Chapter 5.4, Problem 4E

(a)

To determine

The value of $g\left(0\right)$ and $g\left(6\right)$.

Answer to Problem 4E

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

The value of $g(6)$ is $\underset{\_}{0}$

Explanation of Solution

Given information:

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

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

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.

Determine $g\left(0\right)$ using Equation (1).

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 $\underset{\_}{0}$.

Determine $g\left(6\right)$ 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 $g\left(6\right)$ is 0.

(b)

To determine

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

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

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

Calculation:

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 Figure (1).

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

Modify Equation (2).

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

Add 80% of unit square.

Substitute 1 for b, 2 for h.

$\begin{array}{c}g\left(1\right)=bh+80\%\\ =\left(1\right)\left(2\right)+0.8\\ =2.8\end{array}$

Therefore, $g\left(1\right)$ is 2.8.

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 Figure (2).

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

Substitute 2.8 for $g\left(1\right)$ and bh for ${\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}$ in Equation (3).

$\begin{array}{c}g(2)=g\left(1\right)+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}\\ =2.8+bh\end{array}$

Add 90% of unit square.

Substitute 1 for b and 1 for h.

$\begin{array}{c}g(2)=2.8+1+0.9\\ =4.7\end{array}$

Therefore, $g\left(2\right)$ is 4.7.

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 Figure 3.

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

Substitute 4.7 for $g\left(2\right)$ and bh for ${\displaystyle \underset{3}{\overset{2}{\int }}f\left(t\right)dt}$ in Equation (4).

$\begin{array}{c}g\left(3\right)=g\left(2\right)+{\displaystyle \underset{3}{\overset{2}{\int }}f\left(t\right)dt}\\ =4.7+\frac{1}{2}bh\end{array}$

Substitute 1 for b and 1.2 for h.

$\begin{array}{c}g\left(3\right)=g\left(2\right)+{\displaystyle \underset{3}{\overset{2}{\int }}f\left(t\right)dt}\\ =4.7+\frac{1}{2}\left(1\right)\left(1.2\right)\\ =5.3\end{array}$

Therefore, $g\left(3\right)$ is 5.3.

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

Determine $g\left(4\right)$ using Equation (1).

Substitute 4 for x in Equation (1).

$\begin{array}{c}g\left(4\right)={\displaystyle \underset{0}{\overset{4}{\int }}f\left(t\right)dt}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)dt}+{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)dt}\\ =g\left(3\right)+{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)dt}\end{array}$ (5)

Refer Figure 4.

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

Substitute 5.3 for $g\left(3\right)$ and $\left(\frac{1}{2}bh\right)$ for ${\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)dt}$ in Equation (5).

$\begin{array}{c}g\left(4\right)=g\left(3\right)-{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)dt}\\ =5.3-\frac{1}{2}bh\end{array}$

Substitute 1 for b and 1 for h.

$\begin{array}{c}g\left(4\right)=5.3-\frac{1}{2}bh\\ =5.3-\frac{1}{2}\left(1\right)\left(1.2\right)\\ =4.7\end{array}$

Therefore, $g\left(3\right)$ is 5.3.

Draw the graph for calculation of $g\left(5\right)$ as in Figure 5.

Determine $g\left(5\right)$ using Equation (1).

Substitute 5 for x in Equation (1).

$\begin{array}{c}g\left(5\right)={\displaystyle \underset{0}{\overset{5}{\int }}f\left(t\right)dt}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)dt}+{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)dt}+{\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}\\ =g\left(4\right)+{\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}\end{array}$ (6)

Refer to Figure 5.

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

Substitute 4.7 for $g\left(4\right)$ and bh for ${\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}$ in Equation (6).

$\begin{array}{c}g\left(5\right)=g\left(4\right)-{\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}\\ =4.7-bh\end{array}$

Subtract 90% of third unit square.

Substitute 1 for, 1 for h.

$\begin{array}{c}g\left(5\right)=4.7-bh\\ =4.7-\left(1\right)\left(1\right)-0.9\\ =2.8\end{array}$

(c)

To determine

The interval when g is increasing.

Answer to Problem 4E

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

Explanation of Solution

Given information:

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

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

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

(d)

To determine

The location of maximum value of g.

Answer to Problem 4E

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

Explanation of Solution

Given information:

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

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

Calculation:

Refer part (a) calculation

Maximum value of g lies at $x=3$

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

(e)

To determine

To Sketch: The rough graph of g.

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 $g\left(1\right)$, $g\left(2\right)$, $g\left(3\right)$, $g\left(4\right)$, and $g\left(5\right)$ respectively.

Show the graph for function f as in Figure 6.

(f)

To determine

To Sketch: The graph $g^{\prime }\left(x\right)$.

Explanation of Solution

Show the graph for function $g^{\prime }\left(x\right)$ as in Figure 7.

Draw the graph $g^{\prime }\left(x\right)$ as shown in Figure 7.