JEE Main 2024 — 6 April, Shift 1 — Question 19

$y=3 x, 2 y=27-3 x \& y=3 x-x \sqrt{x}$

$A=\int_{0}^{3} 3 x-(3 x-x \sqrt{x}) d x+\int_{3}^{9}\left(\frac{27-3 x}{2}-(3 x-x \sqrt{x})\right) d x$

$A=\int_{0}^{3} x^{3 / 2} d x+\int_{3}^{9} \frac{27}{2}-\frac{9 x}{2}+x^{3 / 2} d x$

$\mathrm{A}=\left[\frac{2 \mathrm{x}^{5 / 2}}{5}\right]_{0}^{3}+\frac{27}{2}[\mathrm{x}]_{3}^{9}-\frac{9}{2}\left[\frac{\mathrm{x}^{2}}{2}\right]_{3}^{9}+\left[\frac{2 \mathrm{x}^{5 / 2}}{5}\right]_{3}^{9}$

$\mathrm{A}=\frac{2}{5}\left(3^{5 / 2}\right)+\frac{27}{2}(6)-\frac{9}{4}(72)+\frac{2}{5}\left(9^{5 / 2}-3^{5 / 2}\right)$

$\mathrm{A}=\frac{2}{5}\left(3^{5 / 2}\right)+81-162+\frac{2}{5} \times 3^{5}-\frac{2}{5} \times 3^{5 / 2}$

$A=\frac{486}{5}-81=\frac{81}{5}$ ;

$10 \mathrm{~A}=162$