A constant

Let $K$ be the smallest positive constant such that $\int_0^1 P(x)\,dx \; \le \; K\mathrm{max}_{x \in [0,1]}P(x)$ for all cubic polynomials $P$ such that $P(0) = 0$ and $0 \le P(x) \le 1$ for all $0 \le x \le 1$ .

If $P(x) = ax^3 + bx^2 + cx + d$ is a cubic polynomial,then $\int_0^1 P(x)\,dx \; = \; \tfrac14a + \tfrac13b + \tfrac12c + d \; = \; \tfrac16P(0) + \tfrac23P(\tfrac12) + \tfrac16P(1)$ Thus, if $P(x)$ is a cubic polynomial such that $P(0)=0$ and $0 \le P(x) \le 1$ for all $0 \le x \le 1$ , then $\int_0^1 |P(x)|\,dx \; = \; \tfrac23P(\tfrac12) + \tfrac16P(1) \; \le \; \tfrac23 + \tfrac16 \; = \; \tfrac56$ and hence $K \le \tfrac56$ . However, if we consider the polynomial $Q_0(x) \; = \; 1 - 4(x-1)(x-\tfrac12)^2$ then $Q_0(x)$ is cubic with $Q_0(0)=0$ and $Q_0(1) = Q_0(\tfrac12) = 1$ , and $Q_0$ has a turning point at $x=\tfrac12$ . The second turning point of $Q_0$ occurs at $x=\tfrac56$ , and $Q_0(\tfrac56) = \tfrac{25}{27} > 0$ . Thus we deduce that $Q_0(x)$ is a cubic polynomial with $Q_0(0)=0$ and $0 \le Q_0(x) \le 1$ for all $0 \le x \le 1$ , and moreover that $\int_0^1 |Q_0(x)|\,dx = \tfrac56$ . Thus we deduce that $K = \tfrac56$ .

If we now assume that the cubic $Q(x)$ is single-signed throughout $[0,1]$ and that either $Q(0) = 0$ or $Q(1) = 0$ , then a single linear change of variable, and scaling $Q$ suitably, shows that $\int_0^1 |Q(x)|\,dx \; \le \; \tfrac56\max_{x \in [0,1]}|Q(x)|$

If $[u,v]$ is a subinterval of $[0,1]$ , and if the cubic $Q(x)$ is single-signed on $[u,v]$ , and vanishes at one of $u$ or $v$ , then a simple linear change of variable shows us that $\int_u^v |Q(x)|\,dx \; \le \; \tfrac56(v-u)\mathrm{max}_{x \in [u,v]}|Q(x)| \; \le \; \tfrac56(v-u)\mathrm{max}_{x \in [0,1]}|Q(x)|$

Suppose now that $P(x)$ is a cubic polynomial which has at least one root in $[0,1]$ . We can then subdivide the interval $[0,1]$ at the zeros of $P(x)$ , and we therefore deduce, considering the integral of $|P(x)|$ on each of these intervals separately, that $\int_0^1 |P(x)|\,dx \;\le \; \tfrac56\mathrm{max}_{x \in [0,1]}|P(x)|$ which tells us that $C \ge \tfrac 56$ . Considering $Q_0(X)$ again tells us that $C = \tfrac56$ . This makes the answer $\big\lfloor 1000 \times \tfrac56\big\rfloor = \boxed{833}$ .