定积分-性质

发表: 5/20/2026 更新: 5/26/2026 字数: 0 字时长: 0 分钟

基本性质

1.规定

(1) $\begin{array}{r} \int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t \end{array}$
(2) $\begin{array}{r} \int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, 特例 : \int_{a}^{a} f (x) d x = 0, \int_{b}^{b} f (x) d x = 0 \end{array}$

注意

对于不定积分: $\int x^{2} d x \neq \int t^{2} d t$
但对于定积分: $\int_{a}^{b} x^{2} d x = \int_{a}^{b} t^{2} d t$
我们可以得出结论:定积分由上下限与被积函数关系决定(因为定积分是一个确定的数值，而不定积分是一族函数，自变量不同，函数就不同).

2.线性性质

(1) $\begin{array}{r} \int_{a}^{b} [f (x) + g (x)] d x = \int_{a}^{b} f (x) d x + \int_{a}^{b} g (x) d x \end{array}$
(2) $\begin{array}{r} \int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x, k 为常数 \end{array}$

3.常数积分性质

$\begin{aligned} \int_{a}^{b} k d x & = k (b - a), k 为常数 \\ \int_{a}^{b} 1 d x & = b - a \end{aligned}$

4.区间可加性

$\begin{array}{r} \int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x \end{array}$

注意

不要求 $a < c < b,$ 只要积分存在就可以使用.

5.比较定理

若在区间 $[a, b]$ 上恒有 $f (x) \geq g (x),$ 则有

\begin{array}{r} \int_{a}^{b} f (x) d x \geq \int_{a}^{b} g (x) d x; \end{array}

若 $f (x), | f (x) |$ 在 $[a, b]$ 上可积，则有

\begin{array}{r} | \int_{a}^{b} f (x) d x | \leq \int_{a}^{b} | f (x) | d x; \end{array}

特例

$\begin{array}{r} f (x) \geq 0, x \in [a, b] \Rightarrow \int_{a}^{b} f (x) d x \geq 0 \end{array}$

6.估值定理

设 $M, m$ 为函数 $f (x)$ 在区间 $[a, b]$ 上的最大值和最小值，则有

\begin{array}{r} m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a); \end{array}

积分中值定理

7.积分中值定理

设函数 $f (x)$ 在区间 $[a, b]$ 上连续，则至少存在一点 $ζ \in [a, b],$ 使得

\int_{a}^{b} f (x) d x = f (ζ) (b - a) .

证:

$\begin{array}{l} f (x) \in c [a, b] \Rightarrow \exists m ， M . 使得 \\ \begin{aligned} m \leq & f (x) \leq M \\ \int_{a}^{b} m d x \leq & \int_{a}^{b} f (x) d x \leq \int_{a}^{b} M d x \\ m (b - a) \leq & \int_{a}^{b} f (x) d x \leq M (b - a) \\ m \leq & \frac{\int_{a}^{b} f (x) d x}{b - a} \leq M \end{aligned} \\ \begin{array}{r} \exists ζ \in [a, b], 使得 f (ζ) = \frac{\int_{a}^{b} f (x) d x}{b - a} \end{array} (介值定理) \\ \begin{array}{r} \Rightarrow \int_{a}^{b} f (x) d x = f (ζ) (b - a), ζ \in [a, b] \end{array} \end{array}$

特殊性质

$f (x) \in c [- a, a],$ 则

\int_{- a}^{a} f (x) d x = \int_{0}^{a} [f (x) + f (- x)] d x

\begin{array}{r} f 奇 函 数 : \int_{- a}^{a} f (x) d x = 0; \end{array}

\begin{array}{r} f 偶 函 数 : \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x . \end{array}

Notes:

① $\begin{array}{r} \int_{- a}^{0} f (x) d x \overset{x = - t}{=} \int_{0}^{a} f (- x) d x \end{array}$
② $\begin{array}{r} \int_{a}^{a + b} f (x) d x \overset{x = t + a}{=} \int_{0}^{b} f (x + a) d x \end{array}$
③ $\begin{array}{r} \int_{a}^{b} f (x) d x \overset{x + t = a + b}{=} \int_{a}^{b} f (a + b - x) d x \end{array}$
④ $\begin{array}{r} \int_{a}^{b} f (x) d x \overset{x = a + (b - a) t}{=} (b - a) \int_{0}^{1} f (a + (b - a) x) d x \end{array}$

设 $f (x) = c [0, 1] .$ 则
(1)

\int_{0}^{\frac{π}{2}} f (\sin x) d x = \int_{0}^{\frac{π}{2}} f (\cos x) d x

根据 1.③ 即可得证

⭐特别地

$\begin{array}{l} \begin{array}{r} \int_{0}^{\frac{π}{2}} \sin^{n} x d x = \int_{0}^{\frac{π}{2}} \cos^{n} x d x ≜ I_{n} (华莱士公式) \end{array} \\ {\begin{aligned} I_{n} & = \frac{n - 1}{n} I_{n - 2} \\ I_{0} & = \frac{π}{2}, I_{1} = 1. \end{aligned} \end{array}$

(2)

\begin{aligned} \int_{0}^{π} f (\sin x) d x & = 2 \int_{0}^{\frac{π}{2}} f (\sin x) d x \\ ⇕ \\ \int_{0}^{\frac{π}{2}} f (\sin x) d x & = \int_{\frac{π}{2}}^{π} f (\sin x) d x \end{aligned}

根据 1.② 和 2.(1) 即可得证

\begin{aligned} \int_{0}^{π} f (| \cos x |) d x & = 2 \int_{0}^{\frac{π}{2}} f (\cos x) d x \\ \int_{0}^{π} f (\cos^{2 n} x) d x & = 2 \int_{0}^{\frac{π}{2}} f (\cos^{2 n} x) d x \end{aligned}

(3)

\int_{0}^{π} x f (\sin x) d x = \frac{π}{2} \int_{0}^{π} f (\sin x) d x

根据 1.③ 即可得证

$\begin{array}{r} 例 1. I = \int_{0}^{1} \frac{d x}{x + \sqrt{1 - x^{2}}} \end{array}$

$\begin{aligned} I & = \int_{0}^{1} \frac{d x}{x + \sqrt{1 - x^{2}}} \overset{令 x = \sin t}{=} \int_{0}^{\frac{π}{2}} \frac{\cos t}{\sin t + \cos t} d t = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\sin x + \cos x} d x \\ I & = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\sin x + \cos x} d x = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\sqrt{1 - \cos^{2} x} + \cos x} d x = \int_{0}^{\frac{π}{2}} \frac{\sin x}{\sqrt{1 - \sin^{2} x} + \sin x} d x \\ = \int_{0}^{\frac{π}{2}} \frac{\sin x}{\cos x + \sin x} d x \\ 2 I & = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\sin x + \cos x} d x + \int_{0}^{\frac{π}{2}} \frac{\sin x}{\cos x + \sin x} d x = \int_{0}^{\frac{π}{2}} 1 \cdot d x = \frac{π}{2} \\ I & = \frac{π}{4} \end{aligned}$

$\begin{array}{r} 例 2. I = \int_{- π}^{π} \frac{\sin^{2} x}{1 + e^{- x}} d x . \end{array}$

$\begin{aligned} I & = \int_{0}^{π} (\frac{\sin^{2} x}{1 + e^{- x}} + \frac{\sin^{2} x}{1 + e^{x}}) d x \\ = \int_{0}^{π} (\frac{1}{1 + e^{- x}} + \frac{1}{1 + e^{x}}) \sin^{2} x d x \\ = \int_{0}^{π} \sin^{2} x d x \\ = 2 \int_{0}^{\frac{π}{2}} \sin^{2} x d x = 2 \cdot I_{2} = 2 \cdot \frac{1}{2} \cdot \frac{π}{2} \\ = \frac{π}{2} \end{aligned}$

设 $f (x)$ 连续且以 $T > 0$ 为周期，a 为常数，则

\begin{matrix} (平移性质) & \begin{aligned} \int_{a}^{a + T} f (x) d x & = \int_{0}^{T} f (x) d x \\ \int_{0}^{n T} f (x) d x & = n \int_{0}^{T} f (x) d x (n \in N) \end{aligned} \end{matrix}

$\begin{aligned} \int_{a}^{a + T} f (x) d x & = \int_{a}^{0} f (x) d x + \int_{0}^{T} f (x) d x + \int_{T}^{a + T} f (x) d x \\ = \int_{a}^{0} f (x) d x + \int_{0}^{T} f (x) d x + \int_{0}^{a} f (x + T) d x \\ = \int_{0}^{T} f (x) d x \end{aligned}$
$\begin{aligned} \int_{0}^{n T} f (x) d x & = \int_{0}^{T} f (x) d x + \int_{T}^{2 T} f (x) d x + \dots + \int_{(n - 1) T}^{n T} f (x) d x \\ = \sum_{k = 1}^{n} \int_{(k - 1) T}^{(k - 1) T + T} f (x) d x = \sum_{k = 1}^{n} \int_{0}^{T} f (x) d x \\ = n \int_{0}^{T} f (x) d x \end{aligned}$

错题集

其他重要的公式定理

定积分-性质

基本性质

积分中值定理

特殊性质

$\begin{array}{r} 例 1. I = \int_{0}^{1} \frac{d x}{x + \sqrt{1 - x^{2}}} \end{array}$

$\begin{array}{r} 例 2. I = \int_{- π}^{π} \frac{\sin^{2} x}{1 + e^{- x}} d x . \end{array}$

定积分-性质 ​

基本性质 ​

积分中值定理 ​

特殊性质 ​

例例1. I=∫01dxx+1−x2

例例2. I=∫−ππsin2⁡x1+e−xdx.

定积分-性质

基本性质

积分中值定理

特殊性质

$\begin{array}{r} 例 1. I = \int_{0}^{1} \frac{d x}{x + \sqrt{1 - x^{2}}} \end{array}$

$\begin{array}{r} 例 2. I = \int_{- π}^{π} \frac{\sin^{2} x}{1 + e^{- x}} d x . \end{array}$