高等数学基础:定积分的定义与性质

定积分定义

设函数$f(x)$定义在$[a,b]$上,若对$[a,b]$的任一种分法$a=x_0<x_1<x_2<···<x_n=b$,令$\Delta x_i=x_i-x_{i-1}$,任取$\xi_i \in [x_i,x_{i-1}]$,只要$\lambda = \max_{1≤i≤n}\{\Delta x_i\} \to 0$时$\sum_{i=1}^n f(\xi_i)\Delta x_i$总趋于确定的极限$I$,则称此极限$I$为函数$f(x)$在区间$[a,b]$上的定积分,记作$\int_a^b f(x)dx$,即

$$
\int_a^b f(x)dx = \lim \limits_{\lambda \to 0} \sum_{i=1}^n f(\xi_i)\Delta x_i
$$

此时称$f(x)$在$[a,b]$上可积

定积分的几何意义

曲边梯形面积

可积的充分条件

定理1:若函数$f(x)$在$[a,b]$上连续,则$f(x)$在$[a,b]$可积
定理2:若函数$f(x)$在$[a,b]$上有界,且只有有限个间断点,则$f(x)$在$[a,b]$可积

例1:利用定义计算定积分$\int_0^1 x^2 dx$

解:将$[0,1]$进行$n$等分,分点为$x_i=\frac{i}{n} , (i=0,1,···,n)$

取$\xi_i = \frac{i}{n}, \Delta x_i = \frac{1}{n} , (i=0,1,···,n)$,则

$$
f(\xi_i)\Delta x_i = \xi_i^2\Delta x_i = \frac{i^2}{n^3}
$$

$$
\begin{align}
\sum_{i=1}^n f(\xi_i)\Delta x_i &= \frac{1}{n^3} \sum_{i=1}^n i^2 \
&= \frac{1}{n^3}·\frac{1}{6}n(n+1)(2n+1) \
&= \frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n})
\end{align}
$$

这里用到公式

$$
\sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)
$$

$$
\begin{align}
\int_0^1 x^2dx &= \lim \limits_{\lambda \to 0} \sum_{i=1}^n \xi_i^2\Delta x_i \
&= \lim \limits_{n \to ∞} \frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n}) \
&= \frac{1}{3}
\end{align}
$$

例2:用定积分表示下列极限

$$
\lim \limits_{n \to ∞} \frac{1}{n} \sum_{i=1}^n \sqrt{1+\frac{i}{n}} = \lim \limits_{n \to ∞} \sum_{i=1}^n \sqrt{1+\frac{i}{n}}·\frac{1}{n} = \int_0^1 \sqrt{1+x}dx
$$

$$
\lim \limits_{n \to ∞} \frac{1^p+2^p+···+n^p}{n^{p+1}} = \lim \limits_{n \to ∞} \sum_{i=1}^n (\frac{i}{n})^p·\frac{1}{n} = \int_0^1 x^pdx
$$

定积分的性质

(1) $\int_a^b f(x)dx = -\int_b^a f(x)dx$

(2) $\int_a^a f(x)dx = 0$

(3) $\int_a^b dx = b-a$

(4) $\int_a^b kf(x)dx = k\int_a^b f(x)dx$

(5) $\int_a^b [f(x) ± g(x)]dx = \int_a^b f(x)dx ± \int_a^b g(x)dx$

(6) $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$

(7) 若在$[a,b]$上$f(x) ≥ 0$,则$\int_a^b f(x)dx ≥ 0$

(8) 若在$[a,b]$上$f(x) ≤ g(x)$,则$\int_a^b f(x)dx ≤ \int_a^b g(x)dx$

(9) $|\int_a^b f(x)dx| ≤ \int_a^b |f(x)|dx$

(10) 设$M=max_{[a,b]}f(x),m=min_{[a,b]}f(x)$,则$m(b-a) ≤ \int_a^b f(x)dx ≤ M(b-a) , (a<b)$

例3:试证$1 ≤ \int_0^{\frac{\pi}{2}} \frac{\sin x}{x} dx ≤ \frac{\pi}{2}$

证:设$f(x) = \frac{\sin x}{x}$,则在$(0,\frac{\pi}{2})$上,有

$f’(x) = \frac{x\cos x - \sin x}{x^2} = \frac{\cos x}{x^2}(x-\tan x) < 0$

$\therefore f(\frac{\pi}{2}) < f(x) < f(0)$

即$\frac{2}{\pi} < f(x) < 1, , x \in (0,\frac{\pi}{2})$

故$\int_0^{\frac{\pi}{2}} \frac{2}{\pi} dx ≤ \int_0^{\frac{\pi}{2}} f(x) dx ≤ \int_0^{\frac{\pi}{2}} 1 dx$

即$1 ≤ \int_0^{\frac{\pi}{2}} \frac{\sin x}{x} dx ≤ \frac{\pi}{2}$

积分中值定理

若$f(x)\in C[a,b]$,则至少存在一点$\xi \in [a,b]$,使

$$
\int_a^b f(x) dx = f(\xi)(b-a)
$$

$$
\frac{\int_a^b f(x) dx}{b-a} = \frac{1}{b-a} \lim \limits_{n \to ∞} \sum_{i=1}^n f(\xi_i)·\frac{b-a}{n} = \lim \limits_{n \to ∞} \frac{1}{n} \sum_{i=1}^n f(\xi_i)
$$

故积分中值定理是有限个数的平均值概念的推广。