Understand the boundary curves

We are given:

• $y=x$
• $x=e$
• $y=\dfrac{1}{x}$
• positive $x$-axis, i.e. $y=0$

We first note the key intersection points:

• $y=x$ and $y=\dfrac{1}{x}$ intersect at $x=\frac{1}{x} \implies x^2=1 \implies x=1$ Since we are on the positive $x$-axis side, the point is $(1,1)$.

• $y=x$ meets the $x$-axis at $(0,0)$.

• $x=e$ meets the curve $y=\dfrac{1}{x}$ at $\left(e,\dfrac{1}{e}\right)$.

• $x=e$ meets the $x$-axis at $(e,0)$.

Describe the enclosed region

The closed region is formed as follows:

• from $(0,0)$ to $(1,1)$ along $y=x$,
• from $(1,1)$ to $\left(e,\dfrac{1}{e}\right)$ along $y=\dfrac{1}{x}$,
• from $\left(e,\dfrac{1}{e}\right)$ to $(e,0)$ along $x=e$,
• from $(e,0)$ back to $(0,0)$ along the $x$-axis.

So the upper boundary is:

• $y=x$ for $0\le x\le 1$,
• $y=\dfrac{1}{x}$ for $1\le x\le e$.

The lower boundary is $y=0$ throughout.

Set up the area integral

Hence area $A=\int_0^1 x\,dx+\int_1^e \frac{1}{x}\,dx$

Evaluate the integrals

First, $\int_0^1 x\,dx=\left[\frac{x^2}{2}\right]_0^1=\frac{1}{2}$

Second, $\int_1^e \frac{1}{x}\,dx=\left[\ln x\right]_1^e=\ln e-\ln 1=1-0=1$

Add the areas

$A=\frac{1}{2}+1=\frac{3}{2}$

Match with the options

$\boxed{\frac{3}{2}}$

So the correct option is B.