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 3E

(a)

To determine

To determine the value of g(0), g(1), g(2), g(3),and g(6).

Expert Solution

Answer to Problem 3E

The value of g(0) is 0.

The value of g(1) is 2_.

The value of g(2) is 5_.

The value of g(3) is 7_.

The value of g(6) is 3_.

Explanation of Solution

Given information:

The integral function is 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.

Substitute 0 for x in Equation (1).

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

Therefore, the g(0) is 0_.

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

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 3E , additional homework tip  1

Determine g(1) using Equation (1).

Substitute 1 for x in Equation (1).

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

Refer to Figure (1).

Modify Equation (2).

g(1)=bh

Substitute 1 for b and 2 for h.

g(1)=bh=(1)(2)=2

Therefore, the g(1) is 2_.

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

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 3E , 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 to figure 2.

The area of shaded triangle and rectangle is the function of t with limits 1 to 2.

Substitute 2 for g(1) and (12bh) for 12f(t)dt in Equation (3).

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

Substitute 1 for b and 2 for h.

g(2)=2+(1)(2)+12(1)(2)=2+2+1=5

Therefore, the g(2) is 5_.

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

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 3E , 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 to Figure 3.

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

Substitute 5 for g(2) and (12bh) for 32f(t)dt in Equation (4).

g(3)=g(2)+32f(t)dt=5+12(b)(h)=12bh

Substitute 1 for b and 4 for h.

g(3)=12bh=5+12(1)(4)=7

Therefore, the g(3) is 7_.

Draw the graph for calculation of g(6) as in Figure 4.

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

Determine g(6) using Equation (1).

g(6)=36f(t)dt=g(3)+34f(t)dt+45f(t)dt+56f(t)dt=g(5)+56f(t)dt (5)

Refer to Figure 4.

Area of shaded portion is the function of t with limits 5 to 6.

g(6)=g(3)+56f(t)dt=7+((12)bh+b1h1)

Substitute 2for b, 2 for h, 1 for b1, and 2 for h1.

g(6)=7+(12(2)(2)+(1)(2))=74=3

Therefore, the g(6) is 3_.

(b)

To determine

The interval g.

Expert Solution

Answer to Problem 3E

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

Explanation of Solution

Given information:

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

Calculation:

Refer 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)_.

(c)

To determine

The maximum value of g.

Expert Solution

Answer to Problem 3E

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

Explanation of Solution

Given information:

The integral function is 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_.

(d)

To determine

To sketch: The rough graph of g.

Expert Solution

Explanation of Solution

The graph for function f is shown in Figure 1.

Plot the graph for function f using the calculated values of 0, 2,5,7,3 for the functions g(0), g(1), g(2), g(3), and g(6) respectively.

Draw the graph for function of f as in Figure (5).

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.4, Problem 3E , additional homework tip  5

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