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

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(A) The given region is defined by the parabola $y^2 = 2x$ and the line $y = 4x - 1$.To find the intersection points,substitute $x = \frac{y+1}{4}$ in

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$\frac{9}{32}$

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(A) The given region is defined by the parabola $y^2 = 2x$ and the line $y = 4x - 1$.To find the intersection points,substitute $x = \frac{y+1}{4}$ into $y^2 = 2x$:$y^2 = 2\left(\frac{y+1}{4}\right) \Rightarrow y^2 = \frac{y+1}{2} \Rightarrow 2y^2 - y - 1 = 0$.Solving the quadratic equation: $(2y+1)(y-1) = 0$,so $y = 1$ and $y = -\frac{1}{2}$.For $y = 1$,$x = \frac{1+1}{4} = \frac{1}{2}$. For $y = -\frac{1}{2}$,$x = \frac{-1/2+1}{4} = \frac{1}{8}$.The points of intersection are $(\frac{1}{2}, 1)$ and $(\frac{1}{8}, -\frac{1}{2})$.The area is given by the integral of (right curve - left curve) with respect to $y$:Area $= \int_{-1/2}^{1} \left( \frac{y+1}{4} - \frac{y^2}{2} \right) dy$$= \left[ \frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6} \right]_{-1/2}^{1}$$= \left( \frac{1}{8} + \frac{1}{4} - \frac{1}{6} \right) - \left( \frac{1/4}{8} - \frac{1/2}{4} - \frac{-1/8}{6} \right)$$= \left( \frac{3+6-4}{24} \right) - \left( \frac{1}{32} - \frac{1}{8} + \frac{1}{48} \right)$$= \frac{5}{24} - \left( \frac{3-12+2}{96} \right) = \frac{5}{24} - \left( -\frac{7}{96} \right) = \frac{20+7}{96} = \frac{27}{96} = \frac{9}{32}$ sq. units.

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

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