The area bounded by the curves y={5-x²} and y=|x-1| is

Correct Answer is : 5π−24\;5π-2/445π−2 sq units

Correct Answer is : 5π−24\;5π-2/445π−2 sq units

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MHT CET 2021 Math Solved Paper

Question : 90 of 100

Marks: +1, -0

y=\sqrt{5-x^{2}}

y=|x-1|

Solution:

Given,

y=\sqrt{5-x^{2}} \Rightarrow y^{2}+x^{2}=5

and

y=|x-1|

\therefore

Required area

\;\; = \int\limits_{-1}^{2} \sqrt{5-x^{2}} d x - \int\limits_{-1}^{1} (1-x) d x - \int\limits_{1}^{2} (x-1) d x

\;\; = \left[\;\frac{x}{2} \sqrt{5-x^{2}} + \;\frac{5}{2} \sin^{-1} \;\frac{x}{\sqrt{5}}\right]_{-1}^{2} - \left[x - \;\frac{x^{2}}{2}\right]_{-1}^{1} - \left[\;\frac{x^{2}}{2} - x\right]_{1}^{2}

\;\; = \left[1 + \;\frac{5}{2} \sin^{-1} \;\frac{2}{\sqrt{5}} + 1 + \;\frac{5}{2} \sin^{-1} \;\frac{1}{\sqrt{5}}\right]

\;\; - \left[1 - \;\frac{1}{2} - \left(-1 - \;\frac{1}{2}\right)\right] - \left[2 - 2 - \left(\;\frac{1}{2} - 1\right)\right]

\;\; = 2 + \;\frac{5}{2} \sin^{-1}\left( \;\frac{2}{\sqrt{5}} \sqrt{1 - \;\frac{1}{5}} + \;\frac{1}{\sqrt{5}} \sqrt{1 - \;\frac{4}{5}} \right) - \;\frac{5}{2}

\;\; = \;\frac{5}{2} \sin^{-1}(1) - \;\frac{1}{2} = \;\frac{5\pi}{4} - \;\frac{1}{2} = \left( \;\frac{5\pi-2}{4} \right) \;\text{ sq units }\;

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