Section01_概念与性质

背景

1. 已知速度与时间的关系$v=v(t), t\in[a,b]$，求路程$S$ $$\begin{array}{lc} \text{Step 01}\\ &a= t_{0} <t_{1}<t_{2}<\cdots < t_{n} = b\\ &[a,b]= [t_{0},t_{1}]\cup[t_{1},t_{2}]\cup\cdots\cup[t_{n-1},t_{n}]\\ &\Delta t_{i} = t_{1}-t_{i-1}\quad i\in\{1,2,\cdots,n\}\\\\ \text{Step 02}\\ &\forall\ \xi\in[t_{i-1},t_{i}]\quad i\in\{1,2,\cdots,n\}\\ &\Delta S_{i} \approx f(\xi)\Delta t_{i} \\ &\displaystyle S\approx \sum_{i=1}^{n}f(\xi_{i})\Delta t_{i} \\\\ \text{Step 03}\\ &\lambda = \max\{\Delta t_{1},\Delta t_{2},\cdots, \Delta t_{n}\} \\ &\displaystyle S = \lim_{\lambda\to 0}\sum_{i=1}^{n} v(\xi_{i})\Delta t_{i} \end{array}$$
2. 已知$y = f(x)\ge 0\quad(x\in[a,b])$ $f(x)$与$x=a,x=b$以及$x$轴围成图形的面积 $$\begin{array}{lc} \text{Step 01}\\ &a= x_{0} <x_{1}<x_{2}<\cdots < x_{n} = b\\ &[a,b]= [x_{0},x_{1}]\cup[x_{1},x_{2}]\cup\cdots\cup[x_{n-1},x_{n}]\\ &\Delta x_{i} = x_{1}-x_{i-1}\quad i\in\{1,2,\cdots,n\}\\\\ \text{Step 02}\\ &\forall\ \xi\in[x_{i-1},x_{i}]\quad i\in\{1,2,\cdots,n\}\\ &\Delta A_{i} \approx f(\xi)\Delta x_{i} \\ &\displaystyle A\approx \sum_{i=1}^{n}f(\xi_{i})\Delta x_{i} \\\\ \text{Step 03}\\ &\lambda = \max\{\Delta x_{1},\Delta x_{2},\cdots, \Delta x_{n}\} \\ &\displaystyle A = \lim_{\lambda\to 0}\sum_{i=1}^{n} v(\xi_{i})\Delta x_{i} \end{array}$$

   定义

3. 设 $f(x)$ 在 $[a,b]$ 上有界 $$\begin{array}{lc} \text{Step 01}\\ &a= x_{0} <x_{1}<x_{2}<\cdots < x_{n} = b\\ &[a,b]= [x_{0},x_{1}]\cup[x_{1},x_{2}]\cup\cdots\cup[x_{n-1},x_{n}]\\ &\Delta x_{i} = x_{1}-x_{i-1}\quad i\in\{1,2,\cdots,n\}\\\\ \text{Step 02}\\ &\exists\ \xi_{i}\in[x_{i-1},x_{i}]\quad i\in\{1,2,\cdots,n\}\text{作}\\ & \displaystyle \sum_{i=1}^{n} f(\xi_{i})\Delta x_{i}\\\\ \text{Step 03}\\ &\lambda = \max\{x_{1},x_{2},\cdots, x_{n}\} \\ \end{array}$$
   • 若$\displaystyle \lim_{\lambda\to 0}\sum_{i=1}^{n}f(\xi_{i})\Delta x_{i}\ \exists$，称 $f(x)$ 在 $[a,b]$ 上可积，极限值称为 $f(x)$ 在 $[a,b]$ 上的定积分，记为 $\displaystyle \int_{a}^{b} f(x)\cdot dx$，即 $\displaystyle \int_{a}^{b} f(x)\cdot dx \overset{\Delta}{=} \lim_{\lambda\to 0}\sum_{i=1}^{n}f(\xi_{i})\Delta x_{i}$

     Notes

     1. $\displaystyle \int_{a}^{a}f(x)\cdot dx = 0 ;\int_{a}^{b} f(x)\cdot dx = -\int_{b}^{a}f(x)\cdot dx$
     2. 当 $f(x)\in C[a,b]$ 或在 $[a,b]$ 上除有限个第一类间断点外处处连续时，$f(x)$ 在 $[a,b]$ 上可积
     3. $\displaystyle \lim_{\lambda \to 0}\sum_{i=1}^{n}f(\xi_{i})\cdot dx$ 与 $\begin{cases}\displaystyle[a,b]\text{的分法} \\ \displaystyle \xi_{i}\text{的取法}\end{cases}$ 无关
     4. $f(x)$ 在 $[a,b]$ 上有界，不一定可积

        Proof $$\begin{array}{ll} & f(x) = \begin{cases} 1, x\in \mathbb{Q} \\ -1, x\in \mathbb{R}\setminus \mathbb{Q}\\ \end{cases} \\ & \displaystyle f(x) \text{在} [a,b] \text{上有界，对于} \lim_{\lambda\to 0}\sum_{i=1}^{n}f(\xi_{i})\cdot \Delta x_{i}\\ \because &\displaystyle \forall\ \xi_{i}\in \mathbb{Q}, \lim_{\lambda\to 0}\sum_{i=1}^{n}f(\xi_{i})\Delta x_{i} = \lim_{\lambda\to 0}\sum_{i=1}^{n}\Delta x_{i} = b-a \\ &\displaystyle \forall\ \xi_{i}\in \mathbb{R}\setminus \mathbb{Q}, \lim_{\lambda\to 0} \sum_{i=1}^{n}f(\xi_{i})\Delta x_{i} = -\lim_{\lambda\to0}\sum_{i=1}^{n}\Delta x_{i} = a-b \\ \therefore & \displaystyle \lim_{\lambda\to 0}\sum_{i=1}^{n}f(\xi_{i})\Delta x_{i}\ \nexists \Rightarrow f(x)\text{在}[a,b]\text{上不可积} \\\end{array}$$

     5. $\lambda\to 0\nLeftarrow\Rightarrow n\to \infty$
     6. 设 $f(x)$ 在 $[0,1]$ 上可积

        $$\begin{array}{ll} & [0,1] = [0,\frac{1}{n}] \cup [\frac{1}{n}, \frac{2}{n}] \cup \cdots \cup [\frac{n-1}{n}, \frac{n}{n}]\\ & \Delta x_{1} = \Delta x_{2} = \cdots = \Delta x_{n} = \frac{1}{n}\\ \therefore & \lambda \to 0 \Leftrightarrow n\to \infty \\ \therefore & \xi_{1} = \frac{1}{n}, \xi_{2} = \frac{2}{n}, \cdots, \xi_{n} = \frac{n}{n} \\ \therefore & \displaystyle \sum_{i=1}^{n} f(\xi_{i})\Delta x_{i} = \sum_{i=1}^{n}\frac{1}{n}\cdot f\bigg(\frac{i}{n}\bigg) \\ \therefore & \displaystyle \int_{0}^{1}f(x)\cdot dx = \lim_{\lambda \to 0} \sum_{i=1}^{n} f(\xi_{i})\Delta x_{i}\\ & \displaystyle = \lim_{n\to \infty} \sum_{i=1}^{n}\frac{1}{n}\cdot f\bigg(\frac{i}{n}\bigg) \end{array}$$

一般性质

1. $\displaystyle \int_{a}^{b}[f(x) + g(x)]\cdot dx = \int_{a}^{b}f(x)\cdot dx + \int_{a}^{b}g(x)\cdot dx$
   1. $\displaystyle \int_{a}^{b}kf(x)\cdot dx = k \int_{a}^{b}f(x)\cdot dx$
   2. $\displaystyle \int_{a}^{b}f(x)\cdot dx = \int_{a}^{c}f(x)\cdot dx + \int_{c}^{b}f(x)\cdot dx$
   3. $\displaystyle \int_{a}^{b}1\cdot dx = b-a$
   4. 不等式
   5. 若$\displaystyle f(x)\ge 0\quad (a\le x\le b)\Rightarrow \int_{a}^{b}f(x)\cdot dx\ge 0$
   6. 若$\displaystyle f(x)\ge g(x)\quad (a\le x\le b)\Rightarrow \int_{a}^{b}f(x)\cdot dx \ge \int_{a}^{b}g(x)\cdot dx$
   7. 重点 若$f(x), \vert f(x) \vert$在$[a,b]$上可积，则 $$\displaystyle \color{#D0104C}{ \left\vert \int_{a}^{b} f(x)\cdot dx \right\vert \le \int_{a}^{b} \vert f(x) \vert\cdot dx}$$
   8. 重点 （积分中值定理） 设 $f(x)\in C[a,b]$ 则 $\exists\ \xi\in[a,b]$，使得 $\displaystyle \int_a^bf(x)\cdot dx = f(\xi)\cdot (b-a)$

Proof $$\begin{array}{ll} \because & f(x)\in C[a,b] \\ & \text{设}\exists\ m, M, m\le f(x)\le M \quad (x\in [a,b])\\ \therefore & \displaystyle m(b-a)\le \int_{a}^{b}f(x)\cdot dx \le M(b-a) \\ & \displaystyle m\le \frac{1}{b-a}\int_{a}^{b}f(x)\cdot dx \le M \\ \therefore &\displaystyle \exists\ \xi\in [a,b], f(\xi) = \frac{1}{b-a}\int_{a}^{b}f(x)\cdot dx \\ \therefore & \displaystyle \exists\ \xi\in[a,b], \int_{a}^{b}f(x)\cdot dx = (b-a)f(\xi) \\\end{array}$$

1. 重点 （积分中值定理的推广） 设 $f(x)\in C[a,b]$ 则 $\exists\ \xi\in(a,b)$，使得$\displaystyle \int_{a}^{b}f(x)\cdot dx = f(\xi)\cdot (b-a)$

Proof $$\begin{array}{ll} & \displaystyle\text{令}F(x) = \int_{a}^{x} f(x) \cdot dx \\ & \displaystyle\text{则} F'(x) = f(x), F(a) = 0, F(b) = \int_{a}^{b}f(x)\cdot dx\\ \because & \exists\ \xi\in (a,b), F(b) - F(a) = (b-a)F'(\xi) \\ \therefore & \displaystyle \int_{a}^{b}f(x)\cdot dx = (b-a)f(\xi) \\\end{array}$$

例题

• 例1 $f(x)\in C[0,1]$，且 $(0,1)$ 内可导，$\displaystyle f(0)=\int_{0}^{1} f(x)\cdot dx$，证 $\exists\ \xi\in(0,1)$，使得 $f'(\xi) =0$ $$\begin{array}{ll} \because & f(x)\in C[a,b] \\ \therefore & \displaystyle\exists\ c\in (0,1), f(c)(1-0) = \int_{0}^{1}f(x)\cdot dx \\ \because & \displaystyle \int_{0}^{1}f(x)\cdot dx= f(0) = f(c) \\ \therefore & \exists\ \xi\in(0,c)\subset(0,1), f'(\xi) = 0 \\ \end{array}$$