Let q be the maximum integral value of p in [0, 10] for which the...

${x}^{2} - px + \dfrac{5p}{4} = 0$

For rational roots, $D$ must be a perfect square.

$D = {p}^{2} - 5p = p\left(p - 5\right)$

Now put values in reverse order as $q$ is the maximum value of $p$

At $p = 10$,

$D = 50 \to \text{not a perfect square}$

At $p = 9$,

$D = 36 \to \text{perfect square}$

So, the maximum integral value of $p$ is $9$

$∴ q = 9$

$0 \le y \le {\left(x - 9\right)}^{2}$

Area of shaded region $= {∫}_{0}^{9}{\left(x - 9\right)}^{2}dx$

$= {\left[\dfrac{{\left(x - 9\right)}^{3}}{3}\right]}_{0}^{9} = 243{\text{unit}}^{2}$