Let the area of the region enclosed by the curves $y=3 x, 2 y=27-3 x$ and $y=3 x-x \sqrt{x}$ be $A$. Then…

Let the area of the region enclosed by the curves $y=3 x, 2 y=27-3 x$ and $y=3 x-x \sqrt{x}$ be $A$. Then $10 A$ is equal to
  1. 172
  2. 162
  3. 154
  4. 184

Solution

$y=3 x, 2 y=27-3 x \& y=3 x-x \sqrt{x}$
$\begin{aligned} & \mathrm{A}=\int_0^3 3 \mathrm{x}-(3 \mathrm{x}-\mathrm{x} \sqrt{\mathrm{x}}) \mathrm{dx}+\int_3^9\left(\frac{27-3 \mathrm{x}}{2}-(3 \mathrm{x}-\mathrm{x} \sqrt{\mathrm{x}})\right) \mathrm{dx} \\ & \mathrm{A}=\int_0^3 \mathrm{x}^{3 / 2} \mathrm{dx}+\int_3^9 \frac{27}{2}-\frac{9 \mathrm{x}}{2}+\mathrm{x}^{3 / 2} \mathrm{dx} \\ & \mathrm{A}=\left[\frac{2 \mathrm{x}^{5 / 2}}{5}\right]_0^3+\frac{27}{2}[\mathrm{x}]_3^9-\frac{9}{2}\left[\frac{\mathrm{x}^2}{2}\right]_3^9+\left[\frac{2 \mathrm{x}^{5 / 2}}{5}\right]_3^9 \\ & \mathrm{~A}=\frac{2}{5}\left(3^{5 / 2}\right)+\frac{27}{2}(6)-\frac{9}{4}(72)+\frac{2}{5}\left(9^{5 / 2}-3^{5 / 2}\right) \\ & \mathrm{A}=\frac{2}{5}\left(3^{5 / 2}\right)+81-162+\frac{2}{5} \times 3^5-\frac{2}{5} \times 3^{5 / 2} \\ & \mathrm{~A}=\frac{486}{5}-81=\frac{81}{5} \\ & 10 \mathrm{~A}=162 \\ & \text {Ans. }=(2)\end{aligned}$

Asked in: JEE Main 2024 (06 Apr Shift 1)

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