\(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {12}{13})\)
\(\frac {(169π)}{4} + \frac {65}{2} - \frac {169}{2} sin^{-1}(\frac {12}{13})\)
\(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {13}{14})\)
\(\frac {(169π)}{4} + \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {13}{14})\)
The correct option is (A): \(\frac {(169π)}{4} - \frac {65}{2} + \frac {169}{2} sin^{-1}(\frac {12}{13})\).
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:
Read More: Area under the curve formula