Let f (x) be a continuous function with period T defined on (- ∞, ∞). It is proved that for any constant a, there is ∫ upper limit a + T lower limit a > F (x) DX = ∫ upper limit t lower limit 0 > F (x) d (x) holds

Let f (x) be a continuous function with period T defined on (- ∞, ∞). It is proved that for any constant a, there is ∫ upper limit a + T lower limit a > F (x) DX = ∫ upper limit t lower limit 0 > F (x) d (x) holds

∫ upper limit a + T lower limit a ＞ f (x) DX = ∫ upper limit 0 lower limit a ＞ f (x) DX + ∫ upper limit t lower limit 0 ＞ f (x) DX + ∫ upper limit a + T lower limit t ＞ f (x) DX = ∫ upper limit a lower limit 0 ＞ f (x + T) DX = - ∫ upper limit 0 lower limit a ＞ f (x) DX, so the above formula is true

The function f (x) satisfies the condition of Rolle's theorem on the interval [a, b], and f (x) is not constant. It is proved that there is at least one point in (a, b) ξ, Make f（ ξ)> 0 can only be used for tools related to the mean value theorem

The fact that f (x) is not constant indicates that there is at least one point C ∈ (a, b) such that f (c) ≠ f (a) = f (b), which can be known from the Lagrange mean value theorem ξ 1 and ξ 2 make
f'( ξ 1)=[f(c)-f(a)]/(c-a)
f'( ξ 2)=[f(c)-f(b)]/(c-b)
Due to f '（ ξ 1)f'( ξ 2)

If the function f (x) is differentiable in the interval (a, b) and there is a constant m so that | f '(x) | is less than or equal to m, it is proved that f (x) is bounded in (a, b)

Use Laplace mean value theorem for f (x) in [C, x], where C is a fixed point in (a, b), X is any point in (a, b), f (x) = f (c) + F '(R) (x-C), where R is a point between X and C, and take the absolute value to obtain the estimation of F (x)

It is known that the nonnegative function y = f (x) is differentiable on [0, positive infinity), and f (2x) is less than or equal to f (x) + 1. It is proved that: (1) there is a constant m such that 18. It is known that the nonnegative function y = f (x) is differentiable on [0, positive infinity), and f (2x)

Note M = max{|f (x): 1

Let the function f (x) be continuous and not constant to zero on [0,1], be differentiable in (0,1), and f (0) = 0, prove that it exists ξ ∈ (0,1), so that f（ ξ) f‘( ξ)> 0

f(x) = f(0) +f'( ξ) x ( Taylor expansion)
put x= ξ
f( ξ) = f'( ξ)ξ
f'( ξ) f( ξ) = [f( ξ)]^ 2/ ξ > 0

It is proved that if f (x) is continuous on [a, b] and differentiable in (a, b), then there is at least one point C in (a, b), so that f (c) + CF '(c) = [BF (b) - AF (a)] / (B-A) Using the mean value theorem,

∵ f (x) is continuous on [a, b] and differentiable in (a, b)
‡ XF (x) is continuous on [a, b] and differentiable in (a, b)
Lagrange mean value theorem
‡ then there is at least one point C in (a, b), so that f (c) + CF '(c) = [BF (b) - AF (a)] / (B-A)