Let f:[-3,1]→ R be given as f(x)= … .If the area bounded by y=f(x) and x-axis is A sq units, then the value of 6A is equal to

Given $f:[-3,1]\rightarrow R,$f:[−3,1]→R, f(x)={\begin{matrix}min{(x+6),{x}^{2}}, & -3\leq x\leq 0 \\ max{\sqrt{x},{x}^{2}}, & 0\leq x\leq 1\end{matrix}

We have $x+6={x}^{2}$x+6=x2

$$\Rightarrow {x}^{2}-x-6=0$$⇒x2−x−6=0

$$\Rightarrow (x-3)(x+2)=0$$⇒(x−3)(x+2)=0

\Rightarrow x=3&-2.

Also, $\sqrt{x}={x}^{2}$x​=x2

$$\Rightarrow {x}^{2}-\sqrt{x}=0$$⇒x2−x​=0

$$\Rightarrow \sqrt{x}({x}^{\frac{3}{2}}-1)=0$$⇒x​(x23​−1)=0

\Rightarrow x=0&x=1.

Hence, the region bounded by $f(x)$f(x) is given by the following graph.

Hence, the area bounded by $y=f(x)$y=f(x) and $x$x-axis is

$$A={\int }_{-3}^{-2}(x+6)dx+{\int }_{-2}^{0}{x}^{2}dx+{\int }_{0}^{1}\sqrt{x}dx$$A=∫−3−2​(x+6)dx+∫−20​x2dx+∫01​x​dx

$$\Rightarrow A={(\frac{{x}^{2}}{2}+6x)}_{-3}^{-2}+{(\frac{{x}^{3}}{3})}_{-2}^{0}+{(\frac{2{x}^{\frac{3}{2}}}{3})}_{0}^{1}$$⇒A=(2x2​+6x)−3−2​+(3x3​)−20​+(32x23​​)01​

$$\Rightarrow A=(\frac{4}{2}-12-(\frac{9}{2}-18))+(\frac{0+8}{3})+(\frac{2\times 1-0}{3})$$⇒A=(24​−12−(29​−18))+(30+8​)+(32×1−0​)

$$\Rightarrow A=-10+\frac{27}{2}+\frac{8}{3}+\frac{2}{3}$$⇒A=−10+227​+38​+32​

$\Rightarrow A=\frac{41}{6}$⇒A=641​ sq units.

$$\Rightarrow 6A=41.$$⇒6A=41.