定理1：积分的可加性
设\(c \in (a,b)\)，函数\(f\)在\([a,c],[c,b]\)上可积，那么\(f\)在\([a,b]\)上也可积，且
\[ \int_a^b f(x) \mathrm{d} x = \int_a^c f(x) \mathrm{d} x + \int_c^b f(x) \mathrm{d} x \]
证：由Lebesgue定理易知\(f\)在\([a,b]\)上可积，取分割
\[
\begin{aligned}
& \pi_1: a < a + \frac{1}{n}(c-a) < a+\frac{2}{n}(c-a)
< \cdots < c \\
& \pi_2: c < c + \frac{1}{n}(b-c) < a+\frac{2}{n}(b-c)
< \cdots < b
\end{aligned}
\]
而
\[
\begin{aligned}
\int_a^c f(x) \mathrm{d} x = \lim_{n \to \infty} \sum_{i=1}^n f(a +
\frac{i}{n}(c-a)) \frac{1}{n} \\
\int_c^b f(x) \mathrm{d} x = \lim_{n \to \infty} \sum_{i=1}^n f(c +
\frac{i}{n}(b-c)) \frac{1}{n}
\end{aligned}
\]
取分割\(\pi = \pi_1 + \pi_2\)，即
\[
\pi: a = x_0 < x_1 < \cdots < x_n < x_{n+1} < \cdots
< x_{2n} = b
\]
其中
\[
x_i = \left\{
\begin{aligned}
&a+\frac{i}{n}(c - a) &(i=1, 2, \cdots, n) \\
&c+\frac{i-n}{n}(b - c) &(i=n+1, \cdots, 2n)
\end{aligned}
\right.
\]
此时有
\[
\int_a^c f(x) \mathrm{d} x + \int_c^b f(x) \mathrm{d} x = \lim_{n
\to \infty} \sum_{i=1}^{2n} f(x_i) \frac{1}{n}
\]
有因为\(f\)在\([a,b]\)上可积，所以
\[
\int_a^b f(x) \mathrm{d} x = \lim_{n \to \infty} \sum_{i=1}^{2n}
f(x_i) \frac{1}{n} = \int_a^c f(x) \mathrm{d} x + \int_c^b f(x)
\mathrm{d} x
\]
Q.E.D.
定义1
为了后续积分的扩展，这里定义两个等式。设函数\(f\)在\([a,b]\)上可积，则定义
（1）\(\displaystyle \int_a^a f(x) \mathrm{d} x = 0\)
（2）\(\displaystyle \int_b^a f(x) \mathrm{d} x = - \int_a^b f(x) \mathrm{d} x\)
定理2
如果\(f\)在\([a,b]\)上连续且非负，但\(f\)不恒等于\(0\)，那么
\[ \int_a^b f(x) \mathrm{d} x > 0 \]
证：设有一点\(x_0 \in
[a,b]\)，使得\(f(x_0) >
0\)，则由连续函数的性质可知，存在一个子区间\([\alpha, \beta]\)，满足\(x_0 \in [\alpha, \beta] \subset
[a,b]\)，使得对一切\(x \in [\alpha,
\beta]\)有
\[
f(x) \ge \frac{f(x_0)}{2}
\]
从而
\[
\begin{aligned}
\int_a^b f(x) \mathrm{d} x &= \int_a^\alpha f(x) \mathrm{d} x +
\int_{\alpha}^{\beta} f(x) \mathrm{d} x + \int_{\beta}^b f(x)
\mathrm{d}x \\
&\ge \int_{\alpha}^{\beta} f(x) \mathrm{d} x \ge
\int_{\alpha}^{\beta} \frac{f(x_0)}{2} \mathrm{d} x = \frac{f(x_0)}{2}
(\beta - \alpha) > 0
\end{aligned}
\]
Q.E.D.
定理3
设\(f\)在\([a,b]\)上可积，那么\(|f|\)也在\([a,b]\)上可积，且
\[ \left| \int_a^b f(x) \mathrm{d} x\right| \le \int_a^b |f(x)| \mathrm{d}x \]
证：由于\(D(|f|) \subset
D(f)\)，所以\(|f|\)在\([a,b]\)上可积，而
\[
- |f(x)| \le f(x) \le |f(x)|
\]
所以
\[
- \int_a^b |f(x)| \mathrm{d}x \le \int_a^b f(x) \mathrm{d}x \le
\int_a^b |f(x)| \mathrm{d}x
\]
Q.E.D.
定理4：积分平均值定理
设函数\(f\)与\(g\)在\([a,b]\)上连续，\(g\)在\([a,b]\)上不改变符号，则存在\(\xi \in (a,b)\)，使得
\[ \int_a^b f(x)g(x) \mathrm{d} x = f(\xi) \int_a^b g(x) \mathrm{d} x \]
证：不妨设当\(x \in [a,b]\)时，\(g(x) \le 0\)但不恒等于0，从而\(\displaystyle \int_a^b g(x) \mathrm{d} x >
0\)。设\(m,M\)分别是\(f\)在\([a,b]\)上的最小值和最大值，则
\[
m \le f(x) \le M \quad (a \le x \le b)
\]
从而
\[
mg(x) \le f(x)g(x) \le M g(x)
\]
求积分可得
\[
m\int_a^b g(x) \mathrm{d} x \le \int_a^b f(x)g(x) \mathrm{d} x \le
M\int_a^b g(x) \mathrm{d}x
\]
所以
\[
m \le \int_a^b f(x)g(x) \mathrm{d} x (\int_a^b g(x) \mathrm{d}
x)^{-1} \le M
\]
由连续函数的介值定理可知，存在一点\(\xi \in
(a,b)\)，使得
\[
f(\xi) = \int_a^b f(x)g(x) \mathrm{d} x (\int_a^b g(x) \mathrm{d}
x)^{-1}
\]
Q.E.D.
定理5
设函数\(f\)在区间\([a,b]\)上连续，则存在一点\(\xi \in [a,b]\)，使得
\[ \int_a^b f(x) \mathrm{d} x = f(\xi)(b - a) \]
证：令定理4中的\(g = 1\)，即可得。
Q.E.D.