.5 ) + 8 = 3 ; ƒ ( 3 ) = − 2 ( 3 ) + 8 = 2 c. Partition [ 0 , 3 ] into three subintervals of equal length 1 and choose u as the right endpoint of each subinterval. The area A is approximated as A = f ( 1 ) 1 + f ( 2 ) 1 + f ( 3 ) 1 = 6 ( 1 ) + 4 ( 1 ) + 2 ( 1 ) = 6 + 4 + 2 = 12 To determine To solve: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] , d. Partition [ 0 , 3 ] into six subintervals of equal length and choose u as the left endpoint of each subinterval. Answer d. 33 2 Explanation Given: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 2 ( 0 ) + 8 = 8 ; ƒ ( 0.5 ) = − 2 ( 0.5 ) + 8 = 7 ; ƒ ( 1 ) = − 2 ( 1 ) + 8 = 6 ; ƒ ( 1.5 ) = − 2 ( 1.5 ) + 8 = 5 ƒ ( 2 ) = − 2 ( 2 ) + 8 = 4 ; ƒ ( 2.5 ) = − 2 ( 2.5 ) + 8 = 3 ; ƒ ( 3 ) = − 2 ( 3 ) + 8 = 2 d. Partition [ 0 , 3 ] into six subintervals of equal length 0.5 and choose u as the left endpoint of each subinterval. The area A is approximated as A = f ( 0 ) ( 0.5 ) + f ( 0.5 ) ( 0.5 ) + f ( 1 ) ( 0.5 ) + f ( 1.5 ) ( 0.5 ) + f ( 2 ) ( 0.5 ) + f ( 2.5 ) ( 0.5 ) = 8 ( 0.5 ) + ( 7 ) ( 0.5 ) + 6 ( 0.5 ) + ( 5 ) ( 0.5 ) + 4 ( 0.5 ) + ( 3 ) ( 0.5 ) = 4 + 3.5 + 3 + 2.5 + 2 + 1.5 = 16.5 = 33 2 To determine To solve: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] , e. Partition [ 0 , 3 ] into six subintervals of equal length and choose u as the right endpoint of each subinterval. Answer e. 27 2 Explanation Given: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 2 ( 0 ) + 8 = 8 ; ƒ ( 0.5 ) = − 2 ( 0.5 ) + 8 = 7 ; ƒ ( 1 ) = − 2 ( 1 ) + 8 = 6 ; ƒ ( 1.5 ) = − 2 ( 1.5 ) + 8 = 5 ƒ ( 2 ) = − 2 ( 2 ) + 8 = 4 ; ƒ ( 2.5 ) = − 2 ( 2.5 ) + 8 = 3 ; ƒ ( 3 ) = − 2 ( 3 ) + 8 = 2 e. Partition [ 0 , 3 ] into six subintervals of equal length 0.5 and choose u as the right endpoint of each subinterval. The area A is approximated as A = f ( 0.5 ) ( 0.5 ) + f ( 1 ) ( 0.5 ) + f ( 1.5 ) ( 0.5 ) + f ( 2 ) ( 0.5 ) + f ( 2.5 ) ( 0.5 ) + f ( 3 ) ( 0.5 ) = ( 7 ) ( 0.5 ) + 6 ( 0.5 ) + ( 5 ) ( 0.5 ) + 4 ( 0.5 ) + ( 3 ) ( 0.5 ) + 2 ( 0.5 ) = 3.5 + 3 + 2.5 + 2 + 1.5 + 1 = 13.5 = 27 2 To determine To solve: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] , f. What is the actual area A ? Answer f. 12 Explanation Given: The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 2 ( 0 ) + 8 = 8 ; ƒ ( 0.5 ) = − 2 ( 0.5 ) + 8 = 7 ; ƒ ( 1 ) = − 2 ( 1 ) + 8 = 6 ; ƒ ( 1.5 ) = − 2 ( 1.5 ) + 8 = 5 ƒ ( 2 ) = − 2 ( 2 ) + 8 = 4 ; ƒ ( 2.5 ) = − 2 ( 2.5 ) + 8 = 3 ; ƒ ( 3 ) = − 2 ( 3 ) + 8 = 2 f. The actual area under the graph of ƒ ( x ) = − 2 x + 8 from 0 to 3 is the area of a right triangle whose base is of length 3 and whose height is 8. The actual area A is Therefore A = 1 2 b a s e × h e i g h t A = 1 2 ( 3 ) ( 8 ) = 12

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ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 14.5, Problem 12AYU

To determine

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

a. Graph $ƒ$ .

In (b)–(e), approximate the area $A$ under $ƒ$ from 0 to 3 as follows:

Expert Solution

Answer to Problem 12AYU

Precalculus, Chapter 14.5, Problem 12AYU , additional homework tip 1

Explanation of Solution

Given:

The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 2 (0) + 8 = 8; ƒ (0.5) = - 2 (0.5) + 8 = 7; ƒ (1) = - 2 (1) + 8 = 6; ƒ (1.5) = - 2 (1.5) + 8 = 5$

$ƒ (2) = - 2 (2) + 8 = 4; ƒ (2.5) = - 2 (2.5) + 8 = 3; ƒ (3) = - 2 (3) + 8 = 2$

a. Graph $ƒ (x) = - 2 x + 8$

Precalculus, Chapter 14.5, Problem 12AYU , additional homework tip 2

To determine

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

b. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Expert Solution

Answer to Problem 12AYU

b. 18

Explanation of Solution

Given:

The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 2 (0) + 8 = 8; ƒ (0.5) = - 2 (0.5) + 8 = 7; ƒ (1) = - 2 (1) + 8 = 6; ƒ (1.5) = - 2 (1.5) + 8 = 5$

$ƒ (2) = - 2 (2) + 8 = 4; ƒ (2.5) = - 2 (2.5) + 8 = 3; ƒ (3) = - 2 (3) + 8 = 2$

b. Partition $[0, 3]$ into three subintervals of equal length 1 and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A = f (0) 1 + f (1) 1 + f (2) 1$

$= 8 (1) + 6 (1) + 4 (1)$

$= 8 + 6 + 4$

$= 18$

To determine

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

c. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Expert Solution

Answer to Problem 12AYU

c. 12

Explanation of Solution

Given:

The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 2 (0) + 8 = 8; ƒ (0.5) = - 2 (0.5) + 8 = 7; ƒ (1) = - 2 (1) + 8 = 6; ƒ (1.5) = - 2 (1.5) + 8 = 5$

$ƒ (2) = - 2 (2) + 8 = 4; ƒ (2.5) = - 2 (2.5) + 8 = 3; ƒ (3) = - 2 (3) + 8 = 2$

c. Partition $[0, 3]$ into three subintervals of equal length 1 and choose $u$ as the right endpoint of each subinterval.

The area $A$ is approximated as

$A = f (1) 1 + f (2) 1 + f (3) 1$

$= 6 (1) + 4 (1) + 2 (1)$

$= 6 + 4 + 2$

$= 12$

To determine

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

d. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Expert Solution

Answer to Problem 12AYU

d. $\frac{33}{2}$

Explanation of Solution

Given:

The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 2 (0) + 8 = 8; ƒ (0.5) = - 2 (0.5) + 8 = 7; ƒ (1) = - 2 (1) + 8 = 6; ƒ (1.5) = - 2 (1.5) + 8 = 5$

$ƒ (2) = - 2 (2) + 8 = 4; ƒ (2.5) = - 2 (2.5) + 8 = 3; ƒ (3) = - 2 (3) + 8 = 2$

d. Partition $[0, 3]$ into six subintervals of equal length $0.5$ and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A = f (0) (0.5) + f (0.5) (0.5) + f (1) (0.5) + f (1.5) (0.5) + f (2) (0.5) + f (2.5) (0.5)$

$= 8 (0.5) + (7) (0.5) + 6 (0.5) + (5) (0.5) + 4 (0.5) + (3) (0.5)$

$= 4 + 3.5 + 3 + 2.5 + 2 + 1.5$

$= 16.5 = \frac{33}{2}$

To determine

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

e. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Expert Solution

Answer to Problem 12AYU

e. $\frac{27}{2}$

Explanation of Solution

Given:

The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 2 (0) + 8 = 8; ƒ (0.5) = - 2 (0.5) + 8 = 7; ƒ (1) = - 2 (1) + 8 = 6; ƒ (1.5) = - 2 (1.5) + 8 = 5$

$ƒ (2) = - 2 (2) + 8 = 4; ƒ (2.5) = - 2 (2.5) + 8 = 3; ƒ (3) = - 2 (3) + 8 = 2$

e. Partition $[0, 3]$ into six subintervals of equal length $0.5$ and choose $u$ as the right endpoint of each subinterval.

The area $A$ is approximated as

$A = f (0.5) (0.5) + f (1) (0.5) + f (1.5) (0.5) + f (2) (0.5) + f (2.5) (0.5) + f (3) (0.5)$

$= (7) (0.5) + 6 (0.5) + (5) (0.5) + 4 (0.5) + (3) (0.5) + 2 (0.5)$

$= 3.5 + 3 + 2.5 + 2 + 1.5 + 1$

$= 13.5 = \frac{27}{2}$

To determine

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

f. What is the actual area $A$ ?

Expert Solution

Answer to Problem 12AYU

f. 12

Explanation of Solution

Given:

The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 2 (0) + 8 = 8; ƒ (0.5) = - 2 (0.5) + 8 = 7; ƒ (1) = - 2 (1) + 8 = 6; ƒ (1.5) = - 2 (1.5) + 8 = 5$

$ƒ (2) = - 2 (2) + 8 = 4; ƒ (2.5) = - 2 (2.5) + 8 = 3; ƒ (3) = - 2 (3) + 8 = 2$

f. The actual area under the graph of $ƒ (x) = - 2 x + 8$ from 0 to 3 is the area of a right triangle whose base is of length 3 and whose height is 8. The actual area $A$ is

Therefore

$A = \frac{1}{2} b a s e \times h e i g h t$

$A = \frac{1}{2} (3) (8) = 12$

Chapter 14.5, Problem 11AYU

Chapter 14.5, Problem 13AYU

Chapter 14 Solutions

Precalculus

Ch. 14.1 - Graph f( x )={ 3x2ifx2 3ifx=2 (pp.100-102)Ch. 14.1 - Prob. 2AYU Ch. 14.1 - Prob. 3AYU Ch. 14.1 - Prob. 4AYU Ch. 14.1 - True or False lim xc f( x )=N may be described by...Ch. 14.1 - Prob. 6AYU Ch. 14.1 - lim x2 ( 4 x 3 )Ch. 14.1 - lim x3 ( 2 x 2 +1 )Ch. 14.1 - lim x0 x+1 x 2 +1 Ch. 14.1 - Prob. 10AYU

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The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] , a. Graph ƒ . In (b)–(e), approximate the area A under ƒ from 0 to 3 as follows:

Solution Summary: The author calculates the area A under from 0 to 3 by dividing the function into three subintervals of equal length.

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To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

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Answer to Problem 12AYU

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To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

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To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

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To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

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The function ƒ ( x ) = − 2 x + 8 is defined on the interval [ 0 , 3 ] , a. Graph ƒ . In (b)–(e), approximate the area A under ƒ from 0 to 3 as follows:

Solution Summary: The author calculates the area A under from 0 to 3 by dividing the function into three subintervals of equal length.

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To solve: The function ƒ( x ) = −2x + 8 is defined on the interval [ 0, 3 ] , a. Graph ƒ . In (b)–(e), approximate the area A under ƒ from 0 to 3 as follows:

Answer to Problem 12AYU

Explanation of Solution

To solve: The function ƒ( x ) = −2x + 8 is defined on the interval [ 0, 3 ] , b. Partition [ 0, 3 ] into three subintervals of equal length and choose u as the left endpoint of each subinterval.

Answer to Problem 12AYU

Explanation of Solution

To solve: The function ƒ( x ) = −2x + 8 is defined on the interval [ 0, 3 ] , c. Partition [ 0, 3 ] into three subintervals of equal length and choose u as the right endpoint of each subinterval.

Answer to Problem 12AYU

Explanation of Solution

To solve: The function ƒ( x ) = −2x + 8 is defined on the interval [ 0, 3 ] , d. Partition [ 0, 3 ] into six subintervals of equal length and choose u as the left endpoint of each subinterval.

Answer to Problem 12AYU

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Answer to Problem 12AYU

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Answer to Problem 12AYU

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Additional Math Textbook Solutions

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

a. Graph $ƒ$ .

In (b)–(e), approximate the area $A$ under $ƒ$ from 0 to 3 as follows:

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

b. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

c. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

d. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

e. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

To solve: The function $ƒ (x) = - 2 x + 8$ is defined on the interval $[0, 3]$ ,

f. What is the actual area $A$ ?