\(|∫_1^3y^ady|=\frac {364}{3}\)
\(|\frac {1}{a+1}(y^{a+1})|_1^3=\frac {364}{3}\)
\(\frac {3a+1−1}{a+1}=±\frac {364}{3}\)
Solving with (+) sign,
\(\frac {3a+1−1}{a+1}=\frac {364}{3}\)
\(a=5\)
Solving with (-) sign,
\(\frac {3a+1−1}{a+1}=-\frac {364}{3}\)
No a exist
\(∴a=5\)
So, the correct option is (B): \(5\)
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:
If the area of the region $$ \{(x, y): |4 - x^2| \leq y \leq x^2, y \geq 0\} $$ is $ \frac{80\sqrt{2}}{\alpha - \beta} $, $ \alpha, \beta \in \mathbb{N} $, then $ \alpha + \beta $ is equal to:
Read More: Area under the curve formula