The area (in sq. units) of the region describ | vedclass

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(B) To find the area of the region bounded by the parabola $y^2 = 2x$ and the line $y = 4x - 1$,we first find their points of intersection.Substitutin

Vedclass · Accepted Answer

$\frac{9}{32}$

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(B) To find the area of the region bounded by the parabola $y^2 = 2x$ and the line $y = 4x - 1$,we first find their points of intersection.Substituting $x = \frac{y^2}{2}$ into the equation of the line $y = 4x - 1$,we get:$y = 4\left(\frac{y^2}{2}ight) - 1$$y = 2y^2 - 1$$2y^2 - y - 1 = 0$$(2y + 1)(y - 1) = 0$Thus,the points of intersection occur at $y = 1$ and $y = -\frac{1}{2}$.The required area is given by the integral of the difference between the line and the parabola with respect to $y$:$	ext{Area} = \int_{-1/2}^{1} \left( \frac{y+1}{4} - \frac{y^2}{2} ight) dy$$= \left[ \frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6} ight]_{-1/2}^{1}$$= \left( \frac{1}{8} + \frac{1}{4} - \frac{1}{6} ight) - \left( \frac{1/4}{8} + \frac{-1/2}{4} - \frac{-1/8}{6} ight)$$= \left( \frac{3+6-4}{24} ight) - \left( \frac{1}{32} - \frac{1}{8} + \frac{1}{48} ight)$$= \frac{5}{24} - \left( \frac{3 - 12 + 2}{96} ight)$$= \frac{5}{24} - \left( -\frac{7}{96} ight) = \frac{20+7}{96} = \frac{27}{96} = \frac{9}{32}$ sq. units.

The area (in sq. units) of the region described by $\{(x, y) : y^2 \leq 2x \text{ and } y \geq 4x - 1\}$ is

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