# 极限和连续性
## 何为极限？
假设对于函数 $f$ 有：
$displaystyle lim _{xto c} f(x)=L$
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：
$$
displaystyle forall varepsilon >0, exists delta >0, s.t. 0<|x−c|< delta Longri
ghtarrow |f(x)-L|< varepsilon
$$
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 $f( x) =begin{cases}
2x & xneq 5\
x & x=5
end{cases}$，证明 $displaystyle lim _{xrightarrow 5} f( x) =10$
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。
$mathnormal{Proof.}$
$$
begin{aligned}
& |x-5| < delta Longrightarrow |2x-10|< varepsilon \
& |2x-10| < 2delta \
& 2delta =varepsilon Rightarrow delta =frac{varepsilon }{2}\
& |2x-10| < varepsilon \
& forall varepsilon >0, exists delta >0 , s.t. |x-5|< delta Longrightarrow |2x
-10|< varepsilon quad Q.E.D.
end{aligned}
$$
---
## 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：
- $g( x) leqslant f( x) leqslant h( x)$
- $displaystyle lim _{xrightarrow c} g( x) =lim _{xrightarrow c} h( x) =L$
则，$displaystyle lim _{xrightarrow c} f( x) =L$。
$g(x)$ 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: $displaystyle lim _{theta rightarrow 0}frac{1-cos theta }{theta } =0$
$mathnormal{Proof.}$
$$
begin{aligned}
lim _{theta rightarrow 0}frac{1-cos theta }{theta } & =lim _{theta rightarrow 0}
frac{( 1-cos theta )( 1+cos theta )}{theta ( 1+cos theta )}\
& =lim _{theta rightarrow 0}frac{sin^{2} theta }{theta ( 1+cos theta )}\
& =lim _{theta rightarrow 0}frac{sin theta }{theta } cdot lim _{theta rightarro
w 0}frac{sin theta }{1+cos theta }\
& =1cdot 0\
& =0
& Q.E.D.
end{aligned}
$$
## 连续性的定义
函数在某一点处连续：$f$ is continuous at $x=cLongleftrightarrow displaystyle lim
_{xrightarrow c} f( x) =f( c)$
函数在开区间连续：$f$ is continuous over $( a, b) Longleftrightarrow f$ is conti
nuous over every point in the interval
函数在闭区间连续：$f$ is continuous over $[ a, b] Longleftrightarrow f$ is conti
nuous over $( a, b)$ and $displaystyle lim _{xrightarrow a^{+}} f( x) =f( a)$, $
displaystyle lim _{xrightarrow 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
$displaystyle f^{prime }( x) =lim _{hrightarrow 0}frac{f( x+h) -f( x)}{h}$
$displaystyle f^{prime }( c) =lim _{xrightarrow c}frac{f( x) -f( c)}{x-c}$
## 可微性
- $f$ is differentiability at $x=cLongrightarrow f$ is continuous at $x=c$
- $f$ is not continuous at $x=cLongrightarrow f$ 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 _{xrightarrow c}frac{f( x) -f( c)}{x
-c}
end{array}
$
$$
begin{aligned}
lim _{xrightarrow c}[ f( x) -f( c)] & =lim _{xrightarrow c}( x-c) cdot frac{f( x
) -f( c)}{x-c}\
& =lim _{xrightarrow c}( x-c) cdot lim _{xrightarrow c}frac{f( x) -f( c)}{x-c}\
& =0cdot f^{prime }( c)\
& =0\
& \
lim _{xrightarrow c}[ f( x) -f( c)] & =0\
lim _{xrightarrow c} f( x) -lim _{xrightarrow c} f( c) & =0\
lim _{xrightarrow c} f( x) -f( c) & =0\
lim _{xrightarrow c} f( x) & =f( c)
& Q.E.D.
end{aligned}
$$
## Justifying the power rule
### Proof: $displaystyle frac{d}{dx}left( x^{n}right) =nx^{n-1}$
$mathnormal{Proof.}$
$displaystyle frac{d}{dx}left( x^{n}right) =lim _{Delta xrightarrow 0}frac{( x+D
elta x)^{n} -x^{n}}{Delta x}$
According to Binomial theorem:
$$
begin{aligned}
displaystyle lim _{Delta xrightarrow 0}frac{( x+Delta x)^{n} -x^{n}}{Delta x} &
=displaystyle lim _{Delta xrightarrow 0}frac{cancel{x^{n}} +binom{n}{1} x^{n-1}
Delta x+binom{n}{2} x^{n-2} Delta x^{2} +...+binom{n}{n} x^{0} Delta x^{n}cancel
{-x^{n}}}{Delta x}\
& =displaystyle lim _{Delta xrightarrow 0}binom{n}{1} x^{n-1} +cancel{binom{n}{
2} x^{n-2} Delta x} +...+cancel{binom{n}{n} Delta x^{n-1}}\
& =displaystyle lim _{Delta xrightarrow 0}binom{n}{1} x^{n-1}\
& =displaystyle lim _{Delta xrightarrow 0}frac{n!}{cancel{1!}( n-1) !} x^{n-1}\
& =displaystyle lim _{Delta xrightarrow 0} nx^{n-1}
& Q.E.D.
end{aligned}
$$
### Proof: $displaystyle frac{d}{dx}left(sqrt{x}right) =frac{1}{2} x^{-frac{1}{2
}}$
$mathnormal{Proof.}$
$$
begin{aligned}
frac{d}{dx}left(sqrt{x}right) & =displaystyle lim _{Delta xrightarrow 0}frac{sqr
t{x+Delta x} -sqrt{x}}{Delta x}\
& =displaystyle lim _{Delta xrightarrow 0}frac{left(sqrt{x+Delta x} -sqrt{x}rig
ht)left(sqrt{x+Delta x} +sqrt{x}right)}{Delta xleft(sqrt{x+Delta x} +sqrt{x}righ
t)}\
& =displaystyle lim _{Delta xrightarrow 0}frac{1}{sqrt{x+Delta x} +sqrt{x}}\
& =displaystyle lim _{Delta xrightarrow 0}frac{1}{2sqrt{x}}\
& =displaystyle lim _{Delta xrightarrow 0}frac{1}{2} x^{-frac{1}{2}}
& Q.E.D.
end{aligned}
$$
## Justifying the basic derivative rules
### Proof: Constant rule ($displaystyle frac{d}{dx} k=0$)
$mathnormal{Proof.}$
$$
begin{array}{l}
because k is constant\
therefore y does not change as x changes\
therefore f( x+h) -f( x) =0\
therefore displaystyle frac{d}{dx} k= lim _{hrightarrow 0}frac{f( x+h) -f( x)}{h
} =lim _{hrightarrow 0}frac{0}{h} =0
end{array}
$$
### Proof: Constant multiple and sum/difference rules
**Constant multiple rule:** $displaystyle dfrac{d}{dx}[kcdot f(x)]=kcdotdfrac{d}
{dx}f(x)$
**Sum rule:** $displaystyle dfrac{d}{dx}[f(x)+g(x)]=dfrac{d}{dx}f(x)+dfrac{d}{dx
}g(x)$
**Difference rule:** $displaystyle dfrac{d}{dx}[f(x)-g(x)]=dfrac{d}{dx}f(x)-dfra
c{d}{dx}g(x)$
$mathnormal{Proof.}$
$displaystyle 1. f( x) =kg( x) Longrightarrow f^{prime }( x) =kg^{prime }( x)$
$$
begin{aligned}
f^{prime }( x) & =displaystyle lim _{hrightarrow 0}frac{f( x+h) -f( x)}{h}\
& =displaystyle lim _{hrightarrow 0}frac{kg( x+h) -kg( x)}{h}\
& =displaystyle lim _{hrightarrow 0} kleft(frac{g( x+h) -g( x)}{h}right)\
& =kdisplaystyle lim _{hrightarrow 0}frac{g( x+h) -g( x)}{h}\
& =kg^{prime }( x)
& Q.E.D.
end{aligned}
$$
$displaystyle 2. f( x) =g( x) pm j( x) Longrightarrow f^{prime }( x) =g^{prime }
( x) pm j^{prime }( x)$
$$
begin{aligned}
f^{prime }( x) & =displaystyle lim _{hrightarrow 0}frac{g( x+h) pm j( x+h) -( g(
x) pm j( x))}{h}\
& =displaystyle lim _{hrightarrow 0}left(frac{g( x+h) -g( x)}{h} pm frac{j( x+h
) -j( x)}{h}right)\
& =displaystyle lim _{hrightarrow 0}frac{g( x+h) -g( x)}{h} pm lim _{hrightarro
w 0}frac{j( x+h) -j( x)}{h}\
& =g^{prime }( x) pm j^{prime }( x)
& Q.E.D.
end{aligned}
$$
## Proof: The derivatives of sin(x) and cos(x)
Known $displaystyle lim _{xrightarrow 0}frac{sin x}{x} =1$ and $displaystyle lim
_{xrightarrow 0}frac{1-cos x}{x} =0$
$mathnormal{Proof.}$
$displaystyle 1. frac{d}{dx}[sin x] =cos x$
$$
begin{aligned}
frac{d}{dx}[sin x] & =displaystyle lim _{Delta xrightarrow 0}frac{sin( x+Delta x
) -sin( x)}{Delta x}\
& =displaystyle lim _{Delta xrightarrow 0}frac{cos xsin Delta x+sin xcos Delta
x-sin x}{Delta x}\
& =displaystyle lim _{Delta xrightarrow 0}left(frac{cos xsin Delta x}{Delta x}
+frac{sin xcos Delta x-sin x}{Delta x}right)\
& =displaystyle lim _{Delta xrightarrow 0}cos xleft(frac{sin Delta x}{Delta x}r
ight) +displaystyle lim _{Delta xrightarrow 0}frac{sin x(cos Delta x-1)}{Delta x
}\
& =cos xdisplaystyle lim _{Delta xrightarrow 0}frac{sin Delta x}{Delta x} -sin
xdisplaystyle lim _{Delta xrightarrow 0}frac{1-cos Delta x}{Delta x}\
& =cos xcdot 1-sin xcdot 0\
& =cos x
& Q.E.D.
end{aligned}
$$
$displaystyle 2. frac{d}{dx}[cos x] =-sin x$
## Proof: The derivative of $e^{x}$ is $e^{x}$
Know the limit definition of $mathbb{e}$ is $e=displaystyle lim _{nrightarrow in
fty }left( 1+frac{1}{n}right)^{n} =displaystyle lim _{nrightarrow 0}( 1+n)^{frac
{1}{n}}$
$mathnormal{Proof.}$
$$
begin{aligned}
frac{d}{dx}left( e^{x}right) & =displaystyle lim _{Delta xrightarrow 0}frac{e^{x
+Delta x} -e^{x}}{Delta x}\
& =e^{x}displaystyle lim _{Delta xrightarrow 0}frac{e^{Delta x} -1}{Delta x}
end{aligned}
$$
$displaystyle Let n=e^{Delta x} -1, we can get n+1=e^{Delta x} , such that Delta
x=ln( n+1) and as Delta xrightarrow 0=nrightarrow 0$
$We can rewrite to:$
$$
begin{aligned}
frac{d}{dx}left( e^{x}right) & =e^{x}displaystyle lim _{nrightarrow 0}frac{n}{ln
( n+1)}\
& =e^{x}displaystyle lim _{nrightarrow 0}frac{frac{1}{n} n}{frac{1}{n}ln( n+1)}
\
& =e^{x}displaystyle lim _{nrightarrow 0}frac{1}{lnleft[( 1+n)^{frac{1}{n}}righ
t]}\
& =e^{x}frac{1}{lnleft[displaystyle lim _{nrightarrow 0}( 1+n)^{frac{1}{n}}righ
t]}\
& =e^{x}
& Q.E.D.
end{aligned}
$$
## Proof: The derivative of $ln( x)$ is $frac{1}{x}$
### Method 1 (Directly from the definition of the derivative as a limit)
$mathnormal{Proof.}$
$$
begin{aligned}
frac{d}{dx}(ln x) & =displaystyle lim _{Delta xrightarrow 0}frac{ln( x+Delta x)
-ln( x)}{Delta x}\
& =displaystyle lim _{Delta xrightarrow 0}frac{lnleft(frac{x+Delta x}{x}right)}
{Delta x}\
& =displaystyle lim _{Delta xrightarrow 0}frac{lnleft( 1+frac{Delta x}{x}right)
}{Delta x}\
& =displaystyle lim _{Delta xrightarrow 0}frac{1}{Delta x}lnleft( 1+frac{Delta
x}{x}right)\
& =displaystyle lim _{Delta xrightarrow 0}lnleft[left( 1+frac{Delta x}{x}right)
^{frac{1}{Delta x}}right]
end{aligned}
$$
$displaystyle Let n=frac{Delta x}{x} , Delta x=nx, frac{1}{Delta x} =frac{1}{n}
cdot frac{1}{x} and as Delta xrightarrow 0=nrightarrow 0$
$We can rewrite to:$
$$
begin{aligned}
displaystyle lim _{Delta xrightarrow 0}lnleft[left( 1+frac{Delta x}{x}right)^{fr
ac{1}{Delta x}}right] & =frac{1}{x}displaystyle lim _{nrightarrow 0}lnleft[( 1+n
)^{frac{1}{n}}right]\
& =frac{1}{x}lnleft[displaystyle lim _{nrightarrow 0}( 1+n)^{frac{1}{n}}right]\
& =frac{1}{x}
& Q.E.D.
end{aligned}
$$
### 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
$mathnormal{Proof.}$
$$
begin{aligned}
frac{d}{dx}[ f( x) g( x)] & =displaystyle lim _{hrightarrow 0}frac{f( x+h) g( x+
h) -f( x+h) g( x) +f( x+h) g( x) -f( x) g( x)}{h}\
& =displaystyle lim _{hrightarrow 0}left[ f( x+h)frac{g( x+h) -g( x)}{h} +g( x)
frac{f( x+h) -f( x)}{h}right]\
& =left[displaystyle lim _{hrightarrow 0} f( x+h)right]left[displaystyle lim _{
hrightarrow 0}frac{g( x+h) -g( x)}{h}right] +left[displaystyle lim _{hrightarrow
0} g( x)right]left[displaystyle lim _{hrightarrow 0}frac{f( x+h) -f( x)}{h}righ
t]\
& =f( x) g^{prime }( x) +g( x) f^{prime }( x)
& Q.E.D.
end{aligned}
$$
## Proof: The derivatives of $tan( x)$、$cos( x)$、$sec( x)$ and $csc( x)$
$mathnormal{Proof.}$
$$
begin{aligned}
frac{d}{dx}(tan x) & =frac{d}{dx}left(frac{sin x}{cos x}right) & frac{d}{dx}(cot
x) & =frac{d}{dx}left(frac{cos x}{sin x}right)\
& =frac{cos^{2} x+sin^{2} x}{cos^{2} x} & & =frac{-left(sin^{2} x+cos^{2} xrig
ht)}{sin^{2} x}\
& =frac{1}{cos^{2} x} & & =-frac{1}{sin^{2} x}\
& =sec^{2} x & & =-csc^{2} x\
frac{d}{dx}(sec x) & =frac{d}{dx}left(frac{1}{cos x}right) & frac{d}{dx}(csc x)
& =frac{d}{dx}left(frac{1}{sin x}right)\
& =frac{0cdot cos x+1cdot sin x}{cos^{2} x} & & =frac{0cdot sin x-1cdot cos x}
{sin^{2} x}\
& =frac{sin x}{cos^{2} x} & & =-frac{cos x}{sin^{2} x}\
& =tan xcdot sec x & & =-cot xcdot csc x
& Q.E.D.
end{aligned}
$$
## 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 $aneq 1$)
$mathnormal{Proof.}$
$displaystyle Known frac{d}{dx}(ln x) =frac{1}{x}$
$$
begin{aligned}
frac{d}{dx}(log_{a} x) & =frac{d}{dx}left(frac{1}{ln a} cdot ln xright)\
& =frac{1}{xln a}
& Q.E.D.
end{aligned}
$$
## Proof: Chain Rule and Quotient Rule
$mathnormal{Chain Rule Proof.}$
$$
begin{aligned}
Known: & 1. If a function is differentiable, then it is also continuous.\
& 2. If function u is continuous at x, then Delta urightarrow 0 as Delta xright
arrow 0
end{aligned}
$$
For why if function $u$ is continuous at $x$, then $Delta urightarrow 0$ as $Del
ta xrightarrow 0$:
$displaystyle The chain rule tell us: frac{d}{dx}[ y( u( x))] =frac{dy}{dx} =fra
c{dy}{du} cdot frac{du}{dx}$
Assuming $y$, $u$ differentiable at $x$. We can get:
$$
begin{aligned}
frac{dy}{dx} & =displaystyle lim _{Delta xrightarrow 0}frac{Delta y}{Delta x}\
& =displaystyle lim _{Delta xrightarrow 0}frac{Delta y}{Delta u} cdot frac{Delt
a u}{Delta x}\
& =left(displaystyle lim _{Delta xrightarrow 0}frac{Delta y}{Delta u}right)left
(displaystyle lim _{Delta xrightarrow 0}frac{Delta u}{Delta x}right)\
& =left(displaystyle lim _{Delta urightarrow 0}frac{Delta y}{Delta u}right)left
(displaystyle lim _{Delta xrightarrow 0}frac{Delta u}{Delta x}right)\
& =frac{dy}{du} cdot frac{du}{dx}
& Q.E.D.
end{aligned}
$$
$mathnormal{Quotient Rule Proof.}$
$$
begin{aligned}
frac{d}{dx}left[frac{f( x)}{g( x)}right] & =frac{d}{dx}left[ f( x) cdot [ g( x)]
^{-1}right]\
& =f^{prime }[ x]( g( x))^{-1} -f[ x]( g( x))^{-2} g^{prime }( x)\
& =frac{f^{prime }( x)}{g( x)} -frac{f( x) g^{prime }( x)}{[ g( x)]^{2}}\
& =frac{f^{prime }( x) g( x) -f( x) g^{prime }( x)}{[ g( x)]^{2}}
& Q.E.D.
end{aligned}
$$
# 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.
$$
f( a) =0, g( a) =0; f^{prime} ( a) exists, g^{prime} ( a) exists Longleftright
arrow displaystyle lim _{xrightarrow a}frac{f( x)}{g( x)} =frac{f^{prime} ( a)}
{g^{prime} ( a)}
$$
$$
begin{aligned}
frac{f^{prime }( a)}{g^{prime }( a)} & =frac{displaystyle lim _{xrightarrow a}fr
ac{f( x) -f( a)}{x-a}}{displaystyle lim _{xrightarrow a}frac{g( x) -g( a)}{x-a}}
& \
& =displaystyle lim _{xrightarrow a}frac{f( x) -f( a)}{g( x) -g( a)} & \
& =displaystyle lim _{xrightarrow a}frac{f( x)}{g( x)} & We know f( a) and g(
a) both equal to zero \
& & Q.E.D
end{aligned}
$$
# Mean Value Theorem
If $f$ is continuous over $[ a, b]$ and every point over $( a, b)$ is differenti
able. Then there exists some $cin ( a, b)$ where $displaystyle frac{Delta y}{Del
ta x} =frac{f( b) -f( a)}{b-a} =f^{prime }( c)$
# Extreme Value Theorem
$f$ continuous over $[ a, b] Longrightarrow exists c, din [ a, b] :f( c) leqsla
nt f( x) leqslant f( d)$ for all $xin [ a, b]$
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 a
s the number of subdivisions approaches infinity.
$$
displaystyle int _{a}^{b} f( x) dx=lim _{nrightarrow infty }sum _{i=1}^{n} f( x_
{i}) Delta x
$$
Where $displaystyle Delta x=frac{b-a}{n}$ and $x_{i} =a+Delta xcdot i$
# Definite integrals properties
**Sum/Difference:**
$$displaystyle int _{a}^{b}[ f( x) pm g( x)] dx=int _{a}^{b} f( x) dxpm int _{a}
^{b} g( x) dx$$
**Constant multiple:**
$$displaystyle int _{a}^{b} kcdot f( x) dx=kint _{a}^{b} f( x) dx$$
**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 th
e 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 int
erval $( a, b)$, 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 c
ontinuous 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
$$displaystyle int x^{n} dx=frac{x^{n+1}}{n+1} +C, nneq -1$$
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：
$$
begin{array}{ l l l }
displaystyle int sec xdx & =displaystyle int frac{1}{cos x} dx & \
& =displaystyle int frac{cos x}{cos^{2} x} dx & \
& =displaystyle int frac{1}{1-sin^{2} x}cos xdx & Let u=sin x\
& =displaystyle int frac{1}{( 1+u)( 1-u)} du & \
& =displaystyle frac{1}{2}int left(frac{1}{1+u} +frac{1}{1-u}right) du & Partia
l fractions\
& =displaystyle frac{1}{2}lnleft| frac{1+u}{1-u}right| +C & \
& =displaystyle frac{1}{2}lnleft| frac{1+sin x}{1-sin x}right| +C &
end{array}
$$
$$
begin{array}{ l l l }
displaystyle int csc xdx & =displaystyle int frac{1}{sin x} dx & \
& =displaystyle int frac{sin x}{sin^{2} x} dx & \
& =displaystyle int frac{1}{1-cos^{2} x}sin xdx & Let u=cos x\
& =displaystyle -int frac{1}{( 1+u)( 1-u)} du & \
& =displaystyle -frac{1}{2}int left(frac{1}{1+u} +frac{1}{1-u}right) du & Parti
al fractions\
& =displaystyle -frac{1}{2}lnleft| frac{1+u}{1-u}right| +C & \
& =displaystyle -frac{1}{2}lnleft| frac{1+cos x}{1-cos x}right| +C &
end{array}
$$
**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**
$$displaystyle int frac{1}{sqrt{a^{2} -x^{2}}} dx=arcsinleft(frac{x}{a}right) +C
$$
$$displaystyle int frac{1}{a^{2} +x^{2}} dx=frac{1}{a}arctanleft(frac{x}{a}right
) +C$$
# Integration by parts
$$displaystyle int uvdx=uint vdx-int left( u^{prime }int vdxright) dx$$
# Integration by reduction formulae
$$displaystyle int sin^{n} xdx=-frac{1}{n}sin^{n-1} xcos x+frac{n-1}{n}int sin^{
n-2} xdx$$
$$displaystyle int cos^{n} xdx=frac{1}{n}cos^{n-1} xsin x+frac{n-1}{n}int cos^{n
-2} xdx$$
$$displaystyle int tan^{n} xdx=frac{1}{n-1}tan^{n-1} x-int tan^{n-2} xdx$$
$$displaystyle int (ln x)^{n} dx=x(ln x)^{n} -nint (ln x)^{n-1} dx$$
