Question

f(x)=6sin^(-1)\sqrt(1-x^(2))

Answer 1

$f(x)=6sin^{-1}(\sqrt{1-x^2})\\ ~~~~~~~~~= \begin{cases} 6cos^{-1}(x) &\text{if }-1\leq x\leq0 \\ 6\pi-6cos^{-1}(x) &\text{if }\;0\leq x\leq1 \end{cases}\\\;\\ \text{Domain of }f(x) \;\text{is}: x\in [-1,1]\\\;\\ \text{Range of }f(x)\;\text{is} : f(x)\in \left[0,3\pi\right]\\\;\\ f'(x)= \begin{cases} -\dfrac{3}{\sqrt{1-x^2}}<0 &\text{if }-1\leq x<0 \\ +\dfrac{3}{\sqrt{1-x^2}}>0 &\text{if }\;0< x\leq1 \end{cases}$

$f(x) \text{ is not differentiable at } x=0$

$\text{Now, }\\ {\displaystyle\int}cos^{-1}\left(x\right)\,\mathrm{d}x =xcos^{-1}\left(x\right)-\sqrt{1-x^2}+C\\ \text{Area bounded by the curve and the x-axis is :-}\\ {{\displaystyle\int}_{1}^{-1}{f(x)}\,\mathrm{d}x =2\times {\displaystyle\int}^1_{0}{f(x)}\,\mathrm{d}x=2\times6=12}$