Chapter 5.4, Problem 2E

Let $g(x)={\displaystyle {\int }_0^{x}f(t)}dt$, where f is the function whose graph is shown.

(a) Evaluate g(x) for x = 0, 1, 2, 3, 4, 5, and 6.

(b) Estimate g(7).

(c) Where does g have a maximum value? Where does it have a minimum value?

(d) Sketch a rough graph of g.

To determine

The value of $g(x)$, for $x=0$, $x=1$, $x=2$, $x=3$, $x=4$, $x=5$, and $x=6$.

Answer to Problem 2E

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

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

The value of $g(2)$ is $\underset{\_}{0}$.

The value of $g(3)$ is $\underset{\_}{-\frac{1}{2}}$.

The value of $g(4)$ is $\underset{\_}{0}$.

The value of $g(5)$ is $\underset{\_}{1.5}$.

The value of $g(6)$ is $\underset{\_}{4}$.

Explanation of Solution

Given information:

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

Draw the graph for the 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).

The area of the shaded triangle is the function of t with limits 0 to 1.

Modify Equation (2).

$g\left(1\right)=\frac{1}{2}bh$

Substitute 1 for b and 1 for h.

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

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

Draw the graph for the 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 the shaded triangle is the function of t with limits 1 to 2.

Substitute $\frac{1}{2}$ 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).

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

Substitute 1 for b and -1 for h.

$\begin{array}{c}g(2)=\frac{1}{2}+\frac{1}{2}bh\\ =\frac{1}{2}+\frac{1}{2}\left(1\right)\left(-1\right)\\ =0\end{array}$

Therefore, $g\left(2\right)$ is $\underset{\_}{0}$.

Draw the graph for the 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 the shaded triangle is the function of t with limits 2 to 3.

Substitute $0$ 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).

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

Substitute 1 for b and -1 for h.

$\begin{array}{c}g\left(3\right)=\frac{1}{2}bh\\ =\frac{1}{2}\left(1\right)\left(-1\right)\end{array}$

Therefore, $g\left(3\right)$ is $\underset{\_}{-\frac{1}{2}}$.

Draw the graph for the 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 to Figure 4.

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

Substitute $-\frac{1}{2}$ 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).

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

Substitute 1 for b and 1 for h.

$\begin{array}{c}g\left(4\right)=-\frac{1}{2}+\frac{1}{2}bh\\ =-\frac{1}{2}+\frac{1}{2}\left(1\right)\left(1\right)\\ =0\end{array}$

Therefore, $g\left(4\right)$ is 0.

Draw the graph for the 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.

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

Substitute $0$ for $g\left(4\right)$ and $\left(b_1{h}_1+\frac{1}{2}{b}_2{h}_2\right)$ for ${\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}$ in Equation (6).$\begin{array}{c}g\left(5\right)=g\left(4\right)+{\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}\\ =0+\left(b_1{h}_1+\frac{1}{2}{b}_2{h}_2\right)\\ =b_1{h}_1+\frac{1}{2}{b}_2{h}_2\end{array}$

Substitute 1 for $b_1$, 1 for $h_1$, 1 for $b_2$, and 1 for $h_2$.

$\begin{array}{c}g\left(5\right)=b_1{h}_1+\frac{1}{2}{b}_2{h}_2\\ =\left(1\times 1\right)+\left(\frac{1}{2}\times 1\times 1\right)\\ =1+\frac{1}{2}\\ =1.5\end{array}$

Therefore, the $g\left(5\right)$ is $\underset{\_}{1.5}$.

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

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

Substitute 6 for x in Equation (1).

$\begin{array}{c}g\left(6\right)={\displaystyle \underset{0}{\overset{5}{\int }}f\left(t\right)dt}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)dt}+{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)dt}+{\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}+{\displaystyle \underset{5}{\overset{6}{\int }}f\left(t\right)dt}\\ =g\left(5\right)+{\displaystyle \underset{5}{\overset{6}{\int }}f\left(t\right)dt}\end{array}$ (7)

Refer to Figure 6.

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

Substitute 1.5 for $g\left(5\right)$ and $\left(b_1{h}_1+b_2{h}_2+\frac{1}{2}{b}_3{h}_3\right)$ for ${\displaystyle \underset{5}{\overset{6}{\int }}f\left(t\right)dt}$ in Equation (7).$\begin{array}{c}g\left(6\right)=g\left(5\right)+{\displaystyle \underset{5}{\overset{6}{\int }}f\left(t\right)dt}\\ =1.5+\left(b_1{h}_1+b_2{h}_2+\frac{1}{2}{b}_3{h}_3\right)\end{array}$

Substitute 1 for $b_1$, 1 for $h_1$, 1 for $b_2$, and 1 for $h_2$, 1 for $b_3$, and 1 for $h_3$.

$\begin{array}{c}g\left(6\right)=1.5+\left(b_1{h}_1+b_2{h}_2+\frac{1}{2}{b}_3{h}_3\right)\\ =1.5+\left[\left(1\times 1\right)+\left(1\times 1\right)+\left(\frac{1}{2}\times 1\times 1\right)\right]\\ =4\end{array}$

Therefore, $g\left(6\right)$ is $\underset{\_}{4}$.

To determine

The value of $g(x)$ at $x=7$.

Answer to Problem 2E

The value of $g(7)$ is $\underset{\_}{6.2}$.

Explanation of Solution

Given information:

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

Calculation:

Draw the graph for the calculation of $g\left(7\right)$ as in Figure (7).

Determine $g\left(7\right)$ using Equation (1).

Substitute 7 for x in Equation (1).

$\begin{array}{c}g\left(7\right)={\displaystyle \underset{0}{\overset{5}{\int }}f\left(t\right)dt}\\ =\left(\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}+{\displaystyle \underset{1}{\overset{2}{\int }}f\left(t\right)dt}+{\displaystyle \underset{2}{\overset{3}{\int }}f\left(t\right)dt}+{\displaystyle \underset{3}{\overset{4}{\int }}f\left(t\right)dt}+{\displaystyle \underset{4}{\overset{5}{\int }}f\left(t\right)dt}\\ +{\displaystyle \underset{5}{\overset{6}{\int }}f\left(t\right)dt+{\displaystyle \underset{6}{\overset{7}{\int }}f\left(t\right)dt}}\end{array}\right)\\ =g\left(6\right)+{\displaystyle \underset{6}{\overset{7}{\int }}f\left(t\right)dt}\end{array}$ (8)

Refer to Figure 7.

The area of the shaded portion is the function of t with limits 6 to 7; it is bounded by a curved shape. Hence, take the approximate value from the graph

Take the approximate area of the shaded portion as 2.2

Substitute 4 for $g\left(6\right)$ and 2.2 for ${\displaystyle \underset{6}{\overset{7}{\int }}f\left(t\right)dt}$ in Equation (8).

$\begin{array}{c}g\left(7\right)=g\left(6\right)+{\displaystyle \underset{6}{\overset{7}{\int }}f\left(t\right)dt}\\ =4+2.2\\ =6.2\end{array}$

Therefore, $g\left(7\right)$ is $\underset{\_}{6.2}$.

To determine

The maximum and minimum value of g.

Answer to Problem 2E

The minimum value of g lies at $\underset{\_}{x=3}$ and maximum value lies at $\underset{\_}{x=7}$.

Explanation of Solution

Given information:

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

Calculation:

From the calculation in Part (A),

The minimum value of function of g is $-\frac{1}{2}$ at $x=3$.

The maximum value of function of g is 6.2 at $x=7$

Therefore, the minimum value of g lies at $\underset{\_}{x=3}$ and maximum value is lies at $\underset{\_}{x=7}$.

To determine

To Sketch: The rough graph of g.

Explanation of Solution

Plot the graph for function f using the calculated values of 0, $\frac{1}{2}$, 0, $-\frac{1}{2}$, 0, 1.5, 4, and for the functions $g\left(0\right)$, $g\left(1\right)$, $g\left(2\right)$, $g\left(3\right)$, $g\left(4\right)$, $g\left(5\right)$, $g\left(6\right)$, and $g\left(7\right)$ respectively.

Draw the graph for the function of f as in Figure (8).