A infimum of integral of a cubic function

Let $f(x)$ be a cubic function defined on $[0,1]$, and $\max_{x \in [0,1]}\left|f(x) \right|=1$. Find the infimum of $$C=\int_0^{1}\left|f(x) \right|\text{d}x.$$

Note that $$C=\inf\int_0^{1}\left|f(x) \right|\text{d}x\,/\max_{x \in [0,1]}\left|f(x) \right|$$ when the condition of maximum is deleted.

I initially guess that $C=\dfrac{1}{4}$ where $f(x)=x^3$, but It is wrong actually.

Anyway, I turned to linear functions and obviously the answer is $C_1=\sqrt{2}-1$, but for quadratic functions it becomes tough. Mathematica told me when $f(x)=x^2-0.705x+0.087,$$C_2$ can reach $0.227774$. And for cubic functions, $C_3$ can reach $0.145263$ where $f(x)=x^3-1.091x^2+0.307x-0.018$.

So my question is can we evaluate the closed form of $C_2,C_3$ and $C_n$?