JEE MAIN - Mathematics (2022 - 28th June Morning Shift - No. 7)
The area of the region S = {(x, y) : y2 $$\le$$ 8x, y $$\ge$$ $$\sqrt2$$x, x $$\ge$$ 1} is
$${{13\sqrt 2 } \over 6}$$
$${{11\sqrt 2 } \over 6}$$
$${{5\sqrt 2 } \over 6}$$
$${{19\sqrt 2 } \over 6}$$
Explanation
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Required area
$$ = \int\limits_1^4 {\left( {\sqrt {8x} - \sqrt 2 x} \right)dx} $$
$$ = \left. {{{2\sqrt 8 } \over 3}{x^{{3 \over 2}}} - {{{x^2}} \over {\sqrt 2 }}} \right|_1^4$$
$$ = {{16\sqrt 3 } \over 3} - {{16} \over {\sqrt 2 }} - {{2\sqrt 8 } \over 3} + {1 \over {\sqrt 2 }}$$
$$ = {{11\sqrt 2 } \over 6}$$ sq. units
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