JEE Main 2023 (Online) 30th January Morning ShiftQuestion 1
Let $\alpha$ be the area of the larger region bounded by the curve $y^2=8x$ and the lines $y=x$ and $x=2$, which lies in the first quadrant. Then the value of $3\alpha$ is equal to ______.
Application of Integrals
Answer$22$
Solution
$y^2=8x$ and $y=x$ meet at $x=0,8$; $x=2$ splits the enclosed region. Left part $\int_0^2(2\sqrt2\sqrt x-x)dx=\frac{10}3$, right part $\int_2^8(2\sqrt2\sqrt x-x)dx=\frac{22}3$. Larger $\alpha=\frac{22}3$, so $3\alpha=22$.