# 急!高等微積分!!!

1.Suppose that the function f:[0,無窮大]--->R is continuous and strictly increasing and that f:[0,無窮大]--->R is differentiable.Moreover, assume f(0)=0 .Consider the formula ∫[0,x]f+∫[0,f(x)]f^-1=xf(x) for all x>=0

Provide a geometric interpretation of this formula in terms of areas.Then prove this formula(Hint:Differentiate the formula and apply the Identity Criterion)

2.Suppose that the function f:[0,無窮大)--->R is continuous and strictly increasing ,with f(0)=0 and f([0,無窮大))=[0,無窮大).Then define F(x)=∫[0,x]f and G(x)=∫[0,x]f^-1 for all x>=0.

Prove Toung`s Inequality: ab<=F(a)+G(b) for all a>=0 and b>=0.

(Hint:A sketch will help,as wll the formula inExercise 1.)

Update:

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• 1 decade ago

f:[0,∞)-->R is continuous and strictly increasing ,so f^(-1) exist

=>∫_[0,x] f(t)dt+∫_[0,f(x)] f^(-1)(t)dt

=∫_[0,x]f(t)dt+∫_[0,f(x)]f^(-1)(f(z))df(z)

=∫_[0,x]f(t)dt+∫_[0,x]zdf(z) (&becaus; f is one to one,f(z)=f(0) f(x)=f(z)

=>z=0,z=x )

=∫_[0,x]f(t)dt+∫_[0,x]z*f'(z)dz

=∫_[0,x]f(t)dt+zf(z)|(0~x)-∫_[0,x]f(z)dz

=xf(x) 得證

幾何意義 你可以畫個圖 一條通過原點的曲線 因為f 是1-1 故對於所有 y>=0 均存在唯一 x>=0 使得 f(x)=y

故f(x)*x 是一矩形面積 剛好就等於 f 在[0,x]下面積和加上f 的反函數在[0,f(x)]下的面積

(2)的部分 只要畫圖就很容易輕易得證

但注意此時 f不可

2009-04-08 12:05:36 補充：

2009-04-08 12:07:02 補充：

(2)的部分等號成立的條件是b=f(a)