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Course note of AMA1130 Calculus for Engineers, The Hong Kong Polytechnic University, 2022.

Mainly focus on functions, limits, continuity, differentiation, and integration.

1 Functions

1.1 Sets

A collection of objects

$\Bbb{N}$: set of positive integers ${1,2,3,4,5,…}$;
$\Bbb{Z}$: set of integers ${…,-3.-2.-1,0,1,2,3,…}$;
$\Bbb{Q}$: set of rational numbers ${\frac{integer}{integer}}$, Q for quotient;
$\Bbb{R}$: set of real numbers;
$\Bbb{C}$: set of complex numbers;

$\Bbb{N}\subset\Bbb{Z}\subset\Bbb{Q}\subset\Bbb{R}\subset\Bbb{C}$

1.2 Functions

1.2.1 Definition

Function is a map from a set $A$ to set $B$, $f:A\to B$.
For each $x \in A$, there is a unique $y \in B$, such that $f(x)=y$
$\begin{cases} f(x_1) = y_1\\ f(x_2) = y_2 \end{cases} \implies y_1=y_2\;\text{[uniqueness]}$

Notation:

$y=f(x)$

$x$: independent variable(argument);
$y$: dependent variable.

$Dom(f)$: domain of the function $f$, the set where $f$ is allowed to take values;
$Range(f)$: range of the function $f$, the set of all possible values of $f(x)$ with $x$ in the $Dom(f)$.

Sum: $(f+g)(x)=f(x)+g(x)$
Difference: $(f-g)(x)=f(x)-g(x)$
Product: $(fg)(x)=f(x)g(x)$
- Domains of above 3 functions:

$Dom(f) \cap Dom(g)$

Quotient: $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}\text{, when }g(x)\ne 0$

Domain of this function:

$\{x\in Dom(f) \cap Dom(g):g(x)\ne 0\}$

1.2.2 Absolute Value

$y=f(x)=|x|=\begin{cases} x,&\text{if\;\;$x\ge 0$,}\\ -x,&\text{if\;\;$x\lt 0$.} \end{cases}$

$Dom(f)=\Bbb{R}$
$Range(f)=[0,\infin)$

1.2.3 Composite Functions

Two functions $f:A\to B$ and $g:C\to D$, then $Range(f)\subset C$ (means $\forall x\in X$, $f(x)\in C$).
The composite function $g\circ f: A\to D$ is:

$(g\circ f)(x)=g(f(x))$

Domain of Composite function:

$Dom(g\circ f) = {x\in Dom(f):f(x)\in Dom(g)} = Dom(f)\cap {x:f(x)\in Dom(g)}$

$Dom(f\circ g) = {x\in Dom(g):g(x)\in Dom(f)} = Dom(g)\cap {x:g(x)\in Dom(f)}$

Find the $Dom(f\circ g)$:

Find $Dom(g)$;

Find $Dom(f)$;

Find $x$ such that in $Dom(g)$ and which $g(x)$ is in $Dom(f)$.

[Example]

[Solution]

1.2.4 Inverse Functions

$(g\circ f)(x)=g(f(x))=x,\; \forall x\in Dom(f)$

, then $g$ is the inverse function of $f$, as $f^{-1}$:

$(f^{-1}\circ f)(x)=x$

A one-to-one function has a inverse function.

1.2.5 One-to-one Functions

The function never takes one the same value twice:

$x_1\ne x_2 \iff f(x_1)\ne f(x_2)$

1.2.5.1 Horizontal Line Test

A function is one-to-one if and only if no horizontal line intersects its graph more than once.

$f$ is a one-to-one function with domain $A$ and range $B$, then $f^{-1}$ has domain $B$ and range $A$:

$f^{-1}(y)=x\iff f(x)=y,\; \forall y \in Bs$

[Example]

for one-to-one function $f$,
$f(1)=2,f(3)=5,f(8)=6$,

then $f^{-1}(2)=1,f^{-1}(5)=3,f^{-1}(6)=8$

Find the inverse function:

Write $y=f(x)$;

Solve above equation for $x$ in terms of $y$;

The above result is $x=f^{-1}(y)$

[Example]

Find the inverse function of $f(x)=x^3+2$

$y=f(x)=x^3+2$;
$x^3=y-2 \implies x=\sqrt[3]{y-2}$;
$f^{-1}(y)=x=\sqrt[3]{y-2}$.

Determine whether is a one-to-one function:

To show it is a one-to-one function:
show $x_1\ne x_2 \iff f(x_1)\ne f(x_2)$ or $x_1= x_2 \iff f(x_1)= f(x_2)$

To show it is NOT a one-to-one function:
show $x_1\ne x_2$, but $f(x_1)= f(x_2)$

[Example]

2 Periodic Functions, Polynomials, Trigonometric Functions, Exponential Functions, Logarithmic Functions

2.1 Periodic Functions

A function with a positive constant $T\gt 0$ such that:

$f(x+T)=f(x),\; \forall x \in Dom(f)$

$T$ is the period.

2.2 Polynomials

A function in thr form:

$P(x)=a_0+a_1x+a_2x^2+...+a_nx^n$

$a_0,a_1,…,a_n\in \Bbb{R}$ are constant numbers, called the coefficients;
$x\in \Bbb{R}$: independent variable;
Degree of $P(x)$: $n$, if $a_n\ne 0$, $deg(P)=n$;
Zero of $P(x)$: the root(or solution) of $P(x)=0$;
Commonly used polynomials:
- degree = 0: constant;
- degree = 1: linear;
- degree = 2: quadratic;
- degree = 3: cubic.

Dividing a polynomial $P(x)$ by $x-a$, the remainder is $P(a)$

[Example]

$P(x)=x^3-1,\;a=2,\text{then }x-a = x-2$
$P(x)=x^3-1=(x-2)(x^2+2x+4)+7=(x-2)(x^2+2x+4)+P(2)$

Some factorization formulas for polynomials

2.3 Trigonometric Functions

2.3.1 Degree and Radian

$360\degree=2\pi\;rad$ $1\degree=\frac{2\pi}{360}rad,\;1\;rad=\Bigl(\frac{360}{3\pi}\Bigr)\degree$

2.3.2 Standard position of angles

Standard position: in the xy-plane, its initial side on the positive x-axis;
Positive Angle: rotating the initial side counterclockwise;
Negative Angles: rotating the initial side clockwise.

2.3.3 Trigonometric Functions

Widely used properties:
- $\sin (-x)=-\sin x \;\;\; \text{odd}$
- $\cos (-x)=\cos x \;\;\; \text{evens}$
- $\tan (-x)=-\tan x \;\;\; \text{odd}$ -
- $\sin (\pi -x)=\sin x$
- $\sin (\pi +x)=-\sin x$
- $\cos (\pi \pm x)=-\cos x$ -
- $\sin (\frac{\pi}{2} \pm x)=\cos x$
- $\cos (\frac{\pi}{2} + x)=-\sin x$
- $\cos (\frac{\pi}{2} - x)=\sin x$ -
- $\sin (x+2n\pi)=\sin x,\;\;\; n=0,\pm 1,\pm 2,...$
- $\cos (x+2n\pi)=\cos x,\;\;\; n=0,\pm 1,\pm 2,...$
- $\tan (x+n\pi)=\tan x,\;\;\; n=0,\pm 1,\pm 2,...$
- $\cot x=\frac{\cos x}{\sin x},\;\sec x=\frac{1}{\cos x},\; \csc x=\frac{1}{\sin x}$
- $\tan^2 x+1=\sec^2x,\;\cot^2 x+1=\csc^2 x$
Compound angle formulas:
- $\sin (A+B)=\sin A\cos B+\cos A \sin B$
- $\sin (A-B)=\sin A\cos B-\cos A \sin B$
- $\cos (A+B)=\cos A\cos B-\sin A \sin B$
- $\cos (A-B)=\cos A\cos B+\sin A \sin B$
- $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$
- $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$
Double angle formulas:
- $\sin 2A=2\sin A\cos A$
- - $=1-2\sin^2 A$
  - $=2\cos^2 A-1$
- $\tan 2A=\frac{2\tan A}{1-\tan^2 A}$
- $\cos^2 A=\frac{1+\cos 2A}{2}$
- $\sin^2 A=\frac{1-\cos 2A}{2}$
Conversion formulas:
- $\sin(x+y)+\sin(x-y)=2\sin x \cos y$
- $\sin(x+y)-\sin(x-y)=2\cos x \sin y$
- $\cos(x+y)+\cos(x-y)=2\cos x \cos y$
- $\cos(x+y)-\cos(x-y)=-2\sin x \sin y$
- $\sin(A)+\sin(B)=2\sin (\frac{A+B}{2}) \cos (\frac{A-B}{2})$
- $\sin(A)-\sin(B)=2\cos (\frac{A+B}{2}) \sin (\frac{A-B}{2})$
- $\cos(A)+\cos(B)=2\cos (\frac{A+B}{2}) \cos (\frac{A-B}{2})$
- $\cos(A)-\cos(B)=-2\sin (\frac{A+B}{2}) \sin (\frac{A-B}{2})$

2.4 Inverse Trigonometric Functions

2.4.1 Inverse of $\sin$

$x=\sin^{-1}(y) \iff y=\sin x,\;-\frac{\pi}{2}\le x \le \frac{\pi}{2}$

$Dom(\sin^{-1})$ = $[-1,1]$
$Range(\sin^{-1})$ = $[-\frac{\pi}{2},\frac{\pi}{2}]$
$\sin^{-1}=\arcsin$

$\sin^{-1} (\sin x)=x,\; \forall x \in [-\frac{\pi}{2},\frac{\pi}{2}]$
$\sin (\sin^{-1}y)=y,\; \forall y \in [-1,1]$

2.4.2 Inverse of $\cos$

$x=\cos^{-1}(y) \iff y=\cos x,\;0\le x \le \pi$

$Dom(\cos^{-1})$ = $[-1,1]$
$Range(\cos^{-1})$ = $[0,\pi]$
$\cos^{-1}=\arccos$
$\cos^{-1} (\cos x)=x,\; \forall x \in [0,\pi]$
$\cos (\cos^{-1}y)=y,\;\forall y \in [-1,1]$

2.4.3 Inverse of $\tan$

$x=\tan^{-1}(y) \iff y=\tan x,\;-\frac{\pi}{2}\le x \le \frac{\pi}{2}$

$Dom(\tan^{-1})$ = $(-\infty,\infty)$
$Range(\tan^{-1})$ = $-\frac{\pi}{2}\lt x \lt \frac{\pi}{2}$
$\tan^{-1}=\arctan$
$\tan^{-1} (\tan x)=x,\; \forall x \in (-\frac{\pi}{2},\frac{\pi}{2})$
$\tan (\tan^{-1}y)=y,\;\forall y \in (-\infty,\infty)$

Widely used properties:

$\cos(\tan^{-1}b)=\frac{1}{\sqrt{1+b^2}}$

2.5 Exponential Functions

$y=a^x$

a to the power x

base: $a>0$
exponent (index,power): $x$

Law of Exponents:

$a^ma^n = a^{m+n}$
$\frac{a^m}{a^n}=a^{m-n}$
$(a^m)^n=a^{mn}$
$a^0=1$
$a^{-1}=\frac{1}{a}$
$a^{-m}=\frac{1}{a^m}$

Natural Exponential Function ($exp$):

$\exp (x)=e^x$

$e=2.718281828459$

2.6 Logarithmic Functions

$y=\log_a x, x\gt 0$

the logarithm of x to the base a

$\log_aa=1,\; \log_a1=0$

Logarithmic Function the inverse of Exponential Function:

Rules of logarithm ($a,b,x,y\in \Bbb{R^+}$):

$\log_a(xy)=\log_ax+\log_ay$
$\log_a(\frac{x}{y})=\log_ax-\log_ay$
$\log_ax=\frac{\log_bx}{\log_ba}$
$\log_a1=0$
$\log_ax^m=m\log_ax,\;m\in \Bbb{R}$
$\log_{a^n}x=\frac{1}{n}\log_ax,\;n\in \Bbb{R}$

when the base is $e=2.718281828459$, the Logarithmic Function is $in$ or $log$;

$x=\ln y \iff y=e^x$

3 Limits

3.1 Limits Definition

$\lim_{x\to a}f(x)=L$

Read as ‘the limit of $f(x)$, as $x$ approaches the point $a$, equals $L$’.

$f(x)$ defined when $x$ around number $a$, which means in a open interval contains $a$ but never consider $x=a$, just near it. Then we can make $f(x)$ very close to $L$ by sending $x$ sufficiently close to $a$, around both sides of $a$.

3.2 One-Side Limits

left-hand limit:

$\lim_{x\to a^-}f(x)=L$

Read as left-hand limit of $f(x)$ as $x$ approaches $a$ is equal to $L$

we can make $f(x)$ very close to $L$ by sending $x$ sufficiently close to $a$, with $x$ less than $a$, from the left of $a$.

right-hand limit:

$\lim_{x\to a^+}f(x)=L$

Read as right-hand limit of $f(x)$ as $x$ approaches $a$ is equal to $L$

we can make $f(x)$ very close to $L$ by sending $x$ sufficiently close to $a$, with $x$ great than $a$, from the right of $a$.

Theorem:

$\lim_{x\to a}f(x)=L\iff \lim_{x\to a^-}f(x)=L \;\text{ and }\; \lim_{x\to a^+}f(x)=L$

3.3 Properties of Limits

$n$: positive integer
$k$: constant
Assume $\lim{x\to a}f(x)$ and $\lim{x\to a}g(x)$ exits.

$\lim_{x\to a}k=k$ $\lim_{x\to a}x=a$ $\lim_{x\to a}kf(x)=k\lim_{x\to a}f(x)$ $\lim_{x\to a}[f(x)\pm g(x)]=\lim_{x\to a}f(x)\pm \lim_{x\to a}g(x)$ $\lim_{x\to a}[\frac{f(x)}{g(x)}]=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ $\lim_{x\to a}[f(x)]^n=[\lim_{x\to a}f(x)]^n$ $\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\to a}f(x)}$

So
$\forall n \in \Bbb{Z}^+$

$\lim_{x\to a}x^n=a^n$ $\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}$

and,

$\lim_{x\to 0}\frac{\sin x}{x}=1$ $\lim_{x\to 0}\frac{x}{\sin x}=1$ $\lim_{x\to 0}\frac{\sin (2x)}{x}=2$

3.4 Composite functions

$\lim_{x\to a}f(x)=A\text{ and }\lim_{u\to A}f(x)=B$ $\implies \lim_{x\to a}g(f(x))=B$

3.5 Squeeze Theorem - Sandwich Principle

For $f(x)\le g(x)\le h(x)$, for all near $a$, expect possibly at $a$ itself:

$\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L$ $\implies \lim_{x\to a}g(x)=L$

Immediate consequence of the Squeeze Theorem:

If $g(x)$ is bounded near $a$, expect possibly ar $a$ itself, which means $|g(x)|\le K$, $K$ is a constant for an open interval containing $a$:

$\lim_{x\to a}f(x)=0\implies\lim_{x\to a}f(x)g(x)=0$

It is also true for ons-side limits.

3.6 Infinite limits

$\lim_{x\to a^+}f(x)=\infin$

the value of $f(x)$ can be bigger than any prescribed positive and large number by taking $x\gt a$ and close enough to $a$, approaches Infinity as x approaches $a$ from the right.

$\frac{1}{0^-}=-\infin$ $\frac{1}{0^+}=\infin$

3.7 Limits at Infinity

$\lim_{x\to \infin}\frac{1}{x}=0$ $\lim_{x\to -\infin}\frac{1}{x}=0$

$\lim_{x\to \infin}\frac{c}{x^r}=0$ $\lim_{x\to -\infin}\frac{c}{x^r}=0$ $\forall r\gt 0$

3.7.1 Limits at infinity for Polynomial

For polynomial:

$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0, a_0\ne 0,$ $\Downarrow$ $\lim_{x\to \infin}P(x)=\lim_{x\to \infin}a_nx^n,$ $\lim_{x\to -\infin}P(x)=\lim_{x\to -\infin}a_nx^n$

3.7.2 Limits at infinity for Rational functions

For Rational functions:

$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0, a_0\ne 0,$ $Q(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots + b_1x+b_0, b_0\ne 0,$ $\Downarrow$ $\lim_{x\to \infin}\frac{P(x)}{Q(x)}=\lim_{x\to \infin}\frac{a_nx^n}{b_mx^m},$ $\lim_{x\to -\infin}\frac{P(x)}{Q(x)}=\lim_{x\to -\infin}\frac{a_nx^n}{b_mx^m}$

3.8 Continuity of functions

$f(x)$ is continuous at a point $a$ if

$\lim_{x\to a}f(x)=f(a)$

$f(x)$ is discontinuous if any of following is true:

$f(a)$ is not defined;
$\lim_{x\to a}f(x)$ does not exist;
$\lim_{x\to a}f(x)\ne f(a)$.

Properties of continuity:

$f(x),g(x)$ are continuous at $a$, $n$ is positive integer, following are also continuous at $a$:

Scalar multiple:

$kf(x)$

Sum & Difference:

$f(x)+g(x),f(x)-g(x)$

Product:

$f(x)g(x)$

Quotient:

$\frac{f(x)}{g(x)},\text{ if }g(a)\ne 0$

Power:

$[f(x)]^n$

Root:

$\sqrt[n]{f(x)}\text{ assume $f(a)\ge 0$ if $n$ is even}$

3.8.1 Continuous at the boundary points

$f(x)$ in a closed interval $[a,b]$:

$f$ is continuous at the left ending point $a$ if

$\lim_{x\to a^+}f(x)=f(a)$

$f$ is continuous at the right ending point $b$ if

$\lim_{x\to b^-}f(x)=f(b)$

3.9 Intermediate Value theorem (IVT)

$f$ is continuous function on closed interval $[a,b]$;
$f(a)\ne f(b)$;
$N$ is a number between $f(a)$ and $f(b)$;

then,

there exits at least one point $c\in(a,b)$ such that $f(c)=N$.

4 Differentiation

4.1 First principle of differentiation

$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

is exists, $f$ is differentiable at the point $a\in I$;
Then

$f^\prime(a):=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

the $f(a)$ must be defined.

Right-hand side derivative:

$f^\prime_+(a):=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}$

Left-hand side derivative:

$f^\prime_-(a):=\lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h}$

If $f$ is differentiable at $a$, then

$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)\text{ :continuous}$ $f^\prime_+(a)=f^\prime_-(a)\text{ :derivative exists}$

4.2 Techniques of Differentiation

Constant:

$\frac{d[cf(x)]}{dx}=c\frac{d[f(x)]}{dx}$

Sum and Difference Rules:

$\frac{d[f(x)\pm g(x)]}{dx}=\frac{d[f(x)]}{dx}\pm \frac{d[g(x)]}{dx}$

Product Rule:

$[f(x)g(x)]^\prime =f^\prime(x)g(x)+f(x)g^\prime(x)$

Quotient Rule:

$\biggl[\frac{f(x)}{g(x)}\biggr]^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{[g(x)]^2},g(x)\ne 0$

Chain Rule:

$\big[f\big(g(x)\big)\big]^\prime=f^\prime\big(g(x)\big)\cdot g^\prime (x)$
$f^\prime(a)=g^\prime(u_0)\times h^\prime(a)=g^\prime(h(a))\times h^\prime(a)$
where $u_0=h(a)$

4.3 Inverse differentiation

$y=f(x)$ differentiable on $(a,b)$, $f^\prime(x_0)$ is nonzero at $x_0\in (a,b)$, and derivative $(f^{-1})^\prime(y_0)$ exits, where $y_0=f(x_0)$, and

$(f^{-1})^\prime(y_0)=\frac{1}{f^\prime(x_0)}$

Steps for finding $(f^{-1})^\prime(y_0)$

solve for $x_0$ from $f(x_0)=y_0$;
Find $f^\prime(x_0)$’;
find $(f^{-1})^\prime(y_0)=\frac{1}{f^\prime(x_0)}$.

[Example]

4.4 Implicit Differentiation

Can not express $y$ explicitly as a function of $x$.

For $x^2+xy+y^2=9$, differentiate with respect to $x$:

$\frac{d(x^2)}{dx}+\frac{d(x\cdot y)}{dx}+\frac{d(y^2)}{dx}=0\text{ (take derivative on both sides)}$ $\implies 2x+(1\cdot y+x\cdot \frac{dy}{dx})+2y\frac{dy}{dx}=0$ $\implies (x+2y)\frac{dy}{dx}=-2x-y$ $\implies \frac{dy}{dx}=\frac{-2x-y}{x+2y}$

4.5 Technique of differentiation of the type $y = f(x)^{g(x)}$

Find the derivative of the function $y=f(x)=x^x$, for $x>0$.

M1:

$y=x^x$ $lny=lnx^x=xlnx$ $\frac{d(lny)}{dx}=\frac{d(lny)}{dy}\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}=lnx+1$ $\frac{dy}{dx}=y(lnx+1)=x^x(lnx+1)$

M2:

$\text{Using }x=e^{lnx}$ $x^x=e^{lnx^x}=e^{xlnx}$ $y^{\prime}={(e^{xlnx})}^{\prime}$ $=e^{xlnx}\cdot (1\cdot lnx+x\cdot \frac{1}{x})$ $=x^x(lnx+1)$

4.6 L’Hopital’s Rule for Finding Limits

4.6.1 Type $\frac{0}{0}$

$f(x),g(x)$ differentiable;
$\lim{x\to a}f(x)=0$, $\lim{x\to a}g(x)=0$
then:

$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^\prime(x)}{f^\prime(x)}$

4.6.2 Type $\frac{\infin}{\infin}$

$f(x),g(x)$ differentiable;
$\lim{x\to a}f(x)=\pm \infin$, $\lim{x\to a}g(x)=\pm \infin$
then:

$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^\prime(x)}{f^\prime(x)}$

Check the form before using the L’Hopital’s rule, type $\frac{0}{1}$ is not applicable.

4.6.3 Using L’Hopital’s Rule to calculus $\lim_{x\to a}f(x)^{g(x)}$

Type $1^\infin$:

$\lim_{x\to 0}(\cos x)^{\frac{1}{x}}$ $=\lim_{x\to 0}e^{ln((\cos x)^{\frac{1}{x}})}$ $=\lim_{x\to 0}e^\frac{ln(\cos x)}{x}$ $=e^{\lim_{x\to 0}\frac{\frac{1}{\cos x}(-\sin x)}{1}}=e^0=1$

Type $\infin^0$:

$\lim_{x\to \infin}(1+x)^{\frac{1}{x}}$ $=\lim_{x\to \infin}e^{ln(1+x)^{\frac{1}{x}}}$ $=\lim_{x\to \infin}e^{\frac{ln(1+x)}{x}}$ $=e^{\lim_{x\to \infin}{\frac{\frac{1}{1+x}}{1}}}=e^0=1$

4.7 Increasing and Decreasing Functions

$f(x)$ is differentiable on open interval $J$:

$f^\prime(x)\gt 0 \text{ on }J\implies$ $f$ is increasing on $J$;
$f^\prime(x)\lt 0 \text{ on }J\implies$ $f$ is decreasing on $J$;

If a function is increasing or decreasing on an interval:

It must be one-to-one function;
It has as inverse function;

Existence of a unique solution: intermediate value theorem(IVT) + monotonicity

4.8 Linear approximation

Approximate a function $y=f(x)$ by a suitable linear function near a given point $a$.

[Example]

4.9 Mean Value Theorem of Differentiation

For $f(x)$:
$f(x)$ is continuous on the closed interval $[a,b]$;
$f(x)$ is differentiable on the open interval $(a,b)$;
$\exist c\in (a,b)$ such that:
$f^\prime (c)=\frac{f(b)-f(a)}{b-a}$
or $f(b)-f(a)=f^\prime (c)(b-a)$

4.10 Higher derivatives

Second-order derivative of $f(x)$:

$f^{\prime \prime}(x)=\frac{df^\prime x}{dx}$

Noted as $y^{\prime \prime},f^{\prime \prime}(x),\frac{d^2y}{dx^2}$

$n^{th}$ derivative noted as:

$y^{(n)},f^{(n)}(x),\frac{d^ny}{dx^n}$ $y^{(0)}=f^{(0)}(x)=f(x)$

Leibniz’s rule:

For $u(x),v(x)$, the $n^{th}$ derivative of $u(x)v(x)$ is:

$(uv)^{(n)}=\sum_{k=0}^{n}\binom{n}{k}u^{(n-k)}v^{(k)},$

$(uv)^{(n)}=\sum_{k=0}^{n}\binom{n}{k}u^{(k)}v^{(n-k)}$

[Example]:

$(uv)^{(3)}=\binom{3}{0}u^{(3)}v+\binom{3}{1}u^{\prime \prime}v^{\prime}+\binom{3}{2}u^{\prime}v^{\prime \prime}+\binom{3}{3}uv^{3}$

4.11 Local maxima and minima

Stationary point (critical point):

$f^{\prime}(a)=0$, then $x=a$ is the stationary point.

4.11.1 First Derivative Test

$f(x)$ is differential in interval $J$ containing $a$, $f^{\prime}(a)=0$:

if $f^{\prime}(x)$ change from $+$ to $-$ with $x$ increasing through $x=a \implies$ $f(x)$ has local maximum at $a$;
if $f^{\prime}(x)$ change from $-$ to $+$ with $x$ increasing through $x=a \implies$ $f(x)$ has local minimum at $a$;

4.11.2 Second Derivative Test

$f(x)$ is twice differential at $a$, and $f^{\prime}(a)=0$:

$f^{\prime \prime}(a)\lt 0\text{ (concave down)} \implies$ local maximum at $a$;
$f^{\prime \prime}(a)\gt 0\text{ (concave down)} \implies$ local minimum at $a$;
$f^{\prime \prime}(a)= 0 \implies$ NO conclusion can be made.

4.12 Global maxima and minima

Closed interval: $J\in [a,b]$

Comparing the $f(x)$ at stationary points $f^{\prime}(c)=0$ and the endpoints $a$ and $b$;

Open interval: $J\in (a,b)$
Comparing the $f(x)$ at stationary points $f^{\prime}(c)=0$ and the limit value at endpoints $x\to a$ and $x\to b$;

if largest(smallest) value is attained in the domain $J\implies$ Global maxima(minima);
if largest(smallest) value is NOT attained in the domain $J\implies$ Global maxima(minima) does NOT exist;

5 Indefinite Integrals

5.1 Definition of indefinite integrals

For

$f(x)=\frac{d}{dx}F(x)$

the $F(x)$ is called the primitive or antiderivative of $f(x)$, $f(x)$ is the derivative of F(x).

Then

$\int f(x) dx=F(x)+C$

$C$ is arbitrary constant.

$\int f(x)dx$ is the indefinite integral of $f(x)$, $f(x)$ is the integrand.

5.2 Table of indefinite integrals

5.3 Basic rules of integration

$\int kf(x)dx=k\int f(x)dx \; \text{ ($k$ is a constant)}$ $\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$

5.4 Techniques of integration: Substitution

If $u=\phi(x)$ with $\phi(x)$ and its derivative $\phi ^\prime(x)$ being continuous, then

$\int f(\phi(x))\phi^\prime(x)dx=\int f(u)du$

[Example]

Find $\int x(x^2+3)^3dx$

$\text{Let } u=x^3+3, \frac{du}{dx}=2x, dx=\frac{du}{2x}$ $\int x(x^2+3)^3dx=\int xu^3\frac{du}{2x}=\frac{1}{2}\int u^3du=\frac{1}{2}\frac{u^4}{4}=\frac{(x^2+3)^4}{8}+C$

[Example]

Find $\int \sqrt{1-4x^2}dx$

5.5 Techniques of integration: Integration by parts

$u(x)$ and $v(x)$ are two differentiable functions, then

$\int u(x)v^\prime (x)dx= u(x)v(x)-\int u^\prime v(x)dx$ $\implies \int u\;dv=uv-\int v\;du$

ILATE order for choosing $v$:

I: $arctan^{-1}x$;
L: $ln(x)$;
A: $x$:
T: $sin(x)$;
E: $e^x$;

lower one to be $v$.

[Example]

Find $\int (x+2)\cos xdx$

$1. \text{ Set }u=x+2, du=dx$ $2.\frac{d\sin x}{dx}=\cos x\implies \cos xdx=d\sin x=dv, v= sin x$ $\therefore \int (x+2)\cos xdx = \int (x+2)d \sin x=(x+2)\sin x- \int \sin x d(x+2)$ $= (x+2)\sin x- \int \sin x dx$ $= (x+2)\sin x+\cos x +C$

5.6 Partial fractions

A proper rational function, with real coefficients, can sometimes be expressed as a sum of two or more proper rational functions, with real coefficients, called partial fractions.

$\frac{x-3}{(2x-1)(x^2+1)}=\frac{-2}{2x-1}+\frac{x+1}{x^2+1}$

[Example]

Resolve $f(x)=\frac{x+3}{(x-1)(x-3)}$

$f(x)=\frac{a}{x-1}+\frac{b}{x-3}=\frac{(a+b)x-3a-b}{(x-1)(x-3)}$ $\implies \begin{cases} a+b=1\\ -3a-b=3 \end{cases}\implies a=-2,b=3$ $\therefore f(x)=\frac{-2}{x-1}+\frac{3}{x-3}$

[Example]

Find $\int \frac{x^2+1}{(x-1)(x-2)(3+3)}dx$

$\frac{x^2+1}{(x-1)(x-2)(3+3)}=\frac{a}{x-1}+\frac{b}{x-2}+\frac{c}{x+3}=\frac{(a+b+c)x^2+(a-2b-3c)x-6a-3b+2c}{(x-1)(x-2)(x+3)}$ $\implies \begin{cases} a+b+c=1\\ a+2b-3c=0\\ -6a-3b+2c=1 \end{cases}\implies a=\frac{-1}{2},b=1,c=\frac{1}{2}$ $\therefore \int \frac{x^2+1}{(x-1)(x-2)(3+3)}dx=\int -\frac{1}{2(x-1)}+\frac{1}{x-2}+\frac{1}{2(x+3)} dx$ $= -\frac{1}{2}\int \frac{1}{x-1}+\int \frac{1}{x-2}+\frac{1}{2} \int \frac{1}{x+3}$ $=-\frac{1}{2}\ln |x-1|+\ln |x-2|+\frac{1}{2}\ln |x+3|+C$

6 Definite Integrals

6.1 Definition of Definite Integrals

$f(x)$ is continuous defined on closed and finite interval $[a,b]$;
$E_i$ is sub-interval of $[a,b]$ with length $\Delta x_i$ and $c_i$ as any point inside;
The Riemann sum of the function $f(x)$ on $[a,b]$ is

$S_N=\sum^N_{i=1}f(c_i)\Delta x_i$

Then the definite integral of $f(x)$ on $[a,b]$ which is $\int^b_af(x)dx$:

$\lim_{\forall \Delta x_i \to 0}S_N=\lim_{\forall \Delta x_i \to 0}\sum^N_{i=1}f(c_i)\Delta x_i$

6.2 Basic Properties of Definite Integrals

$\int^a_bf(x)dx=-\int^b_af(x)dx$
$\int^a_af(x)dx=0$
$\int^b_af(x)dx=\int^b_af(t)dt=\int^b_af(u)du$
Linearity:
- $\int^b_a[\alpha f(x)+\beta g(x)]dx=\alpha\int^b_af(x)dx+\beta\int^b_ag(x)dx$
Additivity over sub-intervals, $a\lt b\lt c$:
- $\int^b_af(x)dx=\int^b_cf(x)dx+\int^c_af(x)dx$