Area bounded by y = g(x), x-axis and the lines x = -2, x = 3, where
$g(x)=\left\{\begin{array}{l} \max i:\{f(t) ;-2 \leq t \leq x\},-2 \leq x<0 \\ \min i:\{f(t) ; 0 \leq t \leq x\}, 0 \leq x<3 \end{array}\right.$
and $f(x)=x^2-|x|$, is equal to
Answer & explanation
Correct answer: option 1
Clearly $g(x)=\left\{\begin{array}{l}2,-2 \leq x<0 \\ x^2-x, 0 \leq x \leq \frac{1}{2} \\ -\frac{1}{4}, \frac{1}{2}<x \leq 3\end{array}\right.$
$\Delta=\int\limits_{-2}^0 2 d x+\int\limits_0^{1 / 2}\left(x-x^2\right) d x+\int\limits_{1 / 2}^3 \frac{1}{4} d x$
$=(2 x)_{-2}^0+\left(\frac{x^2}{2}-\frac{x^3}{3}\right)_0^{1 / 2}+\left(\frac{x}{4}\right)_{1 / 2}^3=\frac{113}{24}$ sq. units