Question

Let $f:[-3,1]\to R$ be given as
$f\left(x\right)=\{\begin{array}{cc}min\left\{\right(x+6),x^2\},& -3\le x\le 0\\ max\{\sqrt{x},x^2\},& 0\le x\le 1.\end{array}$
If the area bounded by $y=f\left(x\right)$ and $x-$ axis is $A,$ then the value of $6A$ is equal to

Answer 1

Area
$A=\underset{-3}{\overset{-2}{\int }}(x+6)dx+\underset{-2}{\overset{0}{\int }}{x}^{2}dx+\underset{0}{\overset{1}{\int }}\sqrt{x}dx\Rightarrow A=\frac{7}{2}+{\left[\frac{x^3}{3}\right]}_{-2}^0+{\left[\frac{2}{3}{x}^{3/2}\right]}_0^1\Rightarrow A=\frac{7}{2}+\frac{8}{3}+\frac{2}{3}=\frac{41}{6}\therefore 6A=41$