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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Let g ( x ) = 0 x f ( t ) d t , where f is the function whose graph is shown.

(a) Evaluate g(0), g(1), g(2), g(3), and g(6).

(b) On what interval is g increasing?

(c) Where does g have a maximum value?

(d) Sketch a rough graph of g.

Chapter 5.3, Problem 3E, Let g(x)=0xf(t)dt, where f is the function whose graph is shown. (a) Evaluate g(0), g(1), g(2),

(a)

To determine

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

Explanation

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.

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.

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.

Determine g(3) using Equation (1)

(b)

To determine

The interval g.

(c)

To determine

The maximum value of g.

(d)

To determine

To sketch: The rough graph of g.

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