通过使用换元法 $x = \frac{1}{u}$，证明 $\int_{2\sqrt{2}}^{4} \frac{1}{x\sqrt{x^2 - 4}} \, \mathrm{d}x = \frac{\pi}{24}$。 $O$ 为原点，$A$ 是曲线 $y = \frac{1}{x\sqrt{x^2 - 4}}$ 上 $x = 2\sqrt{2}$ 处的点。区域 $R$ 由曲线 $y = \frac{1}{x\sqrt{x^2 - 4}}$、直线 $OA$、直线 $x = 4$ 以及 $x$ 轴围成。 求 $R$ 的精确面积。

1. Differentiate the substitution to find an expression for $\mathrm{d}x$ in terms of $\mathrm{d}u$.

$x = \frac{1}{u}$
$\frac{\mathrm{d}x}{\mathrm{d}u} = -\frac{1}{u^2} \implies \mathrm{d}x = -\frac{1}{u^2} \, \mathrm{d}u$

2. Determine the new limits of integration in terms of $u$.

$\text{When } x = 4, \quad u = \frac{1}{4}$
$\text{When } x = 2\sqrt{2}, \quad u = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

3. Substitute into the integral and simplify to a standard integrable form.

$\begin{aligned} \int_{2\sqrt{2}}^{4} \frac{1}{x\sqrt{x^2 - 4}} \, \mathrm{d}x &= \int_{\frac{\sqrt{2}}{4}}^{\frac{1}{4}} \frac{1}{\left(\frac{1}{u}\right)\sqrt{\left(\frac{1}{u}\right)^2 - 4}} \left(-\frac{1}{u^2}\right) \, \mathrm{d}u \\ &= \int_{\frac{\sqrt{2}}{4}}^{\frac{1}{4}} \frac{-1}{u \sqrt{\frac{1 - 4u^2}{u^2}}} \, \mathrm{d}u \\ &= \int_{\frac{\sqrt{2}}{4}}^{\frac{1}{4}} \frac{-1}{u \left(\frac{\sqrt{1 - 4u^2}}{u}\right)} \, \mathrm{d}u \\ &= \int_{\frac{\sqrt{2}}{4}}^{\frac{1}{4}} \frac{-1}{\sqrt{1 - 4u^2}} \, \mathrm{d}u \\ &= \int_{\frac{1}{4}}^{\frac{\sqrt{2}}{4}} \frac{1}{\sqrt{1 - 4u^2}} \, \mathrm{d}u \end{aligned}$

4. Integrate using the standard formula and evaluate the definite integral.

$\begin{aligned} \int_{\frac{1}{4}}^{\frac{\sqrt{2}}{4}} \frac{1}{\sqrt{1 - 4u^2}} \, \mathrm{d}u &= \int_{\frac{1}{4}}^{\frac{\sqrt{2}}{4}} \frac{1}{\sqrt{1 - (2u)^2}} \, \mathrm{d}u \\ &= \left[ \frac{1}{2}\sin^{-1}(2u) \right]_{\frac{1}{4}}^{\frac{\sqrt{2}}{4}} \\ &= \frac{1}{2} \left( \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) - \sin^{-1}\left(\frac{1}{2}\right) \right) \\ &= \frac{1}{2} \left( \frac{\pi}{4} - \frac{\pi}{6} \right) \\ &= \frac{1}{2} \left( \frac{3\pi - 2\pi}{12} \right) \\ &= \frac{\pi}{24} \end{aligned}$

1. Find the $y$-coordinate of point $A$.

At point $A$, $x = 2\sqrt{2}$.
$y = \frac{1}{2\sqrt{2}\sqrt{(2\sqrt{2})^2 - 4}} = \frac{1}{2\sqrt{2}\sqrt{8 - 4}} = \frac{1}{4\sqrt{2}}$

2. Calculate the area of the triangular region bounded by the line $OA$, the $x$-axis, and the vertical line $x = 2\sqrt{2}$.

$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$
$= \frac{1}{2} \times 2\sqrt{2} \times \frac{1}{4\sqrt{2}} = \frac{1}{4}$

3. Calculate the total exact area of $R$ by summing the area of the triangle and the area under the curve from part (i).

$\text{Total Area} = \text{Area of triangle} + \int_{2\sqrt{2}}^{4} \frac{1}{x\sqrt{x^2 - 4}} \, \mathrm{d}x$
Using the result from (i),
$\text{Total Area} = \frac{1}{4} + \frac{\pi}{24}$