The area of the region enclosed between the parabolas y ^ 2 = 2x -...

We have, ${y}^{2} = 2x - 1 . . . \left(1\right)$ and ${y}^{2} = 4x - 3 . . . \left(2\right)$

We first find the points of intersection of the given parabolas by solving the equations $\left(1\right) \text{ and } \left(2\right)$simultaneously.

This gives, $2x - 1 = 4x - 3$

$⇒ 2x = 2$ $⇒ x = 1$

Now, put $x = 1$ in equation \left(1\right)$, we get$y = \pm 1$.</p><p>Now, the equations \left(1\right) \text{ and }$\left(2\right)$can be written as</p><p>${y}^{2} = 2\left(x - \dfrac{1}{2}\right)$</p><p>${y}^{2} = 4\left(x - \dfrac{3}{4}\right)$respectively.</p><p>The graphs of these two curves are shown in figure.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/5c0efdd7-b840-4e55-b17f-11ad4750d950-image.png" style="width: 323px; height: 220px;" /></p><p>We have to find the area of the shaded region.</p><p>Now, the equations$\left(1\right) \text{ and } \left(2\right)$can be written as</p><p>$x = \dfrac{{y}^{2} + 1}{2}&x = \dfrac{{y}^{2} + 3}{4}$respectively.</p><p>It is clear from the figure, required area$A = 2{∫}_{0}^{1}\left({x}_{2} - {x}_{1}\right)dy$($∵$Area is symmetrical about$x -$axis)</p><p>where${x}_{1} = \dfrac{{y}^{2} + 1}{2}$and${x}_{2} = \dfrac{{y}^{2} + 3}{4}$</p><p>$⇒ A = 2{∫}_{0}^{1}\left(\dfrac{{y}^{2} + 3}{4} - \dfrac{{y}^{2} + 1}{2}\right)dy = \dfrac{2}{4}{∫}_{0}^{1}\left(1 - {y}^{2}\right)dy$</p><p>$= \dfrac{1}{2}{\left[y - \dfrac{{y}^{3}}{3}\right]}_{0}^{1} = \dfrac{1}{2}\left[1 - \dfrac{1}{3}\right]$</p><p>$= \dfrac{1}{2}\left(\dfrac{2}{3}\right) = \dfrac{1}{3} sq.units.