JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
Let the area of the region enclosed by the curves $y=3x$, $2y=27-3x$ and $y=3x-x\sqrt{x}$ be $A$. Then $10A$ is equal to
Application of Integrals
Solution
Lines meet at $(3,9)$. Curve $y=3x-x^{3/2}$ meets $y=3x$ at $x=0$ and meets $2y=27-3x$ at $x=9$ (check: $9x-2x^{3/2}=27$ at $x=9$), point $(9,0)$. $A=\int_0^3x^{3/2}dx+\int_3^9\left[\tfrac{27-3x}2-3x+x^{3/2}\right]dx$. Combine the $x^{3/2}$ terms: $\tfrac25\cdot9^{5/2}=\tfrac{2}{5}\cdot243=\tfrac{486}5$; the rest $=\int_3^9\left(\tfrac{27}2-\tfrac{9x}2\right)dx=81-162=-81$. $A=\tfrac{486}5-81=\tfrac{81}5$, so $10A=162$. Answer: (B).