The area (in square units) of the region encl | vedclass

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(A) Given the equation of the parabola $y^2=2x$ $\dots(i)$ and the equation of the line $y=4x-1$ $\dots(ii)$.To find the intersection points, substitu

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$\frac{9}{32}$

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(A) Given the equation of the parabola $y^2=2x$ $\dots(i)$ and the equation of the line $y=4x-1$ $\dots(ii)$.To find the intersection points, substitute $x = \frac{y+1}{4}$ from $(ii)$ into $(i)$:$y^2 = 2\left(\frac{y+1}{4}\right) \implies y^2 = \frac{y+1}{2} \implies 2y^2 - y - 1 = 0$.Solving the quadratic equation: $(2y+1)(y-1) = 0$, so $y = -\frac{1}{2}$ and $y = 1$.The corresponding $x$ values are $x = \frac{(-1/2)+1}{4} = \frac{1}{8}$ and $x = \frac{1+1}{4} = \frac{1}{2}$.The intersection points are $(\frac{1}{8}, -\frac{1}{2})$ and $(\frac{1}{2}, 1)$.The required area is given by the integral with respect to $y$:$Area = \int_{-1/2}^{1} (x_{line} - x_{parabola}) dy = \int_{-1/2}^{1} (\frac{y+1}{4} - \frac{y^2}{2}) dy$.$= \frac{1}{4} \int_{-1/2}^{1} (y+1) dy - \frac{1}{2} \int_{-1/2}^{1} y^2 dy$.$= \frac{1}{4} [\frac{y^2}{2} + y]_{-1/2}^{1} - \frac{1}{2} [\frac{y^3}{3}]_{-1/2}^{1}$.$= \frac{1}{4} [(\frac{1}{2} + 1) - (\frac{1}{8} - \frac{1}{2})] - \frac{1}{6} [1 - (-\frac{1}{8})]$.$= \frac{1}{4} [\frac{3}{2} + \frac{3}{8}] - \frac{1}{6} [\frac{9}{8}] = \frac{1}{4} [\frac{15}{8}] - \frac{3}{16} = \frac{15}{32} - \frac{6}{32} = \frac{9}{32}$ square units.

The area (in square units) of the region enclosed between the parabola $y^2=2x$ and the line $y=4x-1$ is:

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