Integration of the defined function

The derivative of $f(x)$ is $f '(x)=3x^2-3$ , whose zeros are -1 and 1. So $f '(x)$ is positive on the intervals $(-\infty, -1)$ and $(1, \infty)$ and negative on $(-1, 1),$ then the function $f(x)$ is decreasing on the interval $[-1, 1]$ , so $1=f(-1)\geq f(x) \geq f(0)=-1,$ and, consequently, $|f(x)|\leq 1$ for all $x$ such that $-1\leq x \leq t$ if $t\leq 0,$ and $|f(-1)|=1.$ Therefore, $g(t)=1$ for all $t\in [-1, 0].$ Besides that, $-1=f(0) \geq f(x) \geq f(1)=-3$ for all $x$ such that $0\leq x\leq 1$ , therefore $|f(x)|$ is increasing on $[0, 1]$ and $|f(0)|=1$ . Therefore, when $0\leq t\leq 1$ we have $g(t)=\max_{-1\leq x\leq t}|f(x)|=\max_{0\leq x\leq t}|f(x)|=|f(t)|=-t^3+3t+1.$ Then we get $\int_{-1}^{1}g(t) dt=\int_{-1}^{0}g(t) dt+\int_{0}^{1}g(t) dt=\int_{-1}^{0}1 dt+\int_{0}^{1}(-t^3+3t+1) dt=\frac{13}{4}.$ So the answer of the problem is 17.