The answer is 6.

\(\begin{aligned} &Given,\\& x^2+\frac{5}{12}=\frac{2-8x}{3}\\& x^2+\frac{8x}{3}+\frac{5}{12}-2=0\\&\text{Solve for quadratic equation} \\&12 x^2+32 x-19=0 \\ & 12 x^2+38 x-6 x-19=0 \\ & 2 x(6 x+19)-1(6 x+19)=0 \\ & (6 x+19)(2 x-1)=0 \\ & x=\frac{1}{2} \\ &\text{The area a is calculated using definite integration}\\ & \alpha=2 A_1+A_2 \\ & \alpha=2\left(\int_0^{1 / 2} x^2+\frac{5}{12} d x+\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}\right) \\ & \Rightarrow \alpha=2\left[\left(\frac{x^3}{3}+\frac{5 x}{12}\right)_0^{1 / 2}+\frac{1}{12}\right] \\ & \Rightarrow \alpha=2\left[\frac{1}{24}+\frac{5}{24}+\frac{1}{12}\right] \\ & \Rightarrow \alpha=2\left[\frac{1+5+2}{24}\right] \Rightarrow \alpha=2 \times \frac{8}{24} \Rightarrow 9 \alpha=9 \times \frac{8}{12} \\ & \Rightarrow 9 \alpha=6 \end{aligned}\)
S, the answer is 6.
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
There are distinct applications of integrals, out of which some are as follows:
In Maths
Integrals are used to find:
In Physics
Integrals are used to find: