Entropic

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微积分笔记
~ CuB3y0nd

# 极限和连续性

## 何为极限？

假设对于函数 $f$ 有：

\displaystyle \lim _{x\to c} f(x)=L

即：只要 $x$ 无限接近于 $c$ , 则 $f(x)$ 必然无限接近于 $L$ 。

用 $\varepsilon -\delta$ 语言来描述就是：

\displaystyle \forall \varepsilon >0,\ \exists \delta >0,\ s.t.\ 0<|x−c|< \delta \Longrightarrow |f(x)-L|< \varepsilon

说白了就是：无论给定任何一个数字 $\varepsilon (\varepsilon >0)$ ，总能找到一个数 $\delta ( \delta >0)$ 。使当 $x$ 在 $c$ 的 $\delta$ 范围内时， $f(x)$ 在极限 $L$ 的 $\varepsilon$ 范围内。

---

例：已知 $f( x) =\begin{cases} 2x & x\neq 5\\ x & x=5 \end{cases}$ ，证明 $\displaystyle \lim _{x\rightarrow 5} f( x) =10$

根据定义，给定任意 $\varepsilon (\varepsilon >0)$ ，有 $\delta ( \delta >0)$ 。因此，我们本质上是要找到一个 $\delta =function\ of\ \varepsilon$ 的函数。

$\mathnormal{Proof.}$

\begin{aligned} & |x-5| < \delta \Longrightarrow |2x-10|< \varepsilon \\ & |2x-10| < 2\delta \\ & 2\delta =\varepsilon \Rightarrow \delta =\frac{\varepsilon }{2}\\ & |2x-10| < \varepsilon \\ & \forall \varepsilon >0,\ \exists \delta >0 ,\ s.t.\ |x-5|< \delta \Longrightarrow |2x-10|< \varepsilon \ \quad Q.E.D. \end{aligned}

---

## 夹逼定理

设 $I$ 为包含某点 $c$ 的区间， $f, g, h$ 为定义在 $I$ 上的函数。若对于所有属于 $I$ 而不等于 $c$ 的 $x$ ，有：

- $g( x) \leqslant f( x) \leqslant h( x)$ - $\displaystyle \lim _{x\rightarrow c} g( x) =\lim _{x\rightarrow c} h( x) =L$

则， $\displaystyle \lim _{x\rightarrow c} f( x) =L$ 。

$g(x)$ 和 $h(x)$ 分别被称为 $f(x)$ 的下界和上界。

### Proof: $\displaystyle \lim _{\theta \rightarrow 0}\frac{\sin \theta }{\theta } =1$

### Proof: $\displaystyle \lim _{\theta \rightarrow 0}\frac{1-\cos \theta }{\theta } =0$

$\mathnormal{Proof.}$

\begin{aligned} \lim _{\theta \rightarrow 0}\frac{1-\cos \theta }{\theta } & =\lim _{\theta \rightarrow 0}\frac{( 1-\cos \theta )( 1+\cos \theta )}{\theta ( 1+\cos \theta )}\\ & =\lim _{\theta \rightarrow 0}\frac{\sin^{2} \theta }{\theta ( 1+\cos \theta )}\\ & =\lim _{\theta \rightarrow 0}\frac{\sin \theta }{\theta } \cdot \lim _{\theta \rightarrow 0}\frac{\sin \theta }{1+\cos \theta }\\ & =1\cdot 0\\ & =0 & Q.E.D. \end{aligned}

## 连续性的定义

函数在某一点处连续： $f$ is continuous at $x=c\Longleftrightarrow \displaystyle \lim _{x\rightarrow c} f( x) =f( c)$

函数在开区间连续： $f$ is continuous over $( a,\ b) \Longleftrightarrow f$ is continuous over every point in the interval

函数在闭区间连续： $f$ is continuous over $[ a,\ b] \Longleftrightarrow f$ is continuous over $( a,\ b)$ and $\displaystyle \lim _{x\rightarrow a^{+}} f( x) =f( a)$ , $\displaystyle \lim _{x\rightarrow b^{-}} f( x) =f( b)$

## Intermediate Value Theorem

Suppose $f$ is a continuous function at every point of the interval $[ a,\ b]$

- $f$ will take on every value between $f( a)$ and $f( b)$ over the interval - For any $L$ between the values $f( a)$ and $f( b)$ , there exists a number $c$ in $[ a,\ b]$ for which $f( c) =L$

怎么会有这么简单的定理…

# 导数

## 导数的两种定义形式

\displaystyle f^{\prime }( x) =\lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h}

\displaystyle f^{\prime }( c) =\lim _{x\rightarrow c}\frac{f( x) -f( c)}{x-c}

## 可微性

- $f$ is differentiability at $x=c\Longrightarrow f$ is continuous at $x=c$ - $f$ is not continuous at $x=c\Longrightarrow f$ is not differentiability at $x=c$

不可微的三种情况：

1. **not continuous** 2. **vertical tangent** 3. **"sharp turn"**

## Proof: Differentiability implies continuity

$\mathnormal{Proof.}$

Assume: $f$ differentiability at $x=c$

\begin{array}{l} \because f\ differentiability\ at\ x=c\\ \therefore \displaystyle f^{\prime }( c) = \lim _{x\rightarrow c}\frac{f( x) -f( c)}{x-c} \end{array}

\begin{aligned} \lim _{x\rightarrow c}[ f( x) -f( c)] & =\lim _{x\rightarrow c}( x-c) \cdot \frac{f( x) -f( c)}{x-c}\\ & =\lim _{x\rightarrow c}( x-c) \cdot \lim _{x\rightarrow c}\frac{f( x) -f( c)}{x-c}\\ & =0\cdot f^{\prime }( c)\\ & =0\\ & \\ \lim _{x\rightarrow c}[ f( x) -f( c)] & =0\\ \lim _{x\rightarrow c} f( x) -\lim _{x\rightarrow c} f( c) & =0\\ \lim _{x\rightarrow c} f( x) -f( c) & =0\\ \lim _{x\rightarrow c} f( x) & =f( c) & Q.E.D. \end{aligned}

## Justifying the power rule

### Proof: $\displaystyle \frac{d}{dx}\left( x^{n}\right) =nx^{n-1}$

$\mathnormal{Proof.}$

\displaystyle \frac{d}{dx}\left( x^{n}\right) =\lim _{\Delta x\rightarrow 0}\frac{( x+\Delta x)^{n} -x^{n}}{\Delta x}

According to Binomial theorem:

\begin{aligned} \displaystyle \lim _{\Delta x\rightarrow 0}\frac{( x+\Delta x)^{n} -x^{n}}{\Delta x} & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\cancel{x^{n}} +\binom{n}{1} x^{n-1} \Delta x+\binom{n}{2} x^{n-2} \Delta x^{2} +...+\binom{n}{n} x^{0} \Delta x^{n}\cancel{-x^{n}}}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\binom{n}{1} x^{n-1} +\cancel{\binom{n}{2} x^{n-2} \Delta x} +...+\cancel{\binom{n}{n} \Delta x^{n-1}}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\binom{n}{1} x^{n-1}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{n!}{\cancel{1!}( n-1) !} x^{n-1}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0} nx^{n-1} & Q.E.D. \end{aligned}

### Proof: $\displaystyle \frac{d}{dx}\left(\sqrt{x}\right) =\frac{1}{2} x^{-\frac{1}{2}}$

$\mathnormal{Proof.}$

\begin{aligned} \frac{d}{dx}\left(\sqrt{x}\right) & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sqrt{x+\Delta x} -\sqrt{x}}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\left(\sqrt{x+\Delta x} -\sqrt{x}\right)\left(\sqrt{x+\Delta x} +\sqrt{x}\right)}{\Delta x\left(\sqrt{x+\Delta x} +\sqrt{x}\right)}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{\sqrt{x+\Delta x} +\sqrt{x}}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{2\sqrt{x}}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{2} x^{-\frac{1}{2}} & Q.E.D. \end{aligned}

## Justifying the basic derivative rules

### Proof: Constant rule ( $\displaystyle \frac{d}{dx} k=0$ )

$\mathnormal{Proof.}$

\begin{array}{l} \because k\ is\ constant\\ \therefore y\ does\ not\ change\ as\ x\ changes\\ \therefore f( x+h) -f( x) =0\\ \therefore \displaystyle \frac{d}{dx} k= \lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h} =\lim _{h\rightarrow 0}\frac{0}{h} =0 \end{array}

### Proof: Constant multiple and sum/difference rules

**Constant multiple rule:** $\displaystyle \dfrac{d}{dx}[k\cdot f(x)]=k\cdot\dfrac{d}{dx}f(x)$

**Sum rule:** $\displaystyle \dfrac{d}{dx}[f(x)+g(x)]=\dfrac{d}{dx}f(x)+\dfrac{d}{dx}g(x)$

**Difference rule:** $\displaystyle \dfrac{d}{dx}[f(x)-g(x)]=\dfrac{d}{dx}f(x)-\dfrac{d}{dx}g(x)$

$\mathnormal{Proof.}$

\displaystyle 1.\ f( x) =kg( x) \Longrightarrow f^{\prime }( x) =kg^{\prime }( x)

\begin{aligned} f^{\prime }( x) & =\displaystyle \lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h}\\ & =\displaystyle \lim _{h\rightarrow 0}\frac{kg( x+h) -kg( x)}{h}\\ & =\displaystyle \lim _{h\rightarrow 0} k\left(\frac{g( x+h) -g( x)}{h}\right)\\ & =k\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h}\\ & =kg^{\prime }( x) & Q.E.D. \end{aligned}

\displaystyle 2.\ f( x) =g( x) \pm j( x) \Longrightarrow f^{\prime }( x) =g^{\prime }( x) \pm j^{\prime }( x)

\begin{aligned} f^{\prime }( x) & =\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) \pm j( x+h) -( g( x) \pm j( x))}{h}\\ & =\displaystyle \lim _{h\rightarrow 0}\left(\frac{g( x+h) -g( x)}{h} \pm \frac{j( x+h) -j( x)}{h}\right)\\ & =\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h} \pm \lim _{h\rightarrow 0}\frac{j( x+h) -j( x)}{h}\\ & =g^{\prime }( x) \pm j^{\prime }( x) & Q.E.D. \end{aligned}

## Proof: The derivatives of sin(x) and cos(x)

Known $\displaystyle \lim _{x\rightarrow 0}\frac{\sin x}{x} =1$ and $\displaystyle \lim _{x\rightarrow 0}\frac{1-\cos x}{x} =0$

$\mathnormal{Proof.}$

\displaystyle 1.\ \frac{d}{dx}[\sin x] =\cos x

\begin{aligned} \frac{d}{dx}[\sin x] & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin( x+\Delta x) -\sin( x)}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\cos x\sin \Delta x+\sin x\cos \Delta x-\sin x}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\left(\frac{\cos x\sin \Delta x}{\Delta x} +\frac{\sin x\cos \Delta x-\sin x}{\Delta x}\right)\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\cos x\left(\frac{\sin \Delta x}{\Delta x}\right) +\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin x(\cos \Delta x-1)}{\Delta x}\\ & =\cos x\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin \Delta x}{\Delta x} -\sin x\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1-\cos \Delta x}{\Delta x}\\ & =\cos x\cdot 1-\sin x\cdot 0\\ & =\cos x & Q.E.D. \end{aligned}

\displaystyle 2.\ \frac{d}{dx}[\cos x] =-\sin x

## Proof: The derivative of $e^{x}$ is $e^{x}$

Know the limit definition of $\mathbb{e}$ is $e=\displaystyle \lim _{n\rightarrow \infty }\left( 1+\frac{1}{n}\right)^{n} =\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{n}}$

$\mathnormal{Proof.}$

\begin{aligned} \frac{d}{dx}\left( e^{x}\right) & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{e^{x+\Delta x} -e^{x}}{\Delta x}\\ & =e^{x}\displaystyle \lim _{\Delta x\rightarrow 0}\frac{e^{\Delta x} -1}{\Delta x} \end{aligned}

\displaystyle Let\ n=e^{\Delta x} -1,\ we\ can\ get\ n+1=e^{\Delta x} ,\ such\ that\ \Delta x=\ln( n+1) \ and\ as\ \Delta x\rightarrow 0=n\rightarrow 0

$We\ can\ rewrite\ to:$

\begin{aligned} \frac{d}{dx}\left( e^{x}\right) & =e^{x}\displaystyle \lim _{n\rightarrow 0}\frac{n}{\ln( n+1)}\\ & =e^{x}\displaystyle \lim _{n\rightarrow 0}\frac{\frac{1}{n} n}{\frac{1}{n}\ln( n+1)}\\ & =e^{x}\displaystyle \lim _{n\rightarrow 0}\frac{1}{\ln\left[( 1+n)^{\frac{1}{n}}\right]}\\ & =e^{x}\frac{1}{\ln\left[\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{n}}\right]}\\ & =e^{x} & Q.E.D. \end{aligned}

## Proof: The derivative of $\ln( x)$ is $\frac{1}{x}$

### Method 1 (Directly from the definition of the derivative as a limit)

$\mathnormal{Proof.}$

\begin{aligned} \frac{d}{dx}(\ln x) & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln( x+\Delta x) -\ln( x)}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln\left(\frac{x+\Delta x}{x}\right)}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln\left( 1+\frac{\Delta x}{x}\right)}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{\Delta x}\ln\left( 1+\frac{\Delta x}{x}\right)\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\ln\left[\left( 1+\frac{\Delta x}{x}\right)^{\frac{1}{\Delta x}}\right] \end{aligned}

\displaystyle Let\ n=\frac{\Delta x}{x} ,\ \Delta x=nx,\ \frac{1}{\Delta x} =\frac{1}{n} \cdot \frac{1}{x} \ and\ as\ \Delta x\rightarrow 0=n\rightarrow 0

$We\ can\ rewrite\ to:$

\begin{aligned} \displaystyle \lim _{\Delta x\rightarrow 0}\ln\left[\left( 1+\frac{\Delta x}{x}\right)^{\frac{1}{\Delta x}}\right] & =\frac{1}{x}\displaystyle \lim _{n\rightarrow 0}\ln\left[( 1+n)^{\frac{1}{n}}\right]\\ & =\frac{1}{x}\ln\left[\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{n}}\right]\\ & =\frac{1}{x} & Q.E.D. \end{aligned}

### Method 2 (Using the fact that $\displaystyle \frac{d}{dx}\left( e^{x}\right) =e^{x}$ and applying implicit differentiation)

$\mathnormal{Proof.}$

\displaystyle Known\ \frac{d}{dx}\left( e^{x}\right) =e^{x}

\displaystyle Let\ y=\ln( x) ,\ we\ can\ get:

\begin{aligned} \frac{d}{dx}\left( e^{y}\right) & =\frac{d}{dx}( x)\\ e^{y} \cdot \frac{dy}{dx} & =1\\ \frac{dy}{dx} & =\frac{1}{e^{y}}\\ & =\frac{1}{e^{\ln x}}\\ & =\frac{1}{x} & Q.E.D. \end{aligned}

## Proof: The product rule

$\mathnormal{Proof.}$

\begin{aligned} \frac{d}{dx}[ f( x) g( x)] & =\displaystyle \lim _{h\rightarrow 0}\frac{f( x+h) g( x+h) -f( x+h) g( x) +f( x+h) g( x) -f( x) g( x)}{h}\\ & =\displaystyle \lim _{h\rightarrow 0}\left[ f( x+h)\frac{g( x+h) -g( x)}{h} +g( x)\frac{f( x+h) -f( x)}{h}\right]\\ & =\left[\displaystyle \lim _{h\rightarrow 0} f( x+h)\right]\left[\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h}\right] +\left[\displaystyle \lim _{h\rightarrow 0} g( x)\right]\left[\displaystyle \lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h}\right]\\ & =f( x) g^{\prime }( x) +g( x) f^{\prime }( x) & Q.E.D. \end{aligned}

## Proof: The derivatives of $\tan( x)$ 、 $\cos( x)$ 、 $\sec( x)$ and $\csc( x)$

$\mathnormal{Proof.}$

\begin{aligned} \frac{d}{dx}(\tan x) & =\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) & \frac{d}{dx}(\cot x) & =\frac{d}{dx}\left(\frac{\cos x}{\sin x}\right)\\ & =\frac{\cos^{2} x+\sin^{2} x}{\cos^{2} x} & & =\frac{-\left(\sin^{2} x+\cos^{2} x\right)}{\sin^{2} x}\\ & =\frac{1}{\cos^{2} x} & & =-\frac{1}{\sin^{2} x}\\ & =\sec^{2} x & & =-\csc^{2} x\\ \frac{d}{dx}(\sec x) & =\frac{d}{dx}\left(\frac{1}{\cos x}\right) & \frac{d}{dx}(\csc x) & =\frac{d}{dx}\left(\frac{1}{\sin x}\right)\\ & =\frac{0\cdot \cos x+1\cdot \sin x}{\cos^{2} x} & & =\frac{0\cdot \sin x-1\cdot \cos x}{\sin^{2} x}\\ & =\frac{\sin x}{\cos^{2} x} & & =-\frac{\cos x}{\sin^{2} x}\\ & =\tan x\cdot \sec x & & =-\cot x\cdot \csc x & Q.E.D. \end{aligned}

## Proof: The derivatives of $a^{x}$ (For any positive base a)

$\mathnormal{Proof.}$

\displaystyle Known\ \frac{d}{dx}\left( e^{x}\right) =e^{x}

\displaystyle Let\ a=e^{\ln a}

\begin{aligned} \frac{d}{dx}\left( a^{x}\right) & =\frac{d}{dx}\left[\left( e^{\ln a}\right)^{x}\right]\\ & =\frac{d}{dx}\left[ e^{(\ln a) x}\right]\\ & =e^{(\ln a) x} \cdot \ln a\\ & =a^{x} \cdot \ln a & Q.E.D. \end{aligned}

## Proof: The derivatives of $\log_{a} x$ (For any positive base $a\neq 1$ )

$\mathnormal{Proof.}$

\displaystyle Known\ \frac{d}{dx}(\ln x) =\frac{1}{x}

\begin{aligned} \frac{d}{dx}(\log_{a} x) & =\frac{d}{dx}\left(\frac{1}{\ln a} \cdot \ln x\right)\\ & =\frac{1}{x\ln a} & Q.E.D. \end{aligned}

## Proof: Chain Rule and Quotient Rule

$\mathnormal{Chain\ Rule\ Proof.}$

\begin{aligned} Known:\ & 1.\ If\ a\ function\ is\ differentiable,\ then\ it\ is\ also\ continuous.\\ & 2.\ If\ function\ u\ is\ continuous\ at\ x,\ then\ \Delta u\rightarrow 0\ as\ \Delta x\rightarrow 0 \end{aligned}

For why if function $u$ is continuous at $x$ , then $\Delta u\rightarrow 0$ as $\Delta x\rightarrow 0$ :

\displaystyle The\ chain\ rule\ tell\ us:\ \frac{d}{dx}[ y( u( x))] =\frac{dy}{dx} =\frac{dy}{du} \cdot \frac{du}{dx}

Assuming $y$ , $u$ differentiable at $x$ . We can get:

\begin{aligned} \frac{dy}{dx} & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}\\ & =\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta u}\right)\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta u}{\Delta x}\right)\\ & =\left(\displaystyle \lim _{\Delta u\rightarrow 0}\frac{\Delta y}{\Delta u}\right)\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta u}{\Delta x}\right)\\ & =\frac{dy}{du} \cdot \frac{du}{dx} & Q.E.D. \end{aligned}

$\mathnormal{Quotient\ Rule\ Proof.}$

\begin{aligned} \frac{d}{dx}\left[\frac{f( x)}{g( x)}\right] & =\frac{d}{dx}\left[ f( x) \cdot [ g( x)]^{-1}\right]\\ & =f^{\prime }[ x]( g( x))^{-1} -f[ x]( g( x))^{-2} g^{\prime }( x)\\ & =\frac{f^{\prime }( x)}{g( x)} -\frac{f( x) g^{\prime }( x)}{[ g( x)]^{2}}\\ & =\frac{f^{\prime }( x) g( x) -f( x) g^{\prime }( x)}{[ g( x)]^{2}} & Q.E.D. \end{aligned}

# Proof: L'Hôpital's rule

> [!NOTE] > This isn't full proof of L'Hôpital's rule, just a special case. But it should give some intuition for why it works.

f( a) =0,\ g( a) =0;\ f^{\prime} ( a) \ exists,\ g^{\prime} ( a) \ exists\ \Longleftrightarrow \ \displaystyle \lim _{x\rightarrow a}\frac{f( x)}{g( x)} =\frac{f^{\prime} ( a)}{g^{\prime} ( a)}

\begin{aligned} \frac{f^{\prime }( a)}{g^{\prime }( a)} & =\frac{\displaystyle \lim _{x\rightarrow a}\frac{f( x) -f( a)}{x-a}}{\displaystyle \lim _{x\rightarrow a}\frac{g( x) -g( a)}{x-a}} & \\ & =\displaystyle \lim _{x\rightarrow a}\frac{f( x) -f( a)}{g( x) -g( a)} & \\ & =\displaystyle \lim _{x\rightarrow a}\frac{f( x)}{g( x)} & We\ know\ f( a) \ and\ g( a) \ both\ equal\ to\ zero \\ & & Q.E.D \end{aligned}

# Mean Value Theorem

If $f$ is continuous over $[ a,\ b]$ and every point over $( a,\ b)$ is differentiable. Then there exists some $c\in ( a,\ b)$ where $\displaystyle \frac{\Delta y}{\Delta x} =\frac{f( b) -f( a)}{b-a} =f^{\prime }( c)$

# Extreme Value Theorem

$f$ continuous over $[ a,\ b] \Longrightarrow \exists \ c,\ d\in [ a,\ b] :f( c) \leqslant f( x) \leqslant f( d)$ for all $x\in [ a,\ b]$

critical points exists when non endpoint point at

x=a\ \begin{cases} f^{\prime }( a) =0\\ f^{\prime }( a) \ undefined \end{cases}

# Definite Integral & Riemann Sum

The definite integral of a continuous function $f$ over the interval $[ a,\ b]$ , denoted by $\displaystyle \int _{a}^{b} f( x) dx$ , is the limit of a Riemann sum as the number of subdivisions approaches infinity.

\displaystyle \int _{a}^{b} f( x) dx=\lim _{n\rightarrow \infty }\sum _{i=1}^{n} f( x_{i}) \Delta x

Where $\displaystyle \Delta x=\frac{b-a}{n}$ and $x_{i} =a+\Delta x\cdot i$

# Definite integrals properties

**Sum/Difference:**

\displaystyle \int _{a}^{b}[ f( x) \pm g( x)] dx=\int _{a}^{b} f( x) dx\pm \int _{a}^{b} g( x) dx

**Constant multiple:**

\displaystyle \int _{a}^{b} k\cdot f( x) dx=k\int _{a}^{b} f( x) dx

**Reverse interval:**

\displaystyle \int _{a}^{b} f( x) dx=-\int _{b}^{a} f( x) dx

**Zero-length interval:**

\displaystyle \int _{a}^{a} f( x) dx=0

**Adding intervals:**

\displaystyle \int _{a}^{b} f( x) dx+\int _{b}^{c} f( x) dx=\int _{a}^{c} f( x) dx

这么简单的东西相信你一定也知道怎么证。~~那证明就略略略了吧～~~

# First fundamental theorem of calculus

Let $f$ be a continuous real−valued function defined on $[ a,\ b]$ . And $F$ be the function defined, for all $x$ in $[ a,\ b]$ , by $\displaystyle F( x) =\int _{a}^{x} f( t) dt$

Then $F$ is uniformly continuous on $[ a,\ b]$ and differentiable on the open interval $( a,\ b)$ , and $\displaystyle F^{\prime }( x) =f( x)$ for all $x$ in $( a,\ b)$ so $F$ is an antiderivative of $f$ .

# Second fundamental theorem of calculus / Newton–Leibniz theorem

Let $f$ be a continuous real−valued function defined on $[ a,\ b]$ and $F$ is a continuous function on $[ a,\ b]$ which is an antiderivative of $f$ in $( a,\ b)$ : $\displaystyle F^{\prime }( x) =f( x)$

If $f$ is Riemann integrable on $[ a,\ b]$ then $\displaystyle \int _{a}^{b} f( x) dx=F( b) -F( a)$

# Reverse power rule

\displaystyle \int x^{n} dx=\frac{x^{n+1}}{n+1} +C,\ n\neq -1

Yes that just simple!

# Indefinite integration rules

**Polynomials**

\displaystyle \int x^{n} dx=\frac{x^{n+1}}{n+1} +C

**Radicals**

\displaystyle \int \sqrt[m]{x^{n}} dx=\frac{x^{\frac{n}{m} +1}}{\frac{n}{m} +1} +C

**Trigonometric functions**

\displaystyle \int \sin( x) dx=-\cos( x) +C

\displaystyle \int \cos( x) dx=\sin( x) +C

\displaystyle \int \sec^{2}( x) dx=\tan( x) +C

\displaystyle \int \csc^{2}( x) dx=-\cot( x) +C

\displaystyle \int \sec( x)\tan( x) dx=\sec( x) +C

\displaystyle \int \csc( x)\cot( x) dx=-\csc( x) +C

\displaystyle \int \sec xdx=\ln| \sec x+\tan x| +C

（分子分母同乘 $\sec x+\tan x$ ）

\displaystyle \int \csc xdx=\ln| \csc x-\cot x| +C

（分子分母同乘 $\csc x-\cot x$ ）

下面是另一种方法求这两个不定积分：

\begin{array}{ l l l } \displaystyle \int \sec xdx & =\displaystyle \int \frac{1}{\cos x} dx & \\ & =\displaystyle \int \frac{\cos x}{\cos^{2} x} dx & \\ & =\displaystyle \int \frac{1}{1-\sin^{2} x}\cos xdx & Let\ u=\sin x\\ & =\displaystyle \int \frac{1}{( 1+u)( 1-u)} du & \\ & =\displaystyle \frac{1}{2}\int \left(\frac{1}{1+u} +\frac{1}{1-u}\right) du & Partial\ fractions\\ & =\displaystyle \frac{1}{2}\ln\left| \frac{1+u}{1-u}\right| +C & \\ & =\displaystyle \frac{1}{2}\ln\left| \frac{1+\sin x}{1-\sin x}\right| +C & \end{array}

\begin{array}{ l l l } \displaystyle \int \csc xdx & =\displaystyle \int \frac{1}{\sin x} dx & \\ & =\displaystyle \int \frac{\sin x}{\sin^{2} x} dx & \\ & =\displaystyle \int \frac{1}{1-\cos^{2} x}\sin xdx & Let\ u=\cos x\\ & =\displaystyle -\int \frac{1}{( 1+u)( 1-u)} du & \\ & =\displaystyle -\frac{1}{2}\int \left(\frac{1}{1+u} +\frac{1}{1-u}\right) du & Partial\ fractions\\ & =\displaystyle -\frac{1}{2}\ln\left| \frac{1+u}{1-u}\right| +C & \\ & =\displaystyle -\frac{1}{2}\ln\left| \frac{1+\cos x}{1-\cos x}\right| +C & \end{array}

**Exponential functions**

\displaystyle \int e^{x} dx=e^{x} +C

\displaystyle \int a^{x} dx=\frac{a^{x}}{\ln( a)} +C

**Logarithmic functions**

\displaystyle \int \frac{1}{x} dx=\ln |x|+C

**Inverse trigonometric functions**

\displaystyle \int \frac{1}{\sqrt{a^{2} -x^{2}}} dx=\arcsin\left(\frac{x}{a}\right) +C

\displaystyle \int \frac{1}{a^{2} +x^{2}} dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right) +C

# Integration by parts

\displaystyle \int uvdx=u\int vdx-\int \left( u^{\prime }\int vdx\right) dx

# Integration by reduction formulae

\displaystyle \int \sin^{n} xdx=-\frac{1}{n}\sin^{n-1} x\cos x+\frac{n-1}{n}\int \sin^{n-2} xdx

\displaystyle \int \cos^{n} xdx=\frac{1}{n}\cos^{n-1} x\sin x+\frac{n-1}{n}\int \cos^{n-2} xdx

\displaystyle \int \tan^{n} xdx=\frac{1}{n-1}\tan^{n-1} x-\int \tan^{n-2} xdx

\displaystyle \int (\ln x)^{n} dx=x(\ln x)^{n} -n\int (\ln x)^{n-1} dx


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