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.


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