The area (in sq unit) of the region bounded b | vedclass

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(B) The given curve is $2x = y^2 - 1$, which can be rewritten as $x = \frac{y^2 - 1}{2}$.The curve $x = 0$ is the $y$-axis.To find the intersection po

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$\frac{2}{3}$

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(B) The given curve is $2x = y^2 - 1$, which can be rewritten as $x = \frac{y^2 - 1}{2}$.The curve $x = 0$ is the $y$-axis.To find the intersection points of the curve $2x = y^2 - 1$ and $x = 0$, substitute $x = 0$ into the equation:$0 = y^2 - 1 \implies y^2 = 1 \implies y = \pm 1$.Thus, the region is bounded by $y = -1$ to $y = 1$.The area $A$ is given by the integral of the distance between the curves with respect to $y$:$A = \int_{-1}^{1} |x_{	ext{right}} - x_{	ext{left}}| dy = \int_{-1}^{1} |0 - \frac{y^2 - 1}{2}| dy = \int_{-1}^{1} \frac{1 - y^2}{2} dy$.Since the function is symmetric about the $x$-axis, we can write:$A = 2 \int_{0}^{1} \frac{1 - y^2}{2} dy = \int_{0}^{1} (1 - y^2) dy$.$A = [y - \frac{y^3}{3}]_{0}^{1} = (1 - \frac{1}{3}) - (0 - 0) = \frac{2}{3} 	ext{ sq unit}$.

The area (in sq unit) of the region bounded by the curves $2x = y^2 - 1$ and $x = 0$ is

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