Let f be a non-negative function in [0,1] and twice differentiable in (0,1). If ∫ _0^x√1-(f^'(t))^2 dt=∫ _0^xf(t) dt,0≤ x≤ 1 and f(0)=0, then…

Newton - Leibnitz rule

$$\frac{d}{dx}({\int }_{f(x)}^{g(x)}h(x)dx)=g'(x)h(g(x))-f'(x)h(f(x))$$dxd​(∫f(x)g(x)​h(x)dx)=g′(x)h(g(x))−f′(x)h(f(x))

Given that

$${\int }_{0}^{x}\sqrt{1-{({f}^{'}(t))}^{2}}dt={\int }_{0}^{x}f(t)dt$$∫0x​1−(f′(t))2​dt=∫0x​f(t)dt

Differentiate $w.r.t.x$w.r.t.x

$$\sqrt{1-{({f}^{'}(x))}^{2}}=f(x)$$1−(f′(x))2​=f(x)

Squaring on both sides

$$\Rightarrow 1-{({f}^{'}(x))}^{2}=(f(x){)}^{2}$$⇒1−(f′(x))2=(f(x))2

$$\Rightarrow {({f}^{'}(x))}^{2}=1-(f(x){)}^{2}$$⇒(f′(x))2=1−(f(x))2

$$f(x)=y\Rightarrow f'(x)=\frac{dy}{dx}$$f(x)=y⇒f′(x)=dxdy​

$$\Rightarrow {(\frac{dy}{dx})}^{2}=1-{y}^{2}\Rightarrow \frac{dy}{dx}=\pm \sqrt{1-{y}^{2}}$$⇒(dxdy​)2=1−y2⇒dxdy​=±1−y2​

$$\Rightarrow \frac{dy}{\sqrt{1-{y}^{2}}}=\pm dx$$⇒1−y2​dy​=±dx

By using Variable Separable form

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

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

$$\therefore {\mathrm{sin}}^{-1}y=x\Rightarrow y=\mathrm{sin}x$$∴sin−1y=x⇒y=sinx

Now $\displaystyle \underset{x\rightarrow 0}{\mathrm{lim}}\frac{1}{{x}^{2}}{\int }_{0}^{x}\mathrm{sin}tdt$x→0lim​x21​∫0x​sintdt

It is in the form of $\frac{0}{0}$00​ so apply L hospital's rule

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\frac{d}{dx}[{\int }_{0}^{x}\mathrm{sin}tdt]}{\frac{d({x}^{2})}{dx}}$$=x→0lim​dxd(x2)​dxd​[∫0x​sintdt]​

$$=\frac{1}{2}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}$$=21​x→0lim​xsinx​

$$=\frac{1}{2}$$=21​