The area of the region enclosed by the parabola y = 4x - x ^ 2 and...

Given: $y = 4x - {x}^{2}\text{and}3y = {\left(x - 4\right)}^{2}$

$⇒ 4x - {x}^{2} = \dfrac{{\left(x - 4\right)}^{2}}{3}$

$⇒ 12x - 3{x}^{2} = {x}^{2} + 16 - 8x$

$⇒ 4{x}^{2} - 20x + 16 = 0$

$⇒ {x}^{2} - 5x + 4 = 0$

$⇒ \left(x - 4\right)\left(x - 1\right) = 0$

$⇒ x = 4, 1$

$⇒ y = 0, 3$

So, the intersection points are \left(4, 0\right)\text{and}\left(1, 3\right)$.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/2b25f3b7-8514-49aa-b022-e9f96770aab7.png" style="width: 300px; height: 222px;"/></p><p>Area$= {∫}_{1}^{4}\left[\left(4x - {x}^{2}\right) - \dfrac{{\left(x - 4\right)}^{2}}{3}\right]dx$</p><p>Area$= {\left|\dfrac{4{x}^{2}}{2} - \dfrac{{x}^{3}}{3} - \dfrac{{\left(x - 4\right)}^{3}}{9}\right|}_{1}^{4}$</p><p>$= \left|\left(\dfrac{64}{2} - \dfrac{64}{3} - \dfrac{4}{2} + \dfrac{1}{3} - \dfrac{27}{9}\right)\right|$</p><p>$= \left(27 - 21\right) = 6$