Let the maximum value of f(x) on [a, b] be m, where ∫{a, b} f(t)dt ≤ ∫{a, b} Mdt = M(b-a).

However, because f(x) is a very continuous function, the equal sign cannot be established (otherwise, continuity can be used to prove that every point is m).

In addition, if f(x) is nonnegative in [a, b], there is ∫{a, b} f(t)dt ≥ 0.

F(x) is still a very continuous function, and the equal sign cannot be established.

So 0

Let f(x) get the maximum value m on [a, b] at x = e, and f(a) = f(b) = 0.

F(x) is continuous. According to the intermediate value theorem, there are c and d on (a, e) and (e, b) respectively, so that f (c) = (1(b-a)) ∫ {a, b} f (t) dt = f (d).