Chapter 5.4, Problem 3E

(a)

To determine

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

Answer to Problem 3E

The value of $g(0)$ is 0.

The value of $g(1)$ is $\underset{\_}{2}$.

The value of $g(2)$ is $\underset{\_}{5}$.

The value of $g(3)$ is $\underset{\_}{7}$.

The value of $g(6)$ is $\underset{\_}{3}$.

Explanation of Solution

Given information:

The integral function is $g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$.

Calculation:

Show the integral function as below.

$g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$ (1)

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

Substitute 0 for x in Equation (1).

$\begin{array}{c}g\left(0\right)={\displaystyle \underset{0}{\overset{0}{\int }}f\left(t\right)dt}\\ =0\end{array}$

Therefore, the $g\left(0\right)$ is $\underset{\_}{0}$.

Draw the graph for calculation of $g\left(1\right)$ as in Figure 1.

Determine $g\left(1\right)$ using Equation (1).

Substitute 1 for x in Equation (1).

$g\left(1\right)={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}$ (2)

Refer to Figure (1).

Modify Equation (2).

$g\left(1\right)=bh$

Substitute 1 for b and 2 for h.

$\begin{array}{c}g\left(1\right)=bh\\ =\left(1\right)\left(2\right)\\ =2\end{array}$

Therefore, the $g\left(1\right)$ is $\underset{\_}{2}$.

Draw the graph for calculation of $g\left(2\right)$ as in Figure 2.

Determine $g\left(2\right)$ using Equation (1).

Substitute 2 for x in Equation (1).

$\begin{array}{c}g\left(2\right)={\displaystyle \underset{0}{\overset{2}{\int }}f\left(t\right)dt}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}\\ =g\left(1\right)+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}\end{array}$ (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 $g\left(1\right)$ and $\left(\frac{1}{2}bh\right)$ for ${\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}$ in Equation (3).

$\begin{array}{c}g(2)=g\left(1\right)+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}\\ =2+\left(\frac{1}{2}bh\right)\end{array}$

Substitute 1 for b and 2 for h.

$\begin{array}{c}g(2)=2+\left(1\right)\left(2\right)+\frac{1}{2}\left(1\right)\left(2\right)\\ =2+2+1\\ =5\end{array}$

Therefore, the $g\left(2\right)$ is $\underset{\_}{5}$.

Draw the graph for calculation of $g\left(3\right)$ as in Figure 3.

Determine $g\left(3\right)$ using Equation (1).

Substitute 3 for x in Equation (1).

$\begin{array}{c}g\left(3\right)={\displaystyle \underset{0}{\overset{3}{\int }}f\left(t\right)dt}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)dt}\\ =g\left(2\right)+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)dt}\end{array}$ (4)

Refer to Figure 3.

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

Substitute 5 for $g\left(2\right)$ and $\left(\frac{1}{2}bh\right)$ for ${\displaystyle \underset{3}{\overset{2}{\int }}f\left(t\right)dt}$ in Equation (4).

$\begin{array}{c}g\left(3\right)=g\left(2\right)+{\displaystyle \underset{3}{\overset{2}{\int }}f\left(t\right)dt}\\ =5+\frac{1}{2}\left(b\right)\left(h\right)\\ =\frac{1}{2}bh\end{array}$

Substitute 1 for b and 4 for h.

$\begin{array}{c}g\left(3\right)=\frac{1}{2}bh\\ =5+\frac{1}{2}\left(1\right)\left(4\right)\\ =7\end{array}$

Therefore, the $g\left(3\right)$ is $\underset{\_}{7}$.

Draw the graph for calculation of $g\left(6\right)$ as in Figure 4.

Determine $g\left(6\right)$ using Equation (1).

$\begin{array}{c}g\left(6\right)={\displaystyle \underset{3}{\overset{6}{\int }}f\left(t\right)dt}\\ =g\left(3\right)+{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)dt}+{\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}+{\displaystyle \underset{5}{\overset{6}{\int }}f\left(t\right)dt}\\ =g\left(5\right)+{\displaystyle \underset{5}{\overset{6}{\int }}f\left(t\right)dt}\end{array}$ (5)

Refer to Figure 4.

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

$\begin{array}{c}g\left(6\right)=g\left(3\right)+{\displaystyle \underset{5}{\overset{6}{\int }}f\left(t\right)dt}\\ =7+\left(-\left(\frac{1}{2}\right)bh+b_1{h}_1\right)\end{array}$

Substitute 2for $b$, 2 for $h$, 1 for $b_1$, and 2 for $h_1$.

$\begin{array}{c}g\left(6\right)=7+\left(-\frac{1}{2}\left(2\right)\left(2\right)+\left(1\right)\left(2\right)\right)\\ =7-4\\ =3\end{array}$

Therefore, the $g\left(6\right)$ is $\underset{\_}{3}$.

(b)

To determine

The interval g.

Answer to Problem 3E

The function g is increasing at the interval $\underset{\_}{(0,3)}$.

Explanation of Solution

Given information:

The integral function is $g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$.

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 $\underset{\_}{\left(0,3\right)}$.

(c)

To determine

The maximum value of g.

Answer to Problem 3E

The maximum value of g is lies at $\underset{\_}{x=3}$.

Explanation of Solution

Given information:

The integral function is $g\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$.

Calculation:

Refer part (a) calculation

Maximum value of g lies at $x=3$

Therefore, the maximum value of g lies at $\underset{\_}{x=3}$.

(d)

To determine

To sketch: The rough graph of g.

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 $g\left(0\right)$, $g\left(1\right)$, $g\left(2\right)$, $g\left(3\right)$, and $g\left(6\right)$ respectively.

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