JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 12)
The area of the region
$$\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^{2}}\right\}$$ is equal to :
$$\frac{5}{2} \sin ^{-1}\left(\frac{3}{5}\right)-\frac{1}{2}$$
$$\frac{5 \pi}{4}-\frac{3}{2}$$
$$\frac{3 \pi}{4}+\frac{3}{2}$$
$$\frac{5 \pi}{4}-\frac{1}{2}$$
Explanation
$$A = \int\limits_{ - 1}^1 {\left( {\sqrt {5 - {x^2}} - (1 - x)} \right)dx + \int\limits_1^2 {\left( {\sqrt {5 - {x^2}} - (x - 1)} \right)dx} } $$
_29th_July_Morning_Shift_en_12_1.png)
$$\left. {A = 2\left( {{x \over 2}\sqrt {5 - {x^2}} + {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }}} \right) - 2x} \right|_0^1 + \left. {{x \over 2}\sqrt {5 - {x^2}} + {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }} - {{{x^2}} \over 2} + x} \right|_1^2$$
$$ = \left( {{{5\pi } \over 4} - {1 \over 2}} \right)$$ sq. units
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