Chapter C, Problem 1E

(a)

To determine

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

Explanation of Solution

Definition used:

The natural logarithmic function is the function defined by $\ln x={\displaystyle \underset{1}{\overset{x}{\int }}\frac{1}{t}dt},x>0$.

If $x>1$, then $\ln x$ can be interpreted geometrically as the area under the hyperbola $y=\frac{1}{t}$ from t = 1 to t = x.

Calculation:

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

Here, $x>1$.

By using the above mentioned definition, $\ln 1.5$ can be interpreted geometrically as the area under the hyperbola $y=\frac{1}{t}$ from t = 1 to t = 1.5. $\left(1,1\right)$, $\left(1,\frac{2}{3}\right)$$\left(\frac{3}{2},\frac{2}{3}\right)$

Use online graphing calculator, graph of the curve $y=\frac{1}{t}$ as shown in Figure 1.

From Figure 1, it is observed that the area is larger than the area of rectangle BCDE and smaller than the area of trapezoid ABCD.

The area of the rectangle BCDE is calculated as follows.

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

The area of the trapezoid ABCD is calculated as follows.

$\begin{array}{c}\text{Area of }ABCD=\frac{1}{2}\left(a+b\right)h\\ =\frac{1}{2}\left(\frac{2}{3}-0+1\right)\left(1.5-1\right)\\ =\frac{1}{2}\left(\frac{2+3}{3}\right)\left(0.5\right)\\ =\frac{1}{2}\left(\frac{5}{3}\right)\frac{1}{2}\\ =\frac{5}{12}\end{array}$

It is known that the value of $\ln 1.5$ is 0.40546.

The decimal value of the fraction $\frac{1}{3}$ is 0.33333 and the decimal value of the fraction of $\frac{5}{12}$ is 0.41666.

By comparing the above values with $\ln 1.5$, it is noticed that $\frac{1}{3}<\ln 1.5<\frac{5}{12}$.

Hence proved.

(b)

To determine

To show: The function $\ln 1.5$ by using Midpoint Rule with $n=10$.

Answer to Problem 1E

The estimated value of $\ln 1.5$ is approximately equal to $\underset{\_}{0.4054}$.

Explanation of Solution

Definition used:

Midpoint Rule:

Midpoint rule can be defined as $M={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\Delta x\left[f\left(x_1*\right)+f\left(x_2*\right)+\cdots +f\left(x_n*\right)\right]$.

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

Natural logarithmic:

The natural logarithmic function is the function defined by $\ln x={\displaystyle \underset{1}{\overset{x}{\int }}\frac{1}{t}dt},x>0$.

Calculation:

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

Here, $n=10$.

From definition of natural logarithmic, the function $\ln 1.5$ can be defined as $\ln 1.5={\displaystyle \underset{1}{\overset{1.5}{\int }}\frac{1}{t}dt}$.

Here, $a=1,b=1.5$ and $n=10$.

Find $\Delta x=\frac{b-a}{n}$ by using the above values.

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

Estimate the value of $\ln 1.5$ by using the definition of midpoint rule as follows.

$\begin{array}{c}M=\Delta x\left[f\left(x_1*\right)+f\left(x_2*\right)+\cdots +f\left(x_n*\right)\right]\\ =0.05\left[f\left(\frac{1+1.05}{2}\right)+f\left(\frac{1.05+1}{2}\right)+\cdots +\left(\frac{1.45+1.5}{2}\right)\right]\\ =0.05\left[f\left(1.025\right)+f\left(1.075\right)+\cdots +f\left(1.475\right)\right]\\ =0.05\left[\frac{1}{1.025}+\frac{1}{1.075}+\cdots +\frac{1}{1.475}\right]\end{array}$

Further simplification,

$\begin{array}{c}\ln 1.5=0.05\left[8.108\right]\\ \approx 0.4054\end{array}$

Therefore, the estimated value of $\ln 1.5$ is approximately equal to $\underset{\_}{0.4054}$.