Down the rabbit hole ......

Using the Fundamental Theorem of Calculus, we take the derivative of both sides of the equation defining $f(x)$ to find that

$f'(x) = \dfrac{x}{f(x)}$ .

Now let $y = f(x)$ . Then our differential equation becomes

$\dfrac{dy}{dx} = \dfrac{x}{y} \Longrightarrow y dy = x dx \Longrightarrow y^{2} = x^{2} + C$

for some constant $C$ , (noting that I've "absorbed" a factor of $2$ into the constant). But since $f(1) = 0$ , we see that $0 = 1^{2} + C \Longrightarrow C = -1$ , and so, since it was specified that $f(x)$ is non-negative valued, we conclude that $f(x) = \sqrt{x^{2} - 1}$ .

Thus $S = \displaystyle\int_{\sqrt{3}}^{2} \sqrt{x^{2} - 1} dx$ .

This integral can be solved using trigonometric substitution; let $x = \sec(\theta)$ . Then $dx = \sec(\theta)\tan(\theta) d(\theta)$ and $\sqrt{x^{2} - 1} = \tan(\theta)$ . Ignoring the upper and lower bounds for now, the integral then becomes

$\displaystyle\int \sec(\theta)\tan^{2}(\theta) d(\theta)$ .

Now use integration by parts; let $u = \tan(\theta)$ and $dv = \sec(\theta)\tan(\theta) d(\theta)$ .

Then $du = \sec^{2}(\theta) d(\theta)$ and $v = \sec(\theta)$ . So now

$\displaystyle\int \sec(\theta)\tan^{2}(\theta) d(\theta) = \sec(\theta)\tan(\theta) - \displaystyle\int \sec^{3}(\theta) d(\theta) =$

$sec(\theta)\tan(\theta) - \displaystyle\int (\tan^{2}(\theta) + 1)\sec(\theta) d(\theta)$ .

Thus $\displaystyle\int \sec(\theta)\tan^{2}(\theta) d(\theta) = (\frac{1}{2})(\sec(\theta)\tan(\theta) - \ln(\sec(\theta) + \tan(\theta)))$ .

Therefore $S = (\frac{1}{2})(x\sqrt{x^{2} - 1} - \ln(x + \sqrt{x^{2} - 1}))$

evaluated from $x = \sqrt{3}$ to $x = 2$ . Plugging in these values gives us that

$S = (\frac{1}{2})(2\sqrt{3} - \ln(2 + \sqrt{3}) - \sqrt{6} + \ln(\sqrt{2} + \sqrt{3})) = 0.42193....$ ,

and $\lfloor 100*S \rfloor = \boxed{42}$ , the answer to everything and referred to often in 'Alice in Wonderland', (hence the title of this question).

$\begin{aligned} f(x)&=\int_{1}^{x} \dfrac{t}{f(t)} \mathrm{d}t\\ f'(x) &=\dfrac{x}{f(x)} \end{aligned}$

Now let $y=f(x)$ . So,

$\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{x}{y}\\ \int y \mathrm{d}y=\int x \mathrm{d}x \\ y=f(x)=\sqrt {x^2-1}$

Because $f(x)$ is non-negative and $f(1)=0$ .

So now, we need to find $\int_{\sqrt {3}}^{2} \sqrt{x^2-1} \mathrm{d}x$ which can be done by simply substituting $x=\sec{\theta}$ .

The value $\approx 0.4219$ . So, $\lfloor 100S\rfloor=\boxed{42}$

$f'(x)=\dfrac{x}{f(x)}$

$f(x)=\sqrt{x^2-1}$ , since $f(1)=0$

$\displaystyle \int_{\sqrt{3}}^{2} \sqrt{x^2-1} dx=.42195...=S$

Thus, $\lfloor{100S}\rfloor=\boxed{42}$ .