Let f: [ - 3, 1 ] → R be given as f(x) = \ cases min\ ( x + 6 ) , x...

Given $f: [ - 3, 1 ] \to R,$ f\left(x\right) = \left\{\begin{cases}min\left\{( x + 6 ) , {x}^{2}\right\}, & - 3 \le x \le 0 \\ max\left\{\sqrt{x}, {x}^{2}\right\}, & 0 \le x \le 1\end{cases}\right)</p><p>We havex + 6 = {x}^{2}$</p><p>$⇒ {x}^{2} - x - 6 = 0$</p><p>$⇒ \left(x - 3\right)\left(x + 2\right) = 0$</p><p>$⇒ x = 3 \text{ and } - 2 .$</p><p>Also,$\sqrt{x} = {x}^{2}$</p><p>$⇒ {x}^{2} - \sqrt{x} = 0$</p><p>$⇒ \sqrt{x}\left(x\dfrac{3}{2} - 1\right) = 0$</p><p>$⇒ x = 0 \text{ and } x = 1 .$</p><p>Hence, the region bounded by$f\left(x\right)$is given by the following graph.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/ff16e49b-d258-450e-a72a-6da11e5d0936-image.png" style="width: 397px; height: 350px;" /></p><p>Hence, the area bounded by$y = f ( x )$and x$-axis is

$A = {∫}_{- 3}^{- 2} ( x + 6 ) dx + {∫}_{- 2}^{0}{x}^{2}dx + {∫}_{0}^{1}\sqrt{x}dx$

$⇒ A = {\left(\dfrac{{x}^{2}}{2} + 6x\right)}_{- 3}^{- 2} + {\left(\dfrac{{x}^{3}}{3}\right)}_{- 2}^{0} + {\left(\dfrac{2{x}^{\dfrac{3}{2}}}{3}\right)}_{0}^{1}$

$⇒ A = \left(\dfrac{4}{2} - 12 - \left(\dfrac{9}{2} - 18\right)\right) + \left(\dfrac{0 + 8}{3}\right) + \left(\dfrac{2 \times 1 - 0}{3}\right)$

$⇒ A = - 10 + \dfrac{27}{2} + \dfrac{8}{3} + \dfrac{2}{3}$

$⇒ A = \dfrac{41}{6}$ sq units.

$⇒ 6A = 41 .$