Question #153668

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


1
Expert's answer
2021-01-12T13:11:34-0500

f(x)=6sin1(1x2)         ={6cos1(x)if 1x06π6cos1(x)if   0x1  Domain of f(x)  is:x[1,1]  Range of f(x)  is:f(x)[0,3π]  f(x)={31x2<0if 1x<0+31x2>0if   0<x1f(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) is not differentiable at x=0f(x) \text{ is not differentiable at } x=0




Now, cos1(x)dx=xcos1(x)1x2+CArea bounded by the curve and the x-axis is :-11f(x)dx=2×01f(x)dx=2×6=12\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}


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