Chapter G, Problem 1E

(a)

To determine

To show: The inequality $\frac{1}{3}<\ln 1.5<\frac{5}{12}$ by comparing the areas.

Explanation of Solution

Definition used:

Natural logarithmic function:

It is defined as $\ln x={\displaystyle \underset{1}{\overset{x}{\int }}\frac{1}{t}}dt$, where $x>0$.

Geometrically the natural logarithmic function can be interpreted as the area under the curve $y=\frac{1}{t}$ from $t=1$ to $t=x$, where $x>1$.

Calculation:

The given inequality is $\frac{1}{3}<\ln 1.5<\frac{5}{12}$.

Geometrically, the function $\ln 1.5$ can be interpret as $y=\frac{1}{x}$ from 1 to 1.5.

Use online graphing calculator and sketch the graph of the curve as shown below in Figure 1.

From Figure 1, the area under the curve $y=\frac{1}{x}$ from 1 to 1.5 can be interpret by comparing the areas of the rectangle BCDE and trapezoid ABCD.

The area of the rectangle BCDE is computed below.

$\begin{array}{c}\text{Area of BCDE}=bh\\ =\left(\frac{2}{3}-0\right)\left(\frac{1}{2}-1\right)\\ =\frac{2}{3}\times \frac{1}{2}\\ =\frac{1}{3}\end{array}$

The area of the trapezoid ABCD is

$\begin{array}{c}\text{Area of ABCD}=\frac{1}{2}\left(a+b\right)h\\ =\frac{1}{2}\left(\frac{2}{3}+1\right)\frac{1}{2}\\ =\frac{1}{4}\times \frac{5}{3}\\ =\frac{5}{12}\end{array}$

Comparing these two areas with the function $\ln 1.5$.

Thus, it is clear that $\frac{1}{3}<\ln 1.5<\frac{5}{12}$.

(b)

To determine

To estimate: The function $\ln 1.5$ with $n=10$ using the midpoint rule.

Answer to Problem 1E

The estimated value of $\ln 1.5$ is approximately 0.4054.

Explanation of Solution

Definition used:

Midpoint rule:

It is defined as $M_n\approx {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\Delta x\left[f\left(\overline{x_1}\right)+f\left(\overline{x_2}\right)+\cdots +f\left(\overline{x_n}\right)\right]$.

Here, $\Delta x=\frac{b-a}{n}$ and $\overline{x_i}=\frac{1}{2}\left(x_{i-1}+x_i\right)$

Calculation:

The equation of the curve is $f\left(t\right)=\frac{1}{t}$.

Given that $n=10$.

From part (a), the function $\ln 1.5$ can be defined as $\ln 1.5={\displaystyle \underset{1}{\overset{1.5}{\int }}\frac{1}{t}}dt$.

Then,

$\begin{array}{c}\Delta x=\frac{b-a}{n}\\ =\frac{1.5-1}{10}\\ =\frac{0.5}{10}\\ =0.05\end{array}$

Use the definition of midpoint rule to estimate the value of $\ln 1.5$ as follows.

$\begin{array}{c}{M}_n\approx {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\Delta x\left[f\left(\overline{x_1}\right)+f\left(\overline{x_2}\right)+\cdots +f\left(\overline{x_n}\right)\right]\\ \ln 1.5=0.05\left[f\left(\frac{1+1.05}{2}\right)+f\left(\frac{1.05+1.1}{2}\right)+\cdots +f\left(\frac{1.45+1.5}{2}\right)\right]\\ =0.05\left[f\left(1.025\right)+f\left(1.075\right)+\cdots +f\left(1.475\right)\right]\\ =0.05\left[\frac{1}{1.025}+\frac{1}{1.075}+\cdots +\frac{1}{1.475}\right]\end{array}$

On further simplification,

$\begin{array}{c}\ln 1.5=0.05\left[\frac{1}{1.025}+\frac{1}{1.075}+\cdots +\frac{1}{1.475}\right]\\ =0.05\left[8.108\right]\\ \approx 0.4054\end{array}$

Thus the estimated value of $\ln 1.5$ is approximately 0.4054.