Mathematics - Definite Integration Question with Solution | TestHub

Given ${\int }_0^x\sqrt{1-{\left(f^{\prime }(t)\right)}^2}dt={\int }_0^{x}f(t)dt$, $$0\le x\le 1$$ Applying Leibnitz theorem, we get $$\begin{array}{rlrl}& & \sqrt{1-{\left(f^{\prime }(x)\right)}^2}& =f(x)\\ \Rightarrow & & 1-{\left(f^{\prime }(x)\right)}^2& =f^2(x)\\ \Rightarrow & & {\left(f^{\prime }(x)\right)}^2& =1-f^2(x)\\ \Rightarrow & & f^{\prime }(x)& =\pm \sqrt{1-f^2(x)}\\ \Rightarrow & & \frac{dy}{dx}& =\pm \sqrt{1-y^2}\\ & \text{ where }y=f(x)\Rightarrow & \frac{dy}{\sqrt{1-y^2}}=\pm dx& \end{array}$$ On integrating both sides, we get $$\begin{array}{rl}& {\sin }^{-1}(y)=\pm x+C\\ & \because f(0)=0\Rightarrow C=0\Rightarrow y=\pm \sin x\\ & y=\sin x=f(x)\text{ given }f(x)\ge 0\text{ for }\\ & x\in [0,1]\end{array}$$ It is known that $\sin x<x,\forall x\in R^{+}$ $\therefore \sin \left(\frac{1}{2}\right)<\frac{1}{2}\Rightarrow f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $\sin \left(\frac{1}{3}\right)<\frac{1}{3}\Rightarrow f\left(\frac{1}{3}\right)<\frac{1}{3}$