Let a function $f(x)$ be defined in $[-2, 2]$ as $f(x) = \begin{cases} \{x\}, & -2 \leq x < -1 \\ |\text{sgn } x|, & -1 \leq x \leq 1 \\ \{-x\}, & 1 < x \leq 2 \end{cases}$, where $\{x\}$ denotes fractional part, then area bounded by graph of $f(x)$ and $x$-axis is:
Let $f$ be a twice differentiable function defined in $[-3,3]$ such that $f(0) = -4, f'(3) = 0, f'(-3) = 12$ and $f''(x) \geq -2\forall x \in [-3,3]$. If $g(x) = \int_0^x f(t)dt$ then find maximum value of $g(x), x \in [-3,3]$