postgraduate-prep/subjects/math/04_积分.md

175 lines
5.8 KiB
Markdown
Raw Blame History

This file contains ambiguous Unicode characters

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

## 笔记记录
### 要点 01 - 积分与极限求和式的转化
根据公式
$$
\int_a^b f(x)dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x
$$
对于均匀矩形分割的情况,实际上只用分离出 $\frac{1}{n}$
$$
\int_a^b f(x)dx = \lim_{n\to\infty}\sum_{i=1}^n f\left(a + \frac{(b-a) i}{n}\right) \frac{b-a}{n}
$$
---
### 要点 02 - $\frac{1}{x^2 + a^2}$ 型积分
#### 基本公式
$$
\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C \quad (a > 0)
$$
**推导**(换元法):令 $x = a \tan t$,则 $dx = a \sec^2 t \, dt$
$$
\int \frac{dx}{x^2 + a^2} = \int \frac{a \sec^2 t}{a^2 + a^2 \tan^2 t} \, dt = \int \frac{a \sec^2 t}{a^2 \sec^2 t} \, dt = \int \frac{dt}{a} = \frac{t}{a} + C = \frac{1}{a} \arctan \frac{x}{a} + C
$$
#### 推广形式
$$
\begin{aligned}
\int \frac{dx}{x^2 + a^2} &= \frac{1}{a} \arctan \frac{x}{a} + C \\
\int \frac{dx}{b^2 + (x + c)^2} &= \frac{1}{b} \arctan \frac{x + c}{b} + C \\
\int \frac{x \, dx}{x^2 + a^2} &= \frac{1}{2} \ln(x^2 + a^2) + C \\
\int \frac{dx}{(x^2 + a^2)^2} &= \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3} \arctan \frac{x}{a} + C
\end{aligned}
$$
---
### 要点 03 - $\frac{1}{\sqrt{x^2 \pm a^2}}$ 型积分
#### 基本公式
$$
\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C = \operatorname{arsinh} \frac{x}{a} + C
$$
$$
\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C = \operatorname{arcosh} \frac{x}{a} + C \quad (|x| > |a|)
$$
$$
\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C = -\arccos \frac{x}{a} + C \quad (|x| < |a|)
$$
#### 推导方法
**$\sqrt{x^2 + a^2}$ 型**:令 $x = a \sinh t$,则 $dx = a \cosh t \, dt$$\sqrt{x^2 + a^2} = a \cosh t$
$$
\int \frac{dx}{\sqrt{x^2 + a^2}} = \int \frac{a \cosh t}{a \cosh t} \, dt = \int dt = t + C = \ln\left|x + \sqrt{x^2 + a^2}\right| + C
$$
**$\sqrt{x^2 - a^2}$ 型**:令 $x = a \cosh t$$x > a$),则 $dx = a \sinh t \, dt$$\sqrt{x^2 - a^2} = a \sinh t$
$$
\int \frac{dx}{\sqrt{x^2 - a^2}} = \int \frac{a \sinh t}{a \sinh t} \, dt = \int dt = t + C = \ln\left|x + \sqrt{x^2 - a^2}\right| + C
$$
**$\sqrt{a^2 - x^2}$ 型**:令 $x = a \sin t$,则 $dx = a \cos t \, dt$$\sqrt{a^2 - x^2} = a \cos t$
$$
\int \frac{dx}{\sqrt{a^2 - x^2}} = \int \frac{a \cos t}{a \cos t} \, dt = \int dt = t + C = \arcsin \frac{x}{a} + C
$$
#### 等效形式
$$
\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|\frac{x}{a} + \sqrt{\frac{x^2}{a^2} - 1}\right| + C
$$
#### 推广形式
$$
\begin{aligned}
\int \frac{dx}{\sqrt{(x + b)^2 + a^2}} &= \ln\left|x + b + \sqrt{(x + b)^2 + a^2}\right| + C \\
\int \frac{dx}{\sqrt{(x + b)^2 - a^2}} &= \ln\left|x + b + \sqrt{(x + b)^2 - a^2}\right| + C \quad (|x + b| > |a|) \\
\int \frac{x \, dx}{\sqrt{x^2 + a^2}} &= \sqrt{x^2 + a^2} + C \\
\int \frac{x \, dx}{\sqrt{x^2 - a^2}} &= \sqrt{x^2 - a^2} + C \\
\int \sqrt{x^2 + a^2} \, dx &= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C \\
\int \sqrt{x^2 - a^2} \, dx &= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
\end{aligned}
$$
---
### 要点 04 - $\sqrt{x^2 \pm a^2}$ 型积分
#### 基本公式
$$
\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
$$
$$
\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
$$
$$
\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C \quad (|x| < |a|)
$$
#### 推导方法
**$\sqrt{x^2 + a^2}$ 型**:令 $x = a \tan t$,则 $dx = a \sec^2 t \, dt$$\sqrt{x^2 + a^2} = a \sec t$
$$
\begin{aligned}
\int \sqrt{x^2 + a^2} \, dx &= \int a \sec t \cdot a \sec^2 t \, dt = a^2 \int \sec^3 t \, dt \\
&= a^2 \cdot \frac{1}{2}(\sec t \tan t + \ln|\sec t + \tan t|) + C \\
&= \frac{a^2}{2}\left(\frac{x}{a}\sqrt{1 + \frac{x^2}{a^2}} + \ln\left|\frac{x}{a} + \sqrt{1 + \frac{x^2}{a^2}}\right|\right) + C \\
&= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
\end{aligned}
$$
**$\sqrt{x^2 - a^2}$ 型**:令 $x = a \sec t$$x > a$),则 $dx = a \sec t \tan t \, dt$$\sqrt{x^2 - a^2} = a \tan t$
$$
\begin{aligned}
\int \sqrt{x^2 - a^2} \, dx &= \int a \tan t \cdot a \sec t \tan t \, dt = a^2 \int \sec t \tan^2 t \, dt \\
&= a^2 \int \sec t (\sec^2 t - 1) \, dt = a^2 \int (\sec^3 t - \sec t) \, dt \\
&= a^2 \left[\frac{1}{2}(\sec t \tan t + \ln|\sec t + \tan t|) - \ln|\sec t + \tan t|\right] + C \\
&= \frac{a^2}{2}\sec t \tan t - \frac{a^2}{2}\ln|\sec t + \tan t| + C \\
&= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
\end{aligned}
$$
**$\sqrt{a^2 - x^2}$ 型**:令 $x = a \sin t$,则 $dx = a \cos t \, dt$$\sqrt{a^2 - x^2} = a \cos t$
$$
\begin{aligned}
\int \sqrt{a^2 - x^2} \, dx &= \int a \cos t \cdot a \cos t \, dt = a^2 \int \cos^2 t \, dt \\
&= a^2 \int \frac{1 + \cos 2t}{2} \, dt = \frac{a^2}{2}\left(t + \frac{\sin 2t}{2}\right) + C \\
&= \frac{a^2}{2}\left(\arcsin\frac{x}{a} + \frac{x}{a}\sqrt{1 - \frac{x^2}{a^2}}\right) + C \\
&= \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C
\end{aligned}
$$
#### 推广形式
$$
\begin{aligned}
\int (x + b)\sqrt{x^2 + a^2} \, dx &= \frac{1}{3}(x + b)(x^2 + a^2)^{3/2} - \frac{b}{2}x\sqrt{x^2 + a^2} - \frac{ab^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C \\
\int x\sqrt{x^2 + a^2} \, dx &= \frac{1}{3}(x^2 + a^2)^{3/2} + C \\
\int x\sqrt{x^2 - a^2} \, dx &= \frac{1}{3}(x^2 - a^2)^{3/2} + C \\
\int x\sqrt{a^2 - x^2} \, dx &= -\frac{1}{3}(a^2 - x^2)^{3/2} + C
\end{aligned}
$$
---
### 知识点
- 定积分的定义
- 黎曼和与积分的关系
- 均匀分割技巧
- $\frac{1}{x^2 + a^2}$ 型积分公式
- $\frac{1}{\sqrt{x^2 \pm a^2}}$ 型积分公式
- $\sqrt{x^2 \pm a^2}$ 型积分公式