# 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.