The correct answer is : 42
A = required area
\(=\int\limits^{\frac{3}{2}}_0\left[(3-y)-(\frac{2y^2}{3})\right]dy-\pi(\sqrt2)^2.\frac{1}{8}\)
\(⇒\left(3y-\frac{y^2}{2}-\frac{2}{9}y^3\right)\big|^{\frac{3}{2}}_{0}-\frac{\pi}{4}\)
\(⇒3.\frac{3}{2}-\frac{9}{8}-\frac{2}{9}.\frac{27}{8}-\frac{\pi}{4}\)
\(⇒\frac{36-9-6}{8}-\frac{\pi}{4}\)
\(=\frac{21}{8}-\frac{\pi}{4}\)
\(⇒4(\pi+4A)=4(\frac{21}{2})\)
\(=42\)
The area of the region enclosed between the curve \( y = |x| \), x-axis, \( x = -2 \)} and \( x = 2 \) is:
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:
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
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