Mathematics - Area Under Curves Question with Solution | TestHub

Given $f:[-3,1]\to R,$ $f\left(x\right)=\left\{\begin{array}{ll}min\left\{(x+6), x^2\right\},& -3\le x\le 0\\ max\left\{\sqrt{x}, x^2\right\},& 0\le x\le 1\end{array}\right.$

We have $x+6=x^2$

$\Rightarrow x^2-x-6=0$

$\Rightarrow \left(x-3\right)\left(x+2\right)=0$

$\Rightarrow x=3 \& -2.$

Also, $\sqrt{x}=x^2$

$\Rightarrow x^2-\sqrt{x}=0$

$\Rightarrow \sqrt{x}\left(x^{\frac{3}{2}}-1\right)=0$

$\Rightarrow x=0 \& x=1.$

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

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

$A={\int }_{-3}^{-2}(x+6)dx+{\int }_{-2}^0{x}^{2}dx+{\int }_0^1\sqrt{x}dx$

$\Rightarrow A={\left(\frac{x^2}{2}+6x\right)}_{-3}^{-2}+{\left(\frac{x^3}{3}\right)}_{-2}^0+{\left(\frac{2x^{{\displaystyle \frac{3}{2}}}}{3}\right)}_0^1$

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

$\Rightarrow A=-10+\frac{27}{2}+\frac{8}{3}+\frac{2}{3}$

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

$\Rightarrow 6A=41.$