The area (in sq. units) of the region described by [ (x, y) : y ^ 2...

Let us find the points intersections of ${y}^{2} = 2x$ and $y = 4x - 1$

$⇒ {\left(4x - 1\right)}^{2} = 2x$

$⇒ 16{x}^{2} - 10x + 1 = 0$

$⇒ \left(8x - 1\right)\left(2x - 1\right) = 0$

$⇒x = \dfrac{1}{2}, \dfrac{1}{8}$

$x = \dfrac{1}{2} ⇒ y = 4\left(\dfrac{1}{2}\right) - 1 = 1 ⇒ A\left(\dfrac{1}{2}, 1\right)$

$x = \dfrac{1}{8} ⇒ y = 4\left(\dfrac{1}{8}\right) - 1 = \dfrac{- 1}{2} ⇒ B\left(\dfrac{1}{8}, \dfrac{- 1}{2}\right)$

${A}_{1} = {∫}_{0}^{\dfrac{1}{8}}\sqrt{2x}dx + \dfrac{1}{2} \times \dfrac{1}{8} \times \dfrac{1}{2}$

${A}_{1} = \sqrt{2}{\left[\dfrac{{x}^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right]}_{0}^{\dfrac{1}{8}} + \dfrac{1}{32}$

${A}_{1} = \dfrac{1}{24} + \dfrac{1}{32}sq . units$

${A}_{2} = {∫}_{0}^{\dfrac{1}{2}}\sqrt{2x}dx - \dfrac{1}{2} \times 1 \times \dfrac{1}{4}$

${A}_{2} = \sqrt{2}{\left[\dfrac{{x}^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right]}_{0}^{\dfrac{1}{2}} - \dfrac{1}{8}$

${A}_{2} = \dfrac{5}{24}sq . units$

Therefore, the required area is,

$A = \dfrac{1}{24} + \dfrac{1}{32} + \dfrac{5}{24} = \dfrac{9}{32}sq . units$