Let f:[0,1]→Rf:[0,1]\to\mathbb{R}f:[0,1]→R be continuous with ∫01f(t) dt=1\displaystyle\int_0^1 f(t)\,dt = 1∫01f(t)dt=1, and let P(x)=∑k=1nakxk,∑k=1nak=1.P(x) = \sum_{k=1}^n a_k x^k, \qquad \sum_{k=1}^n a_k = 1.P(x)=∑k=1nakxk,∑k=1nak=1. Then there exists c∈(0,1)c \in (0,1)c∈(0,1) such that
f(c)=2∑k=1nakf(c) = 2\sum_{k=1}^n a_kf(c)=2∑k=1nak
f(c)=∫01P(t) dtf(c) = \int_0^1 P(t)\,dtf(c)=∫01P(t)dt
f(c)=P′(c)f(c) = P'(c)f(c)=P′(c)
f(c)=P(c)f(c) = P(c)f(c)=P(c)
