Bài 112062

Question

Cho hàm số $f$ có đạo hàm trên $[a;b]$ thỏa mãn $|f'(x)| \leq 1 , \forall x \in [a;b].$
Chứng minh rằng :
Nếu $ a = x_0<x_1<...<x_n = b, max ${$x_i-x_{i-1}$/$i=\overline{1,n}$}$\leq \frac{1}{b-a}$
và $ \epsilon _i \in [x_{i-1};x_i] $ thì$\left| {\int\limits_{a}^{b} f(x)dx - \sum\limits_{i = 1}^n {f({\varepsilon _i})({x_{i - }}{x_{i - 1}})}} \right|\leq \frac{1}{2} $

score 0 · Answer 1

Ta có : $ \int\limits_{a}^{b} f(x)dx - \sum\limits_{i = 1}^n {f({\varepsilon _i})({x_{i }}-{x_{i - 1}})} = \sum\limits_{i = 1}^n {\int\limits_{x_{i-1}}^{x_i} [f(x)-f(\epsilon _i)]dx }$
Mặt khác : $ | f'(x)| \leq 1 $ nên theo định lí Lagrange:
$|f(x) - f(\epsilon _i) \leq | x- \epsilon _i|, \forall x \in [a;b]$
$\Rightarrow | \int\limits_{a}^{b} f(x) - \sum\limits_{i = 1}^n {f({\varepsilon _i})({x_{i }}-{x_{i - 1}})}| \leq \sum\limits_{i = 1}^n {\int\limits_{x_{i-1}}^{x_i} [f(x)-f(\epsilon _i)]dx } $
$\leq \sum\limits_{i = 1}^n {\int\limits_{x_{i-1}}^{x_i}|x-\epsilon _i|dx } = \sum\limits_{i = 1}^n {(\int\limits_{x_{i-1}}^{\epsilon _i}|x-\epsilon _i|dx + \int\limits_{\epsilon _i}^{x_i}|x-\epsilon _i|dx }$
$ = \sum\limits_{i = 1}^n {(\int\limits_{x_{i-1}}^{\epsilon _i}(\epsilon _i -x )dx + \int\limits_{\epsilon _i}^{x_i} (x-\epsilon _i)dx )}$
$ = \sum ${$ ( -\frac{1}{2} (\epsilon _i - x)^2 \left| \begin{array}{l}
{\varepsilon _i}\\
{x_{i - 1}}
\end{array} \right. + (\frac{1}{2} (x-\epsilon _i)^2)\left| \begin{array}{l}
x_i\\
{\varepsilon _i}
\end{array} \right.$}
$= \sum\limits_{i = 1}^n {[\frac{1}{2} (\epsilon _i - x _{i-1})^2 + \frac{1}{2} (x_i - \epsilon _i)^2]} \leq \sum\limits_{i = 1}^n {\frac{1}{2} (x_i - x_{i-1})^2} $
$ ( vì a^2+ b^2 \leq (a+b)^2, \forall a , b \geq 0).$
$\leq \frac{1}{2(b-a)}\sum\limits_{i = 1}^n {x_i - x_{i-1}} = \frac{1}{2(b-a)}.(b-a) = \frac{1}{2} $

Bài 112062

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Bài 112062

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