Newton - Leibnitz rule 

$\frac{d}{dx}\left({\int }_{f\left(x\right)}^{g\left(x\right)}h\left(x\right)dx\right) = g\text{'}\left(x\right)h\left(g\left(x\right)\right) - f\text{'}\left(x\right)h\left(f\left(x\right)\right)$

Given that

${\int }_0^x\sqrt{1-{\left(f^{\text{'}}\left(t\right)\right)}^2}dt={\int }_0^{x}f\left(t\right)dt$

Differentiate $w.r.t. x$

$\sqrt{1-{\left(f^{\text{'}}\left(x\right)\right)}^2}=f\left(x\right)$

Squaring on both sides
$\Rightarrow 1-{\left(f^{\text{'}}\left(x\right)\right)}^2=\left(f\right(x){)}^2$

$\Rightarrow {\left(f^{\text{'}}\left(x\right)\right)}^2=1-\left(f\right(x){)}^2$

$f\left(x\right) = y \Rightarrow f\text{'}\left(x\right) = \frac{dy}{dx}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}^2=1-y^2 \Rightarrow \frac{dy}{dx}=\pm \sqrt{1-y^2}$

$\Rightarrow \frac{dy}{\sqrt{1-y^2}}=\pm dx$

By using Variable Separable form

${\sin }^{-1}y=\pm x+c$

When $x = 0 , y = 0 \Rightarrow c = 0.$

$\therefore {\sin }^{-1}y=x\Rightarrow y=\sin x$

Now $\underset{x\to 0}{\lim }\frac{1}{x^2}{\int }_0^x\sin tdt$

It is in the form of $\frac{0}{0}$ so apply L hospital's rule

$=\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{d}{dx}}\left[{\int }_0^x\sin tdt\right]}{{\displaystyle \frac{d\left(x^2\right)}{dx}}}$

$= \frac{1}{2}\underset{x\to 0}{\lim }\frac{\sin x}{x}$

$=\frac{1}{2}$