The diagram above shows the curve $C_1$ with equation $y = \frac{\ln x}{x^2}$, where $x \ge 1$.

1. Find the first derivative of $y$ with respect to $x$ using the quotient rule.

$y = \frac{\ln x}{x^2}$
Using the quotient rule:
$\frac{\text{d}y}{\text{d}x} = \frac{x^2 \left(\frac{1}{x}\right) - (\ln x)(2x)}{(x^2)^2}$
$\frac{\text{d}y}{\text{d}x} = \frac{x - 2x \ln x}{x^4}$
$\frac{\text{d}y}{\text{d}x} = \frac{1 - 2\ln x}{x^3}$

2. Equate the derivative to zero to find the $x$-coordinate of the turning point.

At a turning point, $\frac{\text{d}y}{\text{d}x} = 0$:
$\frac{1 - 2\ln x}{x^3} = 0$
Since $x \ge 1$, $x^3 \neq 0$, so the numerator must be zero:
$1 - 2\ln x = 0$
$2\ln x = 1$
$\ln x = \frac{1}{2}$
$x = \text{e}^{\frac{1}{2}} = \sqrt{\text{e}}$

3. Substitute the $x$-coordinate back into the original equation to obtain the corresponding $y$-coordinate.

Substitute $x = \text{e}^{\frac{1}{2}}$ into the original equation for $y$:
$y = \frac{\ln\left(\text{e}^{\frac{1}{2}}\right)}{\left(\text{e}^{\frac{1}{2}}\right)^2}$
$y = \frac{\frac{1}{2}}{\text{e}} = \frac{1}{2\text{e}}$
Thus, the exact coordinates of the turning point are $\left(\sqrt{\text{e}}, \frac{1}{2\text{e}}\right)$ (shown).

1. Find the point(s) of intersection of $C_1$ with the coordinate axes.

For $C_1$: $y = \frac{\ln x}{x^2}$, defined for $x \ge 1$.
Let $y = 0$:
$\frac{\ln x}{x^2} = 0$
$\ln x = 0 \implies x = 1$
The $x$-intercept for $C_1$ is $(1, 0)$.
Since the domain is $x \ge 1$, $x$ cannot be $0$, so there are no $y$-intercepts.

2. Find the point(s) of intersection of $C_2$ with the coordinate axes.

For $C_2$: $\left(x - \sqrt{\mathrm{e}}\right)^2 + (2y)^2 = 1$, where $y \ge 0$.
Let $y = 0$:
$\left(x - \sqrt{\mathrm{e}}\right)^2 = 1$
$x - \sqrt{\mathrm{e}} = \pm 1 \implies x = \sqrt{\mathrm{e}} - 1 \text{ or } x = \sqrt{\mathrm{e}} + 1$
The $x$-intercepts for $C_2$ are $(\sqrt{\mathrm{e}} - 1, 0)$ and $(\sqrt{\mathrm{e}} + 1, 0)$.
Let $x = 0$:
$\left(-\sqrt{\mathrm{e}}\right)^2 + 4y^2 = 1$
$\mathrm{e} + 4y^2 = 1 \implies 4y^2 = 1 - \mathrm{e}$
Since $1 - \mathrm{e} < 0$, there are no real solutions for $y$, so $C_2$ does not intersect the $y$-axis.

3. Identify the shapes to accurately sketch both curves.

$C_1$ starts at $(1, 0)$, climbs to its turning point $\left(\sqrt{\mathrm{e}}, \frac{1}{2\mathrm{e}}\right)$ found in part (i), and asymptotically approaches the $x$-axis.

$C_2$ can be rewritten as:
$\frac{(x - \sqrt{\mathrm{e}})^2}{1^2} + \frac{y^2}{(1/2)^2} = 1$
This represents the upper half ($y \ge 0$) of an ellipse centered at $(\sqrt{\mathrm{e}}, 0)$, with a horizontal semi-axis of $1$ and a vertical semi-axis of $\frac{1}{2}$. Its peak is at $(\sqrt{\mathrm{e}}, \frac{1}{2})$.

Note that $\sqrt{\mathrm{e}} - 1 \approx 0.649 < 1$, so $C_2$ starts to the left of $C_1$.
Also, $\frac{1}{2\mathrm{e}} \approx 0.184 < 0.5$, so the peak of $C_2$ is much higher than the peak of $C_1$.

1. Find the point of intersection between $C_1$ and $C_2$ using a graphing calculator.

From part (ii), $C_1$ intersects the $x$-axis at $x=1$, and $C_2$ intersects the $x$-axis at $x=\sqrt{\text{e}}+1$.
Let $y_1 = \frac{\ln x}{x^2}$ and $y_2 = \frac{1}{2}\sqrt{1 - (x - \sqrt{\text{e}})^2}$ (since $y \ge 0$).
Using a graphing calculator to solve $y_1 = y_2$:
$x \approx 2.6079$

2. Set up the integral for the smaller region bounded by $C_1$, $C_2$, and the $x$-axis.

From the sketch in (ii), the curves enclose two bounded regions with the $x$-axis. The smaller region lies on the right, between $x=1$ and $x=\sqrt{\text{e}}+1$.
It is bounded above by $C_1$ for $x \in [1, 2.6079]$ and by $C_2$ for $x \in [2.6079, \sqrt{\text{e}}+1]$.

$\text{Area} = \int_{1}^{2.6079} \frac{\ln x}{x^2} \,\mathrm{d}x + \int_{2.6079}^{\sqrt{\text{e}}+1} \frac{1}{2}\sqrt{1 - (x - \sqrt{\text{e}})^2} \,\mathrm{d}x$

3. Evaluate the integral numerically using a graphing calculator.

Evaluating the integrals using the GC:
$\text{Area} \approx 0.24900 + 0.00359 = 0.25259$
$\text{Area} = 0.253 \text{ (to 3 s.f.)}$