BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 5.4, Problem 4E

(a)

To determine

The value of g(0) and g(6).

Expert Solution

Answer to Problem 4E

The value of g(0) is 0.

The value of g(6) is 0_

Explanation of Solution

Given information:

The equation is g(x)=0xf(t)dt.

The graph is given for the integral function g(x)=0xf(t)dt.

Calculation:

Show the integral function as below.

g(x)=0xf(t)dt (1)

Here, g(x) is area under the graph of f from a to x and f(t) is function of t.

Determine g(0) using Equation (1).

Substitute 0 for x in Equation (1).

g(0)=00f(t)dt=0

Therefore, the g(0) is 0_.

Determine g(6) using given graph:

Refer the graph.

The curve is symmetrical about the point (3, 0). Hence, the area between the points 0 to 3 and 3 to 6 are equal with alternative sign.

Therefore, the function g(6) is 0.

(b)

To determine

The value of g(x), for x=1,x=2, x=3, x=4, and x=5.

Expert Solution

Answer to Problem 4E

The value of g(1) is 2.8.

The value of g(2) is 4.9.

The value of g(3) is 5.7.

The value of g(4) is 4.9.

The value of g(5) is 2.8.

Explanation of Solution

Given information:

The equation is g(x)=0xf(t)dt.

The graph is given for the integral function g(x)=0xf(t)dt.

Calculation:

Draw the graph for calculation of g(1) as in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 4E , additional homework tip  1

Determine g(1) using Equation (1).

Substitute 1 for x in Equation (1).

g(1)=01f(t)dt (2)

Refer Figure (1).

Area of shaded rectangle is the function of t with limits 0 to 1.

Modify Equation (2).

g(1)=bh

Add 80% of unit square.

Substitute 1 for b, 2 for h.

g(1)=bh+80%=(1)(2)+0.8=2.8

Therefore, g(1) is 2.8.

Draw the graph for calculation of g(2) as in Figure 2.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 4E , additional homework tip  2

Determine g(2) using Equation (1).

Substitute 2 for x in Equation (1).

g(2)=02f(t)dt=01f(t)dt+12f(t)dt=g(1)+12f(t)dt (3)

Refer Figure (2).

Area of shaded rectangle is the function of t with limits 1 to 2.

Substitute 2.8 for g(1) and bh for 12f(t)dt in Equation (3).

g(2)=g(1)+12f(t)dt=2.8+bh

Add 90% of unit square.

Substitute 1 for b and 1 for h.

g(2)=2.8+1+0.9=4.7

Therefore, g(2) is 4.7.

Draw the graph for calculation of g(3) as in Figure 3.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 4E , additional homework tip  3

Determine g(3) using Equation (1).

Substitute 3 for x in Equation (1).

g(3)=03f(t)dt=01f(t)dt+12f(t)dt+23f(t)dt=g(2)+23f(t)dt (4)

Refer Figure 3.

Area of shaded triangle is the function of t with limits 2 to 3.

Substitute 4.7 for g(2) and bh for 32f(t)dt in Equation (4).

g(3)=g(2)+32f(t)dt=4.7+12bh

Substitute 1 for b and 1.2 for h.

g(3)=g(2)+32f(t)dt=4.7+12(1)(1.2)=5.3

Therefore, g(3) is 5.3.

Draw the graph for calculation of g(4) as in Figure (4).

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 4E , additional homework tip  4

Determine g(4) using Equation (1).

Substitute 4 for x in Equation (1).

g(4)=04f(t)dt=01f(t)dt+12f(t)dt+23f(t)dt+34f(t)dt=g(3)+34f(t)dt (5)

Refer Figure 4.

Area of shaded triangle is the function of t with limits 3 to 4.

Substitute 5.3 for g(3) and (12bh) for 34f(t)dt in Equation (5).

g(4)=g(3)34f(t)dt=5.312bh

Substitute 1 for b and 1 for h.

g(4)=5.312bh=5.312(1)(1.2)=4.7

Therefore, g(3) is 5.3.

Draw the graph for calculation of g(5) as in Figure 5.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 4E , additional homework tip  5

Determine g(5) using Equation (1).

Substitute 5 for x in Equation (1).

g(5)=05f(t)dt=01f(t)dt+12f(t)dt+23f(t)dt+34f(t)dt+45f(t)dt=g(4)+45f(t)dt (6)

Refer to Figure 5.

Area of shaded portion is the function of t with limits 4 to 5.

Substitute 4.7 for g(4) and bh for 45f(t)dt in Equation (6).

g(5)=g(4)45f(t)dt=4.7bh

Subtract 90% of third unit square.

Substitute 1 for, 1 for h.

g(5)=4.7bh=4.7(1)(1)0.9=2.8

(c)

To determine

The interval when g is increasing.

Expert Solution

Answer to Problem 4E

The function g is increasing at the interval (0,3)_.

Explanation of Solution

Given information:

The equation is g(x)=0xf(t)dt.

The graph is given for the integral function g(x)=0xf(t)dt.

Calculation:

Refer to Part (a).

The value of g is increasing from the interval 0 to 3.

Therefore, the function g is increasing at the interval (0,3)_.

(d)

To determine

The location of maximum value of g.

Expert Solution

Answer to Problem 4E

The maximum value of g lies at x=3_.

Explanation of Solution

Given information:

The equation is g(x)=0xf(t)dt.

The graph is given for the integral function g(x)=0xf(t)dt.

Calculation:

Refer part (a) calculation

Maximum value of g lies at x=3

Therefore, the maximum value of g lies at x=3.

(e)

To determine

To Sketch: The rough graph of g.

Expert Solution

Explanation of Solution

Plot the graph for function f using the calculated values of 0, 2.8, 4.7, 5.3 and 4.7 for the functions g(1), g(2), g(3), g(4), and g(5) respectively.

Show the graph for function f as in Figure 6.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 4E , additional homework tip  6

(f)

To determine

To Sketch: The graph g(x).

Expert Solution

Explanation of Solution

Show the graph for function g(x) as in Figure 7.

Draw the graph g(x) as shown in Figure 7.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 4E , additional homework tip  7

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