Answer
a) On the interval $[1,3/2]$, the area of the blue rectangular region ($\frac{1}{3}$) is less than the area under the curve ($\ln 1.5$), the latter being less than the area of the trapezoid ($\frac{5}{12}$). See details below.
b) $\color{blue}{0.405}$ (see details below)
Work Step by Step
a)
Consider the graph of the $y=\ln x, x\gt 0,$ over the closed interval $[1,1.5] = [1,3/2]$ (see blue curve in the diagram).
Compare the following areas.
i) The blue rectangular region (see diagram) with vertices at $(1,0),(3/2,0),(3/2,2/3)$, and $(1,2/3)$.
The area of this rectangular region is $1/3$ as the width is $3/2-1=1/2$, the length is $2/3-0 = 2/3$, and $(1/2)(2/3) = 1/3$.
ii) The area under the curve $y= 1/x$ over the closed interval $[1,3/2]$. This area is equal to $\displaystyle \int_1^{3/2} \frac{1}{x}\ dx = \ln (3/2) = \ln 1.5$ by the definition of the $\ln$ function (see Definition G.1). In the diagram this is the areas of the blue and green regions together.
iii) The area inside the trapezoid that is determined by the points $(1,0),(3/2,0),(3/2,2/3)$, and $(1,1)$. The height is $h=3/2-1=1/2$ while the two bases have lengths $b_1=1-0=1$ and $b_2=2/3-0=2/3$. Thus, the area of the trapezoid is $\dfrac{b_1+b_2}{2}\cdot h = \dfrac{1+2/3}{2}\cdot \dfrac{1}{2} = (5/6)(1/2) = 5/12$. This area is equal to the areas of the blue, green, and yellow regions together.
Now, the rectangular area in i) is smaller than the area under the curve in ii), and the latter is smaller than the area of the trapezoid in iii). Thus,
$ \displaystyle \frac{1}{3} \lt \ln 1.5 \lt \dfrac{5}{12}$.
b)
Using the Midpoint Rule,
$\begin{align*}
\int_a^b f(x)\ dx &\approx \sum_{k=0}^{n-1} f\left( a + \frac{\Delta t}{2} + k\Delta t\right)\cdot \Delta t
\end{align*}$
with
$f(x) =1/x, n=10,\ a=1,\ b=3/2 = 1.5, \\ \Delta t = (b-a)/n = (1.5-1)/10 = 1/20,$
we get
$\begin{align*}
\ln 1.5 = \int_1^{1.5} 1/x\ dx &\approx \sum_{k=0}^{9} \frac{1}{\left( 1 + (1/20)/2 + k(1/20)\right)}\cdot (1/20) \\
&= \left(\frac{1}{1 + 1/40} + \frac{1}{1+3/40} + \ldots + \frac{1}{1+ 17/40} + \frac{1}{1+ 19/40}\right) \cdot \frac{1}{20} \\
&= \left(\frac{1}{1.025} + \frac{1}{1.075} + \ldots + \frac{1}{1.425} + \frac{1}{1.475}\right) \cdot \frac{1}{20} \\
\color{blue}{\ln 1.5}\ &\color{blue}{\approx 0.405}
\end{align*}$
