The area bounded by the curve
\(y=|x^2−9| \)
and the line y = 3 is
\(4(2\sqrt3+\sqrt6-4)\)
\(4(4\sqrt3+\sqrt6-4)\)
\(8(4\sqrt3+3\sqrt6-9)\)
\(8(4\sqrt3+\sqrt6-9)\)
The correct asnwer is (D) : \(8(4\sqrt3+\sqrt6-9)\)

\(=2\int^{3}_{0}(\sqrt{9+y}-\sqrt{9-y})dy + 2\int^{9}_{3}(\sqrt{9-y})dy\)
\(=2\int^{3}_{0}((9+y)^{1/2}-(9-y)dy + \int^{9}_{3}({9-y})^{1/2})dy\)
\(=2[\frac{2}{3}[(9+y)^{3/2}]^{3}_{0} + \frac{2}{3}[(9-y)^{3/2}]^{3}_{0}-\frac{2}{3}[(9-y)^{3/2}]^{3}_{0}]\)
\(= \frac{4}{3}[12\sqrt{12}-27+6\sqrt6-27-(0-6\sqrt6)]\)
\(=\frac{4}{3}[24\sqrt3+12\sqrt6-54]\)
\(=8(4\sqrt3+2\sqrt6-9)\)
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Integral calculus is the method that can be used to calculate the area between two curves that fall in between two intersecting curves. Similarly, we can use integration to find the area under two curves where we know the equation of two curves and their intersection points. In the given image, we have two functions f(x) and g(x) where we need to find the area between these two curves given in the shaded portion.

Area Between Two Curves With Respect to Y is
If f(y) and g(y) are continuous on [c, d] and g(y) < f(y) for all y in [c, d], then,
