The correct answer is : 42

A = required area
\(=\int\limits^{\frac{3}{2}}_0\left[(3-y)-(\frac{2y^2}{3})\right]dy-\pi(\sqrt2)^2.\frac{1}{8}\)
\(⇒\left(3y-\frac{y^2}{2}-\frac{2}{9}y^3\right)\big|^{\frac{3}{2}}_{0}-\frac{\pi}{4}\)
\(⇒3.\frac{3}{2}-\frac{9}{8}-\frac{2}{9}.\frac{27}{8}-\frac{\pi}{4}\)
\(⇒\frac{36-9-6}{8}-\frac{\pi}{4}\)
\(=\frac{21}{8}-\frac{\pi}{4}\)
\(⇒4(\pi+4A)=4(\frac{21}{2})\)
\(=42\)
If the area of the region \[ \{(x, y) : |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0\} \] is \( \frac{80\sqrt{2}}{\alpha - \beta} \), where \( \alpha, \beta \in \mathbb{N} \), then \( \alpha + \beta \) is equal to:
Let the area of the region \( \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} \) be \( A \). Then \( 6A \) is equal to:
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: $g = 10\,\text{m s}^{-2}$) 
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
Read More: Area under the curve formula