\(|∫_1^3y^ady|=\frac {364}{3}\)
\(|\frac {1}{a+1}(y^{a+1})|_1^3=\frac {364}{3}\)
\(\frac {3a+1−1}{a+1}=±\frac {364}{3}\)
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:
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:
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
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