Ngô Quốc Anh

October 12, 2007

Bất đẳng thức tích phân

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 20:17

Let f: [0,1]\rightarrow\mathbb{R} is a differentiable and monotone decreasing function. For all x\in(0,1) prove that

x\displaystyle\int_{0}^{1}f(t)dt\leq\int_{0}^{x}f(t)dt.

Proof. Let consider the function

\displaystyle g(x)=\frac{1}{x}\int_0^xf(t)dt

on (0,1]. A simple calculation shows us that

\displaystyle g'(x) = \frac{1}{{{x^2}}}\left( {xf(x) - \int_0^x {f(t)dt} } \right).

Again, let us consider the following function

\displaystyle h(x) = xf(x) - \int_0^x {f(t)dt} ,x \in \left( {0,1} \right].

Clearly,

\displaystyle h'(x) = xf'(x) \leqslant 0

and

h(0)=0

which implies that

h(x) \leqslant 0

for any x \in (0,1]. Thus, we have shown that

g'(x)\leq 0

on (0,1). So g is decreasing on (0,1). This implies for any x\in(0,1) we have

\displaystyle g(x)\geq \lim_{t\to1^{-}}g(t)=g(1)=\int_0^1f(t)dt

which completes our proof.


4 Comments »

  1. rất cám ơn bạn vì bài toán
    dể hiểu và có tính ứng dụng cao

    Comment by hoangdinhnam — February 19, 2010 @ 4:11

  2. cám ơn cái nữa cho thể hiện thiện ý

    Comment by hoangdinhnam — February 19, 2010 @ 4:12

  3. Anh ơi tại sao g'(x) \leqslant 0 trên (0,1) ạ?

    Comment by suatuoi — January 13, 2011 @ 21:41

    • Em đọc lại chứng minh nhé, chứng minh vừa được bổ sung thêm để làm rõ hơn.

      Comment by Ngô Quốc Anh — January 13, 2011 @ 22:16


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