JEE MAIN - Mathematics (2023 - 30th January Morning Shift - No. 14)
Let $$\alpha$$ be the area of the larger region bounded by the curve $$y^{2}=8 x$$ and the lines $$y=x$$ and $$x=2$$, which lies in the first quadrant. Then the value of $$3 \alpha$$ is equal to ___________.
Answer
22
Explanation
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$${A_1} = \int\limits_0^2 {2\sqrt 2 \sqrt x - xdx} $$
$$ = \left. {2\sqrt 2 \times {2 \over 3}\,.\,{x^{{3 \over 2}}} - {{{x^2}} \over 2}} \right]_0^2$$
$$ = {{4\sqrt 2 } \over 3} \times 2\sqrt 2 - 2$$
$$ = {{16} \over 3} - 2 = {{10} \over 3}$$
$${A_2} = \left. {\int\limits_2^8 {2\sqrt 2 \sqrt x - xdx = {{4\sqrt 2 } \over 3}\,.\,{x^{{3 \over 2}}} - {{{x^2}} \over 2}} } \right]_2^8$$
$$ = {{4\sqrt 2 } \over 3}(16\sqrt 2 - 2\sqrt 2 ) - 30$$
$$ = {{112} \over 3} - 30 = {{22} \over 3}$$
$${A_2} > {A_1} \Rightarrow 3\alpha = 22$$
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