From e47e2ce4d1ec8a3150734cddbe434f30418e7d98 Mon Sep 17 00:00:00 2001
From: camera-2018 <2907618001@qq.com>
Date: Wed, 19 Apr 2023 00:01:27 +0800
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--- a/.vitepress/cache/deps/_metadata.json
+++ b/.vitepress/cache/deps/_metadata.json
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+ "browserHash": "659eb1f9",
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"src": "../../../node_modules/vue/dist/vue.runtime.esm-bundler.js",
"file": "vue.js",
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+ "fileHash": "8c80ebf5",
"needsInterop": false
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+++ b/.vitepress/theme/style.css
@@ -105,7 +105,7 @@ mjx-container {
.VPHero .image-bg {
z-index: -1;
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+ opacity: 0.8;
transition: opacity 1s ease;
}
@@ -143,6 +143,8 @@ mjx-container {
/* Safari has a very bad performance on gradient and filter */
.browser-safari, .browser-firefox {
+ --vp-home-hero-name-color: var(--vp-c-brand);
+ --vp-home-hero-name-background: transparent;
--vp-home-hero-image-background-image: transparent;
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本文主要讲解定积分在计算时运用的各种化简技巧。
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# 一：由不定积分计算定积分
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## 1.1原函数成立区间的问题
注意：由不定积分推导出的原函数在定积分的区间不一定成立！！！
对于大多数定积分题目，可直接由代入不定积分的结论。
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# 二：利用对称性化简定积分
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## 2.1奇偶对称和周期性
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命题1：设函数f在区间$[0,a]$可积，且关于区间中点$\frac{a}{2}$为奇函数，即对于$x\in[0,a]$，有$f(x)=-f(a-x)$，则成立
$\int_0^af(x)dx=0$。
命题2：设函数f在区间$[0,a]$可积，且关于区间中点$\frac{a}{2}$为偶函数，即对于$x\in[0,a]$，有$f(x)=f(a-x)$， 则成立
$\int_0^af(x)dx=2\int_0^{\frac{a}{2}}f(x)dx$。
命题3：若函数f为定义在$-\infty$\int_a^{a+T}f(x)dx=\int_0^Tf(x)dx$
**_Proof：_**
将积分区间拆成多段，带入上文对称公式，之后换元即可，证明的本质是统一了积分区间。
_**Example：**
$\int_0^{2\pi}\frac{dx}{sin^4x+cos^4x}$
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+命题1：设函数f在区间$[0,a]$可积，且关于区间中点$\frac{a}{2}$为奇函数，即对于$x\in[0,a]$，有$f(x)=-f(a-x)$，则成立
$\int_0^af(x)dx=0$。
命题2：设函数f在区间$[0,a]$可积，且关于区间中点$\frac{a}{2}$为偶函数，即对于$x\in[0,a]$，有$f(x)=f(a-x)$， 则成立
$\int_0^af(x)dx=2\int_0^{\frac{a}{2}}f(x)dx$。
命题3：若函数f为定义在$-\infty$\int_a^{a+T}f(x)dx=\int_0^Tf(x)dx$
**_Proof：_**
将积分区间拆成多段，带入上文对称公式，之后换元即可，证明的本质是统一了积分区间。
_**Example：**
$\int_0^{2\pi}\frac{dx}{sin^4x+cos^4x}$
利用：$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-\frac{1}{2}sin^2(2x)=\frac{1}{2}(1+cos^2(2x))$，周期为$\pi/2$
两次利用命题3，将积分区间化简为$[0,\frac{\pi}{2}]$。由于被积函数关于$\pi/4$偶对称，利用命题2，可将积分区间化简为$[0,\frac{\pi}{4}]$，故原式等于：
$4\int_0^{\frac{\pi}{2}}\frac{dx}{sin^4x+cos^4x}=8\int_0^{\frac{\pi}{4}}\frac{dx}{sin^4x+cos^4x}=16\int_0^{\frac{\pi}{4}}\frac{dx}{1+cos^2(2x)}=8\int_0^{\frac{\pi}{2}}\frac{dt}{1+cos^2(t)}=8\int_0^{\frac{\pi}{2}}\frac{d(tant)}{2+tan^2t}$
$=8\int_{0}^{+\infty}\frac{du}{u^2+2}=4\sqrt{2}arctan(\frac{u}{\sqrt{2}})|_0^{+\infty}=2\sqrt{2}\pi$_
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## 2.2积分区间的缩小
命题4：设f在$[0,a]$上可积，则成立$\int_0^af(x)dx=\int_0^{\frac{a}{2}}[f(x)+f(a-x)]dx$，此为命题1,2的推广形式。
可推广为：$\int_a^bf(x)dx=\int_a^{\frac{a+b}{2}}[f(x)+f(a+b-x)]dx$，该命题在$f(x)+f(a+b-x)$更好积分时使用。
**_Proof：_
反向利用命题2，$\int_0^af(x)dx=\frac{1}{2}(\int_0^af(x)dx+\int_0^af(a-x)dx)=\int_0^{\frac{a}{2}}[f(x)+f(a-x)]dx$
命题5：若f(x)在闭区间$[0,1]$上连续，则
（a）$\int_0^{\frac{\pi}{2}}f(sinx)dx=\int_0^{\frac{\pi}{2}}f(cosx)dx$
（b）$\int_0^{\pi}xf(sinx)dx=\frac{\pi}{2}\int_0^{\pi}f(sinx)dx$
（c）$\int_0^{\pi}f(sinx)dx=2\int_0^{\frac{\pi}{2}}f(sinx)dx$
**_Proof：_**
（a）$x=\pi/2-t$，（b）$x=\pi-t$，（c）$t=x-\pi/2$
也可由命题4直接推出，如对于（b）
$\int_0^{\pi}(x-\pi/2)f(sinx)dx=\int_0^{\frac{\pi}{2}}(x-\pi/2+\pi-x-\pi/2)f(sinx)dx=0$
_**Example1：**_
（a）$\int_0^{\pi/2}\frac{sinx}{sinx+cosx}dx$ （b）$\int_0^{\pi/2}ln(sinx)dx$ （c）$\int_0^{\pi/2}ln(tanx)dx$
（d）$\int_0^{\pi/2}\frac{1}{1+tan^ax}dx$ （e）$\int_0^{\pi}\frac{xsinx}{1+cos^2x}dx$ （f）$\int_0^1\frac{lnx}{\sqrt{1-x^2}}dx$
例如，对于b，利用命题4，$\int_0^{\frac{\pi}{4}}[ln(sin(x))+ln(cos(x))]dx=\int_0^{\frac{\pi}{4}}[ln(sin(2x)/2]dx$
$=\int_0^{\frac{\pi}{4}}[ln(sin(2x))-ln(2)]dx=\frac{1}{2}\int_0^{\frac{\pi}{2}}ln(sin(x))dx-\frac{\pi}{4}ln2=-\frac{\pi}{2}ln2$
_**Example2：**_
设$f(x)$在$[a,b]$上连续且单调增加，证明：$\int_a^bxf(x)dx\geq\frac{a+b}{2}\int_a^bf(x)dx$
$\int_a^b(x-\frac{a+b}{2})f(x)dx=\int_a^{\frac{a+b}{2}}[(x-\frac{a+b}{2})f(x)+(a+b-x-\frac{a+b}{2})f(a+b-x)]dx$
$=\int_a^{\frac{a+b}{2}}(x-\frac{a+b}{2})[f(x)-f(a+b-x)]dx$
因为$f(x)$单增，所以$f(x)-f(a+b-x)\leq0$，原式大于等于0，证毕。
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# 三：利用递归表达式化简定积分
注意：全部利用分部积分构造递推表达式，使得分部积分脱出积分号的项为0，对积分内指数进行降阶即可。
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## 3.1高阶三角函数
-$I_0=\int_0^{\pi/2}dx=\pi/2$，$I_1=\int_0^{\pi/2}sinxdx=1$，$I_2=\int_0^{\pi/2}sin^2xdx=\pi/4$
思考：如果$I_n$为一数列$I_n=\int_0^{\pi/2}sin^nxdx$，请问
（1）$I_n$的递推表达式为？
（2）$I_n$的通项表达式为？
（3）$I_n(n\to\infty)$的极限为？
（4）求$\lim_{k\to\infty}\frac{I_{2k+1}}{I_{2k}}$？
答案：
（1），（2）
+$I_0=\int_0^{\pi/2}dx=\pi/2$，$I_1=\int_0^{\pi/2}sinxdx=1$，$I_2=\int_0^{\pi/2}sin^2xdx=\pi/4$
思考：如果$I_n$为一数列$I_n=\int_0^{\pi/2}sin^nxdx$，请问
（1）$I_n$的递推表达式为？
（2）$I_n$的通项表达式为？
（3）$I_n(n\to\infty)$的极限为？
（4）求$\lim_{k\to\infty}\frac{I_{2k+1}}{I_{2k}}$？
答案：
（1），（2）
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推论：由命题4得，对cosx做上述操作得到结果相同。
（3）$I_n(n\to\infty)$的极限为0，如图为$sin^n(x)在[0,1]$的函数图像，因为$sin^n(x)
证明：
法一：积分的有界性
这里第二项趋于0是因为$sin^n(x)\to0$，从图中也可以观察出来。
法二：
$if\ n=2k$
$I_{2k}=\frac{(2k-1)!!}{(2k)!!}=\frac{(2k-1)\cdot(2k-3)\cdot\cdot\cdot5\cdot3\cdot1}{(2k)(2k-2)\cdot\cdot\cdot6\cdot4\cdot2}\cdot\frac{\pi}{2}$
$0$\therefore I_{2k}\to0$
$\therefore I_{2k+1}=\int_0^{\pi/2}sinx\cdot sin^{2k}xdx=sinx_0\cdot\int_0^{\pi/2}sin^{2k}xdx\to0$
$\therefore I_{n}\to0$
（4）
维基百科：[沃利斯乘积](https://zh.wikipedia.org/wiki/%E6%B2%83%E5%88%A9%E6%96%AF%E4%B9%98%E7%A7%AF)
3B1B：[沃利斯乘积的几何解释](https://www.bilibili.com/video/av22808876/)
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## 3.2组合三角函数
注意：此类题一律用分部积分画出和原被积函数相关的表达式，但不一定是相等，以实现得到递推关系降阶的目的。
_**Example：**_
求 $L_n=\int_0^{\pi/2}cos^n(x)cos(nx)dx$ 的递归表达式和通项
$=\frac{1}{n}\int_0^{\pi/2}cos^n(x)dsin(nx)=\frac{1}{n}(cos^n(x)sin(nx)|_0^{\pi/2}-\frac{1}{n}\int_0^{\pi/2}nsin(nx)cos^{n-1}(x)(-sin(x))dx$
$=\int_0^{\pi/2}cos^{n-1}(x)sin(nx)sin(x)dx$
$2L_n=\int_0^{\pi/2}[cos^{n-1}(x)sin(nx)sin(x)+cos^n(x)cos(nx)]dx=\int_0^{\pi/2}cos^{n-1}(x)(cos(nx)cos(x)+sin(nx)sin(x))dx$$=\int_0^{\pi/2}cos^{n-1}(x)cos((n-1)x)dx=L_{n-1}$
所以 $L_n=\frac{1}{2}L_{n-1}$ 而$L_1=\int_0^{\pi/2}cos(x)cos(x)dx=\pi/4$
所以数列的通项表达式为：$L_n=\frac{\pi}{2^{n+1}}$
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## 3.3其他函数
_**Example：**_
求 $I_n=\int_0^1x^m(ln(x))^ndx$ 的递推表达式和通项。
_**Exercise:**_
_****（3）求$a_n$的极限
（3）结果为0，证法和高阶三角函数一样。
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# 四：利用级数对定积分进行估值
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## 4.1常用的几种级数
基本的思路是：把函数做泰勒展开，交换积分和求和符号，对每一项幂函数进行积分，最后合并即可。
下面给出三个重要常数：
@@ -54,55 +58,55 @@ _**Example：**_
求 $I_n=\int_0^1x^m(ln(x))^ndx$ 的递推表达式和
在计算积分时，通常通过上述思路把积分化为上述重要常数，进而直接得出结果。
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## 4.2将定积分转换为级数
_**Example1：重要常数（a）**_
$\int_0^1ln(x)ln(1-x)dx$
答案：
[视频链接](https://www.youtube.com/watch?v=q9jbmEGClSk&list=PLj7p5OoL6vGzJxJU1seMmj-5CBNhMYStQ&index=8)
_**Example2：重要常数（c）**_
（a） $\int_0^{\pi/4}ln(tan(x))dx$ （b） $\int_0^{\pi/4}ln(cot(x))dx$
对于（a），令$t=tan(x)$ ，换元后可得：
$\int_0^1ln(t)d arctan(t)=[ln(t)arctan(t)]|_0^1-\int_0^1arctant\frac{dt}{t}=-\int_0^1arctant\frac{dt}{t}=-G\approx-0.92$
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# 五：特殊积分
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## 5.1伽马函数
维基百科点击下方链接：
[WIKI百科](https://zh.wikipedia.org/wiki/%CE%93%E5%87%BD%E6%95%B0)
下面为详细的视频教程
1：定义伽马函数
[视频链接](https://www.youtube.com/watch?v=L4Trz6pFut4&t=21s)
2：$(\frac{1}{2})!$
[视频链接](https://www.youtube.com/watch?v=dBmApl6gkII&t=5s)
3：$(-1)!$
[视频链接](https://www.youtube.com/watch?v=MfP2FQPBdHI&t=294s)
4：$n!!$
[视频链接](https://www.youtube.com/watch?v=imcooZyo4vE)
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## 5.2分段函数积分
此类题目的关键是原函数的连续性要求，通过待定系数求解，具体见不定积分中的[特殊积分部分](https://www.yuque.com/uhhida/tcgb6r/dg7du1)。
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## 5.3变限积分
形如$I(x)=\int_a^xf(t)dt$ 的积分叫做变限积分，他的本质是一个自变量在积分限的函数。
变限积分的求导法则尤其重要。此类题目需要扎实的数学分析思想，极其易错，具体内容见[积分证明框架](https://www.yuque.com/uhhida/tcgb6r/pcc7t2)。
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## 5.4反常积分
具体内容见[积分证明框架](https://www.yuque.com/uhhida/tcgb6r/pcc7t2)
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# 六：定积分的应用
我们可以看到定积分的本质就是微元法取极限，所以一些几何和物理问题都可以用微元法的思想解决。
_**Example1：**_
已知一段半圆弧电荷密度为+1，计算圆心处的电场强度。
注意，这类题目直接用几何解法更快，由库仑定律得，两段对称的微弧对圆心的电场作用为：
$\Delta E=2\Delta E'cos\theta=\frac{2k\Delta lcos\theta}{R^2}=\frac{2k\Delta r}{R^2}$
$\Delta r$代表上图小三角形垂直部分，$\sum \Delta r=R$ ，所以：
$E=\sum\Delta E=\frac{2k}{R^2}\sum\Delta r=\frac{2k}{R}$
_**Example2：**_
我们可以利用微元法找到极坐标的面积公式
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## 6.1几何问题
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#### 6.1.1向量簇
请下载：[New horizons in geometry(几何新视野).pdf](https://www.yuque.com/attachments/yuque/0/2019/pdf/422826/1564708063142-a6a603c9-ef19-42c5-bbac-3a746583ac46.pdf?_lake_card=%7B%22uid%22%3A%22rc-upload-1564707962520-4%22%2C%22src%22%3A%22https%3A%2F%2Fwww.yuque.com%2Fattachments%2Fyuque%2F0%2F2019%2Fpdf%2F422826%2F1564708063142-a6a603c9-ef19-42c5-bbac-3a746583ac46.pdf%22%2C%22name%22%3A%22New%20horizons%20in%20geometry(%E5%87%A0%E4%BD%95%E6%96%B0%E8%A7%86%E9%87%8E).pdf%22%2C%22size%22%3A9849775%2C%22type%22%3A%22application%2Fpdf%22%2C%22ext%22%3A%22pdf%22%2C%22progress%22%3A%7B%22percent%22%3A0%7D%2C%22status%22%3A%22done%22%2C%22percent%22%3A0%2C%22id%22%3A%227fZdG%22%2C%22card%22%3A%22file%22%7D)阅读第一章。
这里只讲一个拓扑定理：切线向量扫过的面积等于将这些向量在不改变方向和长度的情况下，将所有的向量尾部接到一起时所组成的图形的面积相等。即切线扫过的面积等于切线簇的面积，如图所示。
想象一辆自行车，其尾部车轮始终与某一闭合轨迹相切，也就是骑了一圈，那么无论骑行轨迹如何，自行车车身划过的面积始终不变。如上图所示，将一簇切线在不改变方向和长度的情况下尾部连接在一起，他们所组成的形状面积和原来扫过的面积相同，且原来图形的形状可以任意。
和开普勒第二定律及椭圆面积的关系：[视频链接](https://www.bilibili.com/video/av28012188)
_**Example3：**_
利用该定理证明$\int_{-\infty}^{x}e^{t/b}dt=be^{x/b}$
显然，所有切线构成的三角形底边长度都是1，把所有的切线全部放到黄色三角形内，尾部连接在一起，就得到积分的面积是两倍三角形的面积，即$be^{x/b}$。
_**Example4：**_
利用该定理求摆线下方的面积。
摆线相关知识点击：[维基百科](https://zh.wikipedia.org/wiki/%E6%91%86%E7%BA%BF)
由于PT垂直于PC（在圆内），又因为点P在每个瞬时相对于点C做圆周运动，所以PT必然是点P的切向运动方向，所以PT也是摆线的切线，DOT的面积（及切线扫过的面积）等于右图中切线簇的面积（将所有切线平移到点T）。更具体的说，圆在运动时，点P划过了一段圆弧，在每一个中间点，都可以做线段PT，线段PT既扫过了DOT，也同时覆盖了T'P'C'，故两者面积相等。
当圆旋转一周时，切线划过了整个摆线上方的面积，这面积等于切线簇的面积$\pi R^2$，而整个矩形的面积是$2\pi R\cdot 2R=4\pi R^2$，所以摆线下方的面积为$4\pi R^2-\pi R^2=3\pi R^2$。
利用参数方程的证明方法见下文。
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#### 6.1.2体积
这里依然是微元法，我们讲两种求旋转体体积的方法：切片法和圆柱法
**切片法：**
A(x)就是x处物体的面积，$\Delta x$就是高度，两者相乘就是和x有关的一个体积微元，求和或积分后就得到了整个体积。
_**Example5：**_
求$y=x$和$y=x^2$中间的部分绕$y=2$旋转一周所得到的图像的体积。
本题的关键是千万不要把正负号弄错，x和y的正负号永远是对原始图像而言的！！！
_**Example6：**_
求$y=x$和$y=x^2$中间的部分绕$x=-1$旋转一周所得到的图像的体积。
**圆柱法：**
如图所示，$x_L和x_R$比较难求，当边界无法被解析式表达时，我们对垂直于旋转轴的底面做圆柱分解。
一小段圆柱的体积=底面积：$2\pi xdx$ 乘以 高：$f(x)$，之后对自变量求积分。
_**Example7：**_
_****_
_****_
注意：在本题中，圆柱法比切片法简单的多，由于不用求边界，对于具体的题目，根据情况选择两种方法即可。_
-
+
#### 6.1.3弧长和表面积
求弧长的方法十分显然，该方法也适用于极坐标，只需用微元法得到公式。
求表面积的方法也是微元法，如下图所示：
已知圆锥表面积是$\pi rl$，现在试图求解一段圆台的表面积，它等于两个圆锥相减。
如上图所示，将任意旋转体分成无数小圆台，进行求和即可（这里为什么不能用圆柱做微元法？）。
最后我们得到了表面积的表达式：
_**Example8：**_
-
+
#### 6.1.4极坐标和参数方程
**参数方程：**
参数方程本质就是把x和y想成某个变量t的函数，利用链式法则可以把微积分的计算用参数方程实现。
微分：
求面积：
求长度：
_**Example9：**_
我们以摆线为例，将摆线方程建模为关于转动角度的函数，如下图所示。
摆线的面积：方法只是简单的换元。
摆线的周长：
摆线相关知识点击：[维基百科](https://zh.wikipedia.org/wiki/%E6%91%86%E7%BA%BF)
摆线的本质是最速降线，请看3B1B的视频：[视频链接](https://www.bilibili.com/video/av6385842)
另一个相关视频：[费马小定理](https://www.bilibili.com/video/av4283063)
**极坐标：**
极坐标求面积公式前面已给出：
求弧长的公式推导依然是微元法：
更直观的方法是：弧长显然等于径向速度与法向速度之和。$\Delta l=\sqrt{(dr)^2+(rd\theta)^2}$
_**Example10：**_
_**Example11：**_
本题请务必自己手算一遍。
_****_
_****_
-
+
## 6.2物理问题
-
+
#### 6.2.1质心
质心和势相关，是力学中研究旋转的概念，质心和势一般来说是对于某个转轴的定义。
质心是一个点，在这点上施加外力不会引起系统的旋转。
势是物体旋转的惯性，表示在施加外力后系统旋转的能力。
如上图所示，区域R具有恒定密度，它相对于y轴和x轴的势计算方法如下：
（相当于每个矩形到y轴距离x乘以每个矩形面积f(x)dx）
（相当于每个矩形中点到x轴距离f(x)/2乘每个矩形面积f(x)dx）
上式除以总质量就得到了质心：
_**Example12：**_
-
+
#### 6.2.2电磁场
_**Example13：**_
一根电荷密度为+1的，长度为L的直导体棒，求距离其中点a的一点的电场强度。
提示：第一种做法是通过微元法转换成积分，第二种做法是找出与其等价的圆弧。
答案：[点击链接](https://www.bilibili.com/video/av55298726)
-
+
## 6.3一题十解
_**Example14：**_
求$Ax^2+2Bxy+Cy^2=1(A>0,AC-B^2>0)$所围图形的面积。
解1：
diff --git "a/8.\345\237\272\347\241\200\345\255\246\347\247\221/8.\345\237\272\347\241\200\345\255\246\347\247\221.md" "b/8.\345\237\272\347\241\200\345\255\246\347\247\221/8.\345\237\272\347\241\200\345\255\246\347\247\221.md"
index 67ea11d0..7923bf02 100644
--- "a/8.\345\237\272\347\241\200\345\255\246\347\247\221/8.\345\237\272\347\241\200\345\255\246\347\247\221.md"
+++ "b/8.\345\237\272\347\241\200\345\255\246\347\247\221/8.\345\237\272\347\241\200\345\255\246\347\247\221.md"
@@ -1 +1 @@
-# 8. 外行社科
+# 8. 基础学科
\ No newline at end of file
diff --git "a/8.\345\237\272\347\241\200\345\255\246\347\247\221/static/lake_card_mindmap.png" "b/8.\345\237\272\347\241\200\345\255\246\347\247\221/static/lake_card_mindmap.png"
new file mode 100644
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