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# MATH 138 Notes by Raphael Koh

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# 1 Review of Integration

## Riemann Integral

Let $f(x)$ be a continuous (or piecewise continuous) function defined over $[a,b]$.
1. We partition $[a,b]$ into $n$ intervals such that each interval has has width $\Delta x=\frac{b-a}{n}$ with partition points $x_k=a+k\Delta x$, $0\leq k\leq n$. Note that the endpoints are $x_0=a$ and $x_n=b$.
2. For each interval $I_k=[x_{k-1}, x_k]$, $1\leq k\leq n$, pick a sample point $x_k^*\in I_k$. This sample point can be the left endpoint ($x_{k-1}$), the right endpoint ($x_k$) or midpoint.
3. Construct Riemann Sum: $$S_n=\sum_{k=1}^n f(x_k^*)\Delta x$$
4. Take the limit: $$S=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n f(x_k^*)\Delta x=\int_a^b f(x) dx$$

## Properties of Integrals

1. $\int_a^b c dx=c(b-a)$
2. $\int_a^b [f(x)+g(x)] dx=\int_a^b f(x) dx+\int_a^b g(x) dx$
3. $\int_a^b cf(x) dx=c\int_a^b f(x) dx$
4. $\int_a^c f(x) dx+\int_c^b f(x) dx=\int_a^b f(x) dx$

## Fundamental Theorem of Calculus 1

Let $f(x)$ be continuous on $[a,b]$. Given the function $$g(x)=\int_a^x f(t)dt,\quad a\leq x\leq b$$ Then $g(x)$ is continuous on $[a,b]$ and $$g'(x)=f(x),\quad a\leq x\leq b$$

## Fundamental Theorem of Calculus 2

Let $F(x)$ be any antiderivative of $f(x)$, then $$\int_a^b f(x)dx = F(b) - F(a)$$

# 2 Integration Techniques

## Integration by Substitution

e.g. $\int \sqrt{5x+4}dx$.

Let $u=5x+4$, $du=5dx$,

\begin{align*} \int \sqrt{5x+4}dx&=\frac{1}{5}\int u^{\frac{1}{2}}du\\ &=\frac{2}{3}\frac{1}{5}u^{\frac{3}{2}} + C\\ &=\frac{2}{15}(5x+4)^{\frac{3}{2}} + C\\ \end{align*}

## Integration By Parts

$$\int udv=uv-\int vdu$$
e.g. $\int xe^xdx$

Let $u=x$, $dv=e^x$, $du=dx$, $v=e^x$. $$\int xe^xdx=xe^x-\int e^xdx=xe^x-e^x+C$$

Strategy for Integration by Parts: "DETAIL" determines which term should be $dv$ in order of whichever comes first in the list.
• D - $dv$
• E - Exponential
• T - Trigonometric
• A - Algebraic
• I - Inverse Trigonometric
• L - Logarithmic

## Average of a Function

Given a function continuous over $[a,b]$,
1. Partition $[a,b]$ into $n$ intervals of width $\Delta x=\frac{b-a}{n}$ and $x_n=a+k\Delta x$.
2. Pick a sample point $x_k^*\in [x_{k-1},x_k]$.
3. \begin{align*} \overline{f(x)}&\approx \frac{1}{n}\sum_{k=1}^nf(x_k^*)\\ &=\frac{1}{n\Delta x}\sum_{k=1}^nf(x_k^*)\Delta x\\ &=\frac{1}{n(\frac{b-a}{n})}\sum_{k=1}^nf(x_k^*)\Delta x\\ &=\frac{1}{b-a}\sum_{k=1}^nf(x_k^*)\Delta x\\ \end{align*}
4. Taking the limit as $n\to\infty$, \begin{align*} \overline{f(x)}&=\frac{1}{b-a}\lim_{n\to\infty}\sum_{k=1}^nf(x_k^*)\Delta x\\ &=\frac{1}{b-a}\int_a^b f(x)dx\\ \end{align*}

## Trigonometric Functions

### Strategy for Evaluating $\int \sin^mx\ \cos^nx\ dx$

• If power of cosine is odd, $n=2k+1$, save one cosine and use $\cos^2x=1-\sin^2x$ to get $$\int \sin^mx\ \cos^{2k+1}x\ dx=\int \sin^mx(1-\sin^2x)^k\cos x\ dx$$ Then, subsitute $u=\sin x$.
• If power of sine is odd, similar process as above.
• If both powers are even, use double angle identities: $$\sin^2x=\frac{1}{2}(1-cos2x)\quad\quad\cos^2x=\frac{1}{2}(1+cos2x)\quad\quad\sin x\cos x=\frac{1}{2}\sin 2x$$

### Strategy for Evaluating $\int \tan^mx\ \sec^nx\ dx$

• If power of secant is even, $n=2k, k\geq 2$, save one $\sec^2x$ and use $\sec^2x=1+\tan^2x$ to get $$\int \tan^mx\ \sec^{2k}x\ dx=\int \tan^mx(1+\tan^2x)^{k-1}\sec^2 x\ dx$$ Then, subsitute $u=\tan x$.
• If power of tangent is odd, $m=2k+1$, save one $\sec x\tan x$ and use $\tan^2x=\sec^2x-1$ to get $$\int \tan^{2k+1}x\ \sec^nx\ dx=\int (\sec^2x-1)^k\sec^{n-1}x\sec x\tan x\ dx$$ Then, subsitute $u=\sec x$.

## Trigonometric Substitution

This technique is used when you find: $$\sqrt{A-x^2}\quad\quad\sqrt{A+x^2}\quad\quad\sqrt{x^2-A}$$ We use the following trig id's: $$\sin^2x+\cos^2x=1\quad\quad\tan^2x+1=\sec^2x$$

### Integrals with $\sqrt{a^2-x^2}$

e.g. $\int\sqrt{a^2-x^2}$

Let $x=a\sin u$, $dx=a\cos u\ du$.

$\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2u}=a\sqrt{1-\sin^2u}=a\cos u$

\begin{align*} \int\sqrt{a^2-x^2}dx&=\int a^2\cos^2u\ du\\ &=\frac{a^2}{2}\int (1+\cos 2u)du\\ &=\frac{a^2}{2}(u+\frac{\sin2u}{2})+C\\ &=\frac{a^2}{2}(u+\sin u\cos u)+C\\ &=\frac{a^2}{2}\big(\sin^{-1}(\frac{x}{a})+\frac{x}{a}\frac{\sqrt{a^2-x^2}}{a}\big)+C\\ &=\frac{a^2}{2}\sin^{-1}(\frac{x}{a})+\frac{x}{2}\sqrt{a^2-x^2}+C \end{align*}

### Integrals with $\sqrt{a^2+x^2}$

Substitute $x=a\tan u$. Similar method as above.

### Integrals with $\sqrt{x^2-a^2}$

Substitute $x=a\sec u$. Similar method as above.

## Partial Fractions

Used to integrate rational functions by expressing it as a sum of simple fractions. Given a rational function, $\frac{P(x)}{Q(x)}$, if $\deg (P) < \deg (Q)$ then it can be expressed as a partial fraction. If not, perform polynomial division to get $\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)}$ such that $\deg (R) < \deg (Q)$. Split the fractions into the form $\frac{A}{ax+b}+\frac{B}{cx+d}$.
e.g. \begin{align*} \int\frac{4x+1}{x^2-x-2}dx&=\int\frac{4x+1}{(x+1)(x-2)}dx\\ &=\int(\frac{3}{x-2}+\frac{1}{x+1}) dx\\ &=3\ln|x-2|+\ln|x+1|+C \end{align*}

### Repeated Roots

If a root is repeated $k$ times, we must repeat that term $k$ times such that $\frac{P(x)}{(ax+b)^k}=\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\cdots+\frac{A_k}{(ax+b)^k}$.

### Complex Roots

If a function has $Q(x)$ with irreducible quadratic factors, the partial fraction will be of the form $\frac{Ax+B}{ax^2+bx+c}$. Complete the square and use $$\int\frac{dx}{x^2-a^2}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\big)+C$$

### Repeated Complex Roots

$$\frac{P(x)}{Q(x)^k}=\frac{A_1x+B_1}{Q(x)}+\frac{A_2x+B_2}{Q(x)^2}+\cdots+\frac{A_kx+B_k}{Q(x)^k}$$

## Conservative Forces & Potential Energy

$\vec{F}=f(x)\hat{i}$ is conservative if there exists a function $U(x)$ such that $f(x)=-U'(x)$, where $U(x)$ is the potential energy function. So, $U(x)=-\int f(x)dx$.
Consequence: If $\vec{F}$ acts on a mass $m$ to move it from $x=a$ to $x=b$, the work done is: $$\int_a^bf(x)dx=-\int_a^bU'(x)dx=-[U(b)-U(a)]=-\Delta U$$ Alternatively, we can let $x=a$ be a reference point and define $U(x)=-\int_a^xf(s)ds$. We define $U(a)=0$. If $f(x)=-mg$, $U(x)=-\int_a^x-mg\ ds=mgx-mga=mgx$.

Physical Interpretation: $U(x)$ is the work done against the force when $m$ is moved from $a$ to $x$. If $U(x)>0$, we have to work against the force to move $m$ from $a$ to $x$ (i.e. causes a "buildup" of PE).

:
Suppose that a mass $m$ is moving on the $x$-axis under the influence of a conservative force $\vec{F}(x)=f(x)\hat{i}$ according to Newton's Second Law of Motion, $\vec{F}=m\vec{a}$. Then, the total mechanical energy of the mass is constant in time.
:
Total energy $E=KE+PE$. Let $x(t)$ be the trajectory of the mass. Then, $E(t)=\frac{1}{2}mv(t)^2+U(x(t))$ where $U'(x)=-f(x)$. \begin{align*} E'(t)=\frac{dE}{dt}&=\frac{1}{2}m\frac{d}{dt}v(t)^2+\frac{d}{dt}U(x(t))\\ &=mv(t)v'(t)+U'(x(t))x'(t)\\ &=mv(t)a(t)-f(x(t))v(t)\\ &=v(t)[ma(t)-f(x(t))]\\ &=v(t)[ma(t)-ma(t)]\\ &=0\quad\square \end{align*}

# 3 Applications of Integration

### Centre of Mass

$$\bar{x}=\frac{\sum_{i=0}^nm_ix_i}{\sum_{i=0}^nm_i}$$ Consider an object of length $L$ with a continuous mass distribution, $\rho(x)$, lying on the $x$-axis with one end on the origin. $$\Delta m\approx\rho(x_k^*)\Delta x$$ $$M=\int_0^L\rho(x)dx$$ $$\bar{x}\approx\frac{\sum_{i=1}^n\Delta m_ix_i^*}{\sum_{i=1}^n\Delta m_i}=\frac{\sum_{i=1}^n\rho(x_i^*)x_i^*\Delta x}{\sum_{i=1}^n\rho(x_i^*)\Delta x}$$ $$\bar{x}=\lim_{n\to\infty}\bigg(\frac{\sum_{i=1}^n\rho(x_i^*)x_i^*\Delta x}{\sum_{i=1}^n\rho(x_i^*)\Delta x}\bigg)=\frac{\int_0^L\rho(x)x dx}{\int_0^L\rho(x)dx}=\frac{1}{M}\int_0^Lx\rho(x)dx$$

### Centre of Mass of Centroids

Given $f(x)$ and $g(x)$ where $f(x)>g(x)$ and are continuous over $(a,b)$. Then $$dA=[f(x)-g(x)]dx$$ First moment w.r.t $y$-axis: $dM_y=xdA$. Net moment w.r.t $y$-axis: $M_y=\int_a^bxdA=\int_a^bx[f(x)-g(x)]dx$
The $x$-coordinate of the Centre of Mass is determimed by $$\int_a^b(x-\bar{x})dA=0$$ $$\int_a^bx\ dA=\bar{x}\int_a^bdA$$ $$\bar{x}=\frac{\int_a^bx\ dA}{\int_a^bdA}=\frac{M_y}{A}=\frac{\int_a^bx[f(x)-g(x)]dx}{\int_a^b[f(x)-g(x)]dx}$$ Similarly, the $y$-coordinate of the Centre of Mass can be found by: $$\bar{y}=\frac{M_x}{A}=\frac{\int_a^by\ dA}{\int_a^bdA}$$ Note that $\bar{x}$ can be determined from $y$ and $\bar{y}$ from $x$ by using the midpoint. $$M_x=\frac{1}{2}\int[f(x)+g(x)]dA$$

### Computation of Arclength

Given a function $f(x)$ continuous over $[a,b]$, we want to find its arclength.
1. Partition into $n$ sub-intervals such that for each sub-interval $[x_{k-1},x_k]$ has an arclength of $C_k\approx||\overrightarrow{P_{k-1}P_k}||=\sqrt{\Delta x^2+\Delta y^2}=\sqrt{\Delta x^2+[f(x_k)-f(x_{k-1})]^2}=\Delta x\sqrt{1+\bigg[\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}\bigg]^2}$
By the Mean Value Theorem, there exists a $x_k^*\in[x_{k-1},x_k]$ such that $f'(x_k^*)=\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}$, so $C_k\approx\Delta x\sqrt{1+f'(x_k^*)^2}$
2. Sum over all intervals. $$L\approx\sum_{k=1}^nC_k=\sum_{k=1}^n\sqrt{1+f'(x_k^*)^2}\Delta x$$
3. Take $n\to\infty$. $$L=\lim_{n\to\infty}\sum_{k=1}^n\sqrt{1+f'(x_k^*)^2}\Delta x=\int\sqrt{1+f'(x)^2}dx$$

Similarly, given a curve $C$, $$L=\int_Cds=\int_C\sqrt{dx^2+dy^2}=\int_Cdx\sqrt{1+\frac{dy}{dx}^2}=\int_a^b\sqrt{1+f'(x)^2}dx$$

### Computation of Centroids of Curves

Consider a function $f(x)$ continuous over $[a,b]$. Let $C$ represent the curve with linear density $\rho_0=1$, so $dm=\rho_0ds=ds$.
Moment w.r.t $y$-axis: $dM_y=xds\ \rightarrow\ M_y=\int_CdM_y=\int_Cxds$
Moment w.r.t $x$-axis: $dM_x=yds\ \rightarrow\ M_x=\int_CdM_x=\int_Cyds$

Recall that for centroid($\bar{x},\bar{y}$):
$$\int_C(x-\bar{x})ds=0\quad\text{Net moment w.r.t \bar{x}}$$ $$\int_C(y-\bar{y})ds=0\quad\text{Net moment w.r.t \bar{y}}$$ So, $$\int_Cxds=\bar{x}\int_Cds\ \rightarrow\ \bar{x}=\frac{\int_Cxds}{\int_Cds}=\frac{\int_Cxds}{L}$$ Similarly, $$\bar{y}=\frac{\int_Cyds}{L}$$

# 4 Improper Integral

### Type I - Integrals over infinite domains

$$\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^af(x)dx+\int_{a}^\infty f(x)dx=\lim_{b\to\infty}I_1(b)+\lim_{b\to\infty}I_2(b)=\lim_{b\to\infty}\int_{-b}^af(x)dx+\lim_{b\to\infty}\int_{a}^b f(x)dx$$

### Important Class of Improper Integrals

Consider $I_p=\int_1^\infty \frac{1}{x^p}dx,\ p\in\mathbb{R}$. $I_p$ converges for $p>1$ and diverges for $p\leq 1$. This class of integrals is useful for "comparison tests" to determine whether the improper integral converges/diverges.

### Comparison Test

Given functions $f(x), g(x)$ such that $f(x)\geq g(x)\geq 0$ for all $x\geq a$, then $\int_a^bf(x)dx\geq\int_a^bg(x)dx$.
1. If $\int_a^\infty f(x)dx$ converges, then $\int_a^\infty g(x)dx$ converges.
2. If $\int_a^\infty g(x)dx$ diverges, then $\int_a^\infty f(x)dx$ diverges.

### Type II - Discontinuous and (possibly) unbounded at a point $c\in[a,b]$

1. $f(x)$ is continuous on $[a,b)$ and discontinuous at $b$, then $\int_a^b f(x)dx=\lim_{t\to b^-}\int_a^t f(x)dx$
2. $f(x)$ is continuous on $(a,b]$ and discontinuous at $a$, then $\int_a^b f(x)dx=\lim_{t\to a^+}\int_t^b f(x)dx$
3. $f(x)$ is continuous on $[a,b]$ except at $c\in(a,b)$, then $\int_a^b f(x)dx=\lim_{t\to c^-}\int_a^t f(x)dx+\lim_{t\to c^+}\int_t^b f(x)dx$

# 5 Differential Equations

A differential equation involves derivaties of a function and possibly the function itself.
Definition: The order of a DE is the order of the highest derivative in the DE.
Definition: A DE is linear if the dependent variable or derivative is not multiplied by itself or each other (i.e. $y + \frac{dy}{dx}$).

### Separable First-Order DEs

Definition: A DE is separable if it can be written in the form $\frac{dy}{dx}=g(x)f(y)$.
e.g.
1. $\frac{dy}{dx}=xy\quad\checkmark$
2. $\frac{dy}{dx}=(x+1)(y^2+1)\quad\checkmark$
3. $\frac{dy}{dx}=x+y\quad\times$
To solve a separable DE:
1. Divide by $f(y)$. $$\frac{1}{f(y)}\frac{dy}{dx}=g(x)\quad *f(y)\neq 0$$ If $f(y^*)=0$, then $y(x)=y^*$ is a constant solution to the DE.
2. Let $h(y)=\frac{1}{f(y)}$, so $h(y)\frac{dy}{dx}=g(x)$.
3. Integrate both sides and solve. $$\int h(y)dy=\int g(x)dx$$
e.g. $\frac{dy}{dx}=xy$
1. Divide by $y$. $$\frac{1}{y}\frac{dy}{dx}=x\quad\textbf{*Check: } y=0\rightarrow\frac{dy}{dx}=x\cdot 0=0\ \rightarrow y(x)=0\text{ is a solution.}$$
2. Integrate. $$\int\frac{1}{y}dy=\int xdx$$
3. Solve. \begin{align*} \ln|y|&=\frac{1}{2}x^2+C\\ |y|&=e^{\frac{1}{2}x^2+C}\\ y&=\pm e^Ce^{\frac{1}{2}x^2}\\ y&=Ae^{\frac{x^2}{2}}\quad A\neq 0\\ \text{But, y=0 is a solution, }y&=Ce^{\frac{1}{2}x^2}\quad , C\in\mathbb{R} \end{align*}

#### Logistic Model for Population Growth

$$\frac{dP}{dt}=kP(1-\frac{P}{M})\quad\text{where P is the population, M is the carrying capacity, and k is the relative growth rate.}$$ The solution to the logistic equation is: $$P(t)=\frac{M}{1+Ae^{-kt}}\quad\text{where }A=\frac{M-P_0}{P_0}$$

### Linear First-Order DE

$$\text{General Form: }\frac{dy}{dx}+P(x)y=Q(x)$$ To solve,
1. Multiply both sides by the integrating factor, $I(x)=e^{\int P(x)dx}$ $$e^{\int P(x)dx}\frac{dy}{dx}+e^{\int P(x)dx}P(x)y=Q(x)e^{\int P(x)dx}$$ $$\frac{d}{dx}[ye^{\int P(x)dx}]=Q(x)e^{\int P(x)dx}$$
2. Integrate both sides. $$ye^{\int P(x)dx}+C=\int Q(x)ye^{\int P(x)dx}dx$$
3. Solve for y. $$y(x)=e^{-\int P(x)dx}\int Q(x)ye^{\int P(x)dx}dx-Ce^{-\int P(x)dx}$$
e.g. Solve $\frac{dy}{dx}+2y=x$.
$P(x)=2,\ Q(x)=x,\ I(x)=e^{\int 2dx}=e^{2x}$
1. Multiply by $I(x)$. $$e^{2x}\frac{dy}{dx}+2ye^{2x}=xe^{2x}$$ $$\frac{d}{dx}[ye^{2x}]=xe^{2x}$$
2. Integrate. $$ye^{2x}+C_1=\int xe^{2x}dx$$ Let $u=x,\ du=dx,\ dv=e^{2x}dx,\ v=\frac{1}{2}e^{2x}$. $$ye^{2x}=\frac{1}{2}xe^{2x}-\int \frac{1}{2}e^{2x}dx +C_2$$ $$ye^{2x}=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x} +C$$
3. Solve for y. $$y=\frac{1}{2}x-\frac{1}{4} +Ce^{-2x}\quad\square$$
e.g. Projectile Motion with Air Resistance
$$m\frac{dv}{dt}=-mg-kv_y\ \rightarrow\ \frac{dv}{dt}=-g-\frac{k}{m}v_y$$ Let's rewrite this in terms of a first-order linear DE. $$\frac{dv}{dt}+\frac{k}{m}v=-g\quad P(t)=\frac{k}{m},\ Q(t)=-g,\ I(t)=e^{\int \frac{k}{m}dt}=e^{\frac{kt}{m}}$$ We multiply by $I(t)$: $$e^{\frac{kt}{m}}\frac{dv}{dt}+\frac{k}{m}ve^{\frac{kt}{m}}=-ge^{\frac{kt}{m}}$$ $$\frac{d}{dt}[ve^{\frac{kt}{m}}]=-ge^{\frac{kt}{m}}$$ Integrate both sides: $$ve^{\frac{kt}{m}}+C_1=-\int ge^{\frac{kt}{m}}$$ $$ve^{\frac{kt}{m}}=-\frac{mg}{k}e^{\frac{kt}{m}}+C$$ Solve for v. $$v=-\frac{mg}{k}+Ce^{-\frac{kt}{m}}$$ If we impose initial condition such that $v_y(0)=v_0\sin\theta$, $v(0)=v_0\sin\theta=-\frac{mg}{k}+Ce^{-\frac{k(0))}{m}}=-\frac{mg}{k}+C$ So, $C=v_0\sin\theta+\frac{mg}{k}$. $$\therefore v(t)=(v_0\sin\theta+\frac{mg}{k})e^{-\frac{kt}{m}}-\frac{mg}{k}$$

# 6 Parametric Representations of Curves

i.e. Projectile Motion. These are the parametric representations of the trajectory of a projectile: $$x(t)=(v_0\cos\theta)t\quad y(t)=(v_0\sin\theta)t-\frac{1}{2}gt^2$$

### Calculus with Parametric Curves

#### Tangents

Given $x=f(t)$ and $y=g(t)$, $$\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$$ So, $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\quad\frac{dx}{dt}\neq 0$$

#### Areas

Given a curve defined by $x=f(t)$ and $y=g(t)$, the area under the curve as $x$ goes from $a$ to $b$ or as $t$ goes from $\alpha$ to $\beta$ can be found by: $$A=\int_a^b ydx=\int_\alpha^\beta g(t)f'(t)dt$$

### Arclength Using Parametric Representation

Given a curve $C$ described by $x=f(t)$ and $y=g(t)$ and $f',\ g'$ are continuous over $[a,b]$, and $C$ is traversed exactly once as $t$ goes from $a$ to $b$, find the arclength of the curve in the interval $[a,b]$.
1. Partition $[a,b]$ into $n$ sub-intervals of width $\Delta t=\frac{b-a}{n}$
2. \begin{align*} L_k=||\overrightarrow{Q_kQ_{k+1}}||&\approx\sqrt{[f(t_{k+1})-f(t_k)]^2+[g(t_{k+1})-g(t_k))]^2}\\ &=\Delta t\sqrt{\bigg[\frac{f(t_{k+1})-f(t_k))}{\Delta t}\bigg]^2+\bigg[\frac{g(t_{k+1})-g(t_k))}{\Delta t}\bigg]^2}\\ &\approx\sqrt{f'(t_k^*)^2+g'(t_k^*)^2}\Delta t \end{align*}
3. Take the Riemann Sum. $$L=\lim_{n\to\infty}\sum_{k=1}^nL_k=\int_a^b\sqrt{f'(t)^2+g'(t)^2}dt$$
Alternatively, we can derive this formula from the formula for arclength of a function $y=f(x)$. $$L=\int_a^b\sqrt{1+\big(\frac{dy}{dx}\big)^2}dx=\int_a^b\sqrt{1+\bigg(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\bigg)^2}dx=\int_a^b\sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}dx\cdot\frac{dt}{dx}=\int_a^b\sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}dt$$

# 7 Sequences

Definition: An infinite set of real numbers ordered $a_1, a_2,\ldots\$ Denoted as $\{a_n\}$ or $\{a_n\}_{n=1}^\infty$
Definition: A sequence has the limit, L, which we write as $$\lim_{n=\infty}a_n=L\quad\text{or}\quad a_n\to L\text{ as } n\to\infty$$ if $\ \forall\epsilon > 0,\ \exists N > 0$ such that $|a_n-L|<\epsilon\quad\forall n>N$
If the limit exists, then $\{a_n\}$ converges. Otherwise, it diverges.
If $\{a_n\}$ and $\{b_n\}$ are convergent sequences and $c\in\mathbb{R}$, then
1. $\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_b+\lim_{n\to\infty}b_n$
2. $\lim_{n\to\infty}ca_n=c\lim_{n\to\infty}a_n$
3. $\lim_{n\to\infty}(a_nb_n)=\lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}b_n$
4. $\lim_{n\to\infty}a_n^p=[\lim_{n\to\infty}a_n]^p\quad$if $p>0$ and $a_n>0$

### Increasing/Decreasing Sequences

Given $\{a_n\}$,
1. Increasing: $a_n < a_{n+1}$ for all $n\geq 1$
2. Decreasing: $a_n > a_{n+1}$ for all $n\geq 1$
3. Monotone: if either increasing or decreasing
Definition: $\{a_n\}$ is bounded above if $\ \exists M$ such that $a_n\leq M\ \ \forall n\geq 1$
Definition: $\{a_n\}$ is bounded below if $\ \exists m$ such that $a_n\geq m\ \ \forall n\geq 1$
Definition: $\{a_n\}$ is bounded if $\ \exists m,\ M$ such that $m\leq a_n\leq M\ \ \forall n\geq 1$
Completeness Axiom: Let $S\subset\mathbb{R}$ be bounded above such that there exists an $M\in\mathbb{R}$, $x\leq M$ for all $x\in S$. Then, $S$ has a least upper bound $b$ such that if $M$ is any other upper bound, then $b\leq M$. **This axiom is used to express that there is no hole or gap in the real number line.

:
A bounded monotone sequence is convergent (i.e. has a limit)

# 8 Infinite Series

Given a sequence $\{a_n\}_{n=1}^\infty$, we consider the series $\sum_{n=1}^\infty a_n$
Partial Sum: $S_n=a_1+a_2+\cdots+a_n=\sum_{k=1}^n a_k$

These partial sums for a sequence $\{S_n\}$.
If $\{S_n\}$ converges and $\lim_{n\to\infty}S_n=S$ exists and is finite, then the series $\sum a_n$ is convergent such that $\sum_{n=1}^\infty a_n=S$, where $S$ is the sum of the series. If $\{S_n\}$ is divergent, then the series is divergent.

e.g. Geometric Series
Consider the geometric series $a+ar+ar^2+\cdots$ where $a_n=ar^{n-1}$, $n\geq 1$.
Its partial sum is \begin{align*} S_n&=a+ar+\cdots+ar^{n-1}\\ rS_n&=ar+ar^2+\cdots+ar^n\\ S_n(1-r)&=a-ar^n=a(1-r^n)\\ S_n&=\frac{a(1-r^n)}{1-r}\quad r\neq 1 \end{align*}
1. If $|r|< 1$, then $r^n\to 0$ as $n\to\infty\ \rightarrow\ \lim_{n\to\infty}S_n=\frac{a}{1-r}\quad$(sum of a geometric series)
2. If $|r|> 1$, then $|S_n|\to\infty$ so the series diverges.
3. $r=1$, $a+a+a+\cdots$ is divergent.
4. $r=-1$, $a-a+a-a+\cdots$ oscillates and is divergent
Thus, geometric series converge to $\frac{a}{1-r}$ for $|r|< 1$ and diverge for $|r|\geq 1$.

### Convergence and Divergence of Series

:
If $\sum a_n$ is convergent, then $\lim_{n\to\infty}a_n=0$
:
Consider the partial sum $S_n=a_1+\cdots+a_n=S_{n-1}+a_n$. We can rewrite this as $a_n=S_n-S_{n-1}$. We suppose that $\sum a_n$ is convergent, so we $\lim_{n\to\infty}S_n=s=\lim_{n\to\infty}S_{n-1}$. Thus, $\lim_{n\to\infty}a_n=\lim_{n\to\infty}(S_n-S_{n-1})=\lim_{n\to\infty}S_n-\lim_{n\to\infty}S_{n-1}=s-s=0\quad\square$
Note: $\lim_{n\to\infty}a_n=0$ does not imply that $\sum a_n$ is convergent. (i.e. Harmonic Series)
:
If $\lim_{n\to\infty}a_n\neq 0$, then $\sum a_n$ is divergent.
:
If $\sum a_n$ and $\sum b_n$ are convergent, then $\sum (a_n+b_n)$ is convergent.

### Convergence Tests

#### 1. Integral Test

If $\sum a_n=f(n)$ where $f(x)$ is continuous, positive and decreasing on $[b,\infty)$, then $$\sum_{n=b}^\infty a_n\text{ is convergent }\Longleftrightarrow\ \int_b^\infty f(x)dx\ \text{ is convergent }$$ Estimating Sums
Given $\sum a_n$ converges and its sum is $S$, its partial sums $S_n$ are approximations to $S$. For $n\geq 1$, $S=S_n+R_n$ where $R_n$ is the remainder (or the error in approximation). We can view $R_n$ as the "tail" of the series, $R_n=a_{n+1}+a_{n+2}+\cdots$ $$R_n\leq \int_n^\infty f(x)dx$$

#### 2. Comparison Test

Given $\sum a_n$ and $\sum b_n$ where $a_n,b_n>0$ then
1. If $\sum b_n$ is convergent and $a_n\leq b_n$ for all $n\geq 1$, then $\sum a_n$ is convergent.
2. If $\sum b_n$ is divergent and $a_n\geq b_n$ for all $n\geq 1$, then $\sum a_n$ is divergent.
Tip: Determine what $a_n$ converges to as $n\to\infty$. If $a_n\to 0$ then find $\sum b_n$ such that $b_n\geq a_n$ for all $n\geq 1$.

e.g. Determine whether $\sum_{n=1}^\infty \frac{5}{2n^2+4n+3}$ converges or diverges.
We note that as $n\to\infty$, $a_n\to\frac{5}{2n^2}=0$. Thus, we know it will converge, so we choose $b_n=\frac{5}{2n^2}$. Since $\frac{5}{2n^2+4n+3}<\frac{5}{2n^2}$ and $b_n$ converges (p-series where $p>1$), thus $\sum_{n=1}^\infty \frac{5}{2n^2+4n+3}$ converges.

Estimating Sums
If we have used the Comparison Test to show that a series $\sum a_n$ converges by comparison with $\sum b_n$, then we can compare the remainders.
Let $R_n$ be the remainder for $\sum a_n$ such that $R_n=s-s_n=a_{n+1}+a_{n+2}+\cdots$
Let $T_n$ be the remainder for $\sum b_n$ such that $T_n=t-t_n=b_{n+1}+b_{n+2}+\cdots$
Since $a_b\leq b_n$ for all $n\geq 1$, then $R_n\leq T_n$.
• If $\sum b_n$ is a geometric series, $T_n=\frac{a}{1-r}$
• If $\sum b_n$ is a p-series, we use the Remainder Estimate for the Integral Test shown above.

#### 3. Limit Comparison Test

Given $\sum a_n$ and $\sum b_n$ where $a_n,b_n>0$, if $\lim_{n\to\infty}\frac{a_n}{b_n}=c$, where $c$ is finite and $c>0$, then either both converge or both diverge.
Proof: For any $\epsilon > 0$, there exists an $N$ such that $$\bigg|\frac{a_n}{b_n}-c\bigg|<\epsilon\quad\forall n > N$$ $$-\epsilon < \frac{a_n}{b_n}-c < \epsilon$$ $$c-\epsilon < \frac{a_n}{b_n} < c+\epsilon$$ Letting $m=c-\epsilon$ and $M=c+\epsilon$, $$mb_n < a_n < Mb_n$$ Thus, we observe that if $\sum b_n$ converges, then by the comparison test, since $Mb_n > a_n$ and $Mb_n$ converges, $\sum a_n$ must converge. If $\sum b_n$ diverges, then by the comparison test, since $Mb_n < a_n$ and $mb_n$ diverges, $\sum a_n$ must diverge. Likewise, by the comparison test, if $\sum a_n$ converges, $b_n$ must converge and if $\sum a_n$ diverges, $b_n$ must diverge.

Tip: Determine the dominant term in both the numerator and denominator and use that as $b_n$.

e.g. Determine whether the series $\sum_{n=1}^\infty\frac{2n^3+3n}{\sqrt{5+n^5}}$ converges or diverges.
Note that the dominant term in the numerator is $2n^2$ and the dominant term in the denominator is $\sqrt{n^5}$. Thus, we take $b_n=\frac{2n^2}{n^{\frac{5}{2}}}=\frac{2}{n^{\frac{1}{2}}}$. \begin{align*} \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{2n^3+3n}{\sqrt{5+n^5}}\cdot\frac{n^{\frac{1}{2}}}{2}=\lim_{n\to\infty}\frac{2n^{\frac{5}{2}}+3n^{\frac{3}{2}}}{2\sqrt{5+n^5}}=\lim_{n\to\infty}\frac{2+\frac{3}{n}}{2\sqrt{\frac{5}{n^5}+1}}=\frac{2}{2}=1 \end{align*} Since $c$ is a positive, finite number, $b_n$ is divergent (p-series with $p< 1$), the given series is divergent by the Limit Comparison Test.

### Alternating Series

#### 4. Alternating Series Test

If an alternating series $\sum_{n=1}^\infty (-1)^{n-1}b_n$ satisfies
1. $b_{n+1}\leq b_n$ for all $n$
2. $\lim_{n\to\infty}b_n=0$
then the series is convergent.

Proof: Consider the even partial sums $\{S_{2n}\}$. We want to show that it is convergent, i.e. that it is increasing and bounded above. $S_{2n}=S_{2n-2}+(b_{2n-1}-b_{2n})\geq S_{2n-2}$. Since $b_{2n}\leq b_{2n-1}$. Thus, the sequence of even partial sums $\{S_{2n}\}$ is increasing. We can also rewrite the sum as $S_{2n}=b_1-(b_2-b_3)-\ldots-(b_{2n-2}-b_{2n-1})-b_{2n}$. Each term in parantheses is positive, so $S_{2n}\leq b_1$ for all $n$. Therefore, the sequence $\{S_{2n}\}$ is increasing and bounded above and is convergent by the Bounded Monotone Sequence Theorem. We let $\lim_{n\to\infty}S_{2n}=s$.
We now compute the odd partial sums. $$\lim_{n\to\infty}S_{2n+1}=\lim_{n\to\infty}(S_{2n}+b_{2n+1})=\lim_{n\to\infty}S_{2n}+\lim_{n\to\infty}b_{2n+1}=s+0=s$$ Since both even and odd partial sums converge to $s$, we have that $\lim_{n\to\infty}S_n=s$ and thus the series is convergent.
:
If $\sum (-1)^{n-1}b_n$, $b_n>0$, satisfies the conditions for an alternating series, then $|R_n|=|s-s_n|\leq b_{n+1}$.

### Absolute Convergence

Definition: A series $\sum a_n$ is called absolutely convergent if the series $\sum|a_n|$ is convergent.
Definition: A series $\sum a_n$ is called conditionally convergent if it is convergent but not absolutely convergent.
:
If a series $\sum a_n$ is absolutely convergent, then it is convergent.
:
Observe the following inequality: $0\leq a_n + |a_n|\leq 2|a_n|$.
Suppose $\sum a_n$ is absolutely convergent, then $\sum |a_n|$ is convergent. By the Comparison Test, $\sum (a_n+|a_n|)$ is convergent. Since $\sum a_n=\sum(a_n+|a_n|)-\sum |a_n|$ is a difference of two convergent series, then $\sum a_n$ is convergent. $\quad\square$

#### 5. Ratio Test

Given a series $\sum a_n$, let $\lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg| = L$
1. If $L< 1$, then $\sum a_n$ is absolutely convergent (and therefore convergent).
2. If $L>1$ or $L=\infty$, then $\sum a_n$ is divergent.
3. If $L=1$, test is inconclusive.
Proof:
1. If $L< 1$, we can choose an $r$ such that $0 < L < r < 1$, so $\bigg|\frac{a_{n+1}}{a_n}\bigg|< r$ for a large enough $n\geq N$. This give \begin{align*} |a_{N+1}| &< r|a_N|\\ |a_{N+2}| &< r|a_{N+1}| < r^2|a_N|\\ &\vdots\\ r|a_{N+k}| &< r^kr|a_N| \end{align*} The series $\sum |a_N|r^k$ converges (geometric series with $r< 1$), so by the Comparison Test, $\sum|a_{N+k}|$ converges. Thus $\sum|a_k|$ converges absolutely.
2. If $L > 1$, there exists an integer $N$ such that $\bigg|\frac{a_{n+1}}{a_n}\bigg| > 1$ for $n\geq N$ so $|a_{n+1}| > |a_n|$. Thus, $\lim_{n\to\infty}a_n\neq 0$ so $\sum a_n$ diverges by the Divergence Test.

#### 6. Root Test

This is used when $n$th power occurs.
Given a series $\sum a_n$, let $\lim_{n\to\infty}\sqrt[n]{|a_n|} = L$
1. If $L< 1$, then $\sum a_n$ is absolutely convergent (and therefore convergent).
2. If $L>1$ or $L=\infty$, then $\sum a_n$ is divergent.
3. If $L=1$, test is inconclusive.

# 9 Power Series

A power series about a is a series that can be written in the form of $$\sum_{n=0}^{\infty} c_{n}(x-a)^{n}$$ Where $a, c_{n} \in \mathbb{R}$. Some things to note are the $c_{n}$'s are often called the coefficients of the series, and that the series is a function of $x$.
:
For a given power series $\sum_{n=0}^\infty c_n(x-a)^n$, there are only three possibilities for convergence:
1. Converges only at $x=a$.
2. Converges for all $x$.
3. Converges if $|x-a|< R$ and diverges if $|x-a| > R$ for $R>0$. There are four possibilities: $$(a-R, a+R)\quad\quad[a-R, a+R]\quad\quad(a-R, a+R]\quad\quad[a-R, a+R)$$
To find the radius of convergence, apply the ratio test and test at endpoints.
e.g. Find the radius of convergence and interval of convergence of $\sum_{n=0}^\infty \frac{(-3)^nx^n}{\sqrt{n+1}}$.

Let $a_n=\frac{(-3)^nx^n}{\sqrt{n+1}}$. Applying the ratio test, we get $$\lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=\bigg|\frac{(-3)^{n+1}x^{n+1}}{\sqrt{n+2}}\cdot\frac{\sqrt{n+1}}{(-3)^nx^n}\bigg|=\bigg|-3x\sqrt{\frac{n+1}{n+2}}\bigg|=3|x|\sqrt{\frac{1+\frac{1}{n}}{1+\frac{2}{n}}}\rightarrow 3|x|\quad\text{as }n\to\infty$$ By the ratio test, the series converges if $3|x|< 1$ and diverges if $3|x| > 1$. Thus, it converges if $x<|\frac{1}{3}|$ and diverges if $x>|\frac{1}{3}|$. So, the radius of convergence is $R=\frac{1}{3}$.
We now test for the endpoints. When $x=-\frac{1}{3}$, the series becomes $\sum_{n=0}^\infty \frac{(-3)^n(-\frac{1}{3})^n}{\sqrt{n+1}}=\sum_{n=0}^\infty \frac{1}{\sqrt{n+1}}$, which diverges (p-series with $p = \frac{1}{2} < 1$). When $x=\frac{1}{3}$, the series becomes $\sum_{n=0}^\infty \frac{(-1)^n}{\sqrt{n+1}}$ which converges by the Alternating Series Test. Thus, the interval of convergence is $\big(-\frac{1}{3}, \frac{1}{3}\big]$.

### Representations of Functions as Power Series

Functions can be modelled with a power series by manipulating the geometric series or by differentiating or integrating the series.
Useful for: Integrating functions without an elementary antiderivative, solving differential equations, and approximating functions by polynomials.

#### Representation of Functions with the Geometric Series

$$\frac{1}{1-x}=1+x+x^2+\cdots=\sum_{n=0}^\infty x^n\quad |x|< 1$$

#### Manipulating Series

You can manipulate series by:
1. Substituting into $x$
2. Multiplying
e.g. Find a power series representation of $\frac{x^3}{x+2}$.
We note that this is very similar to the function represented by a geometric series as shown above. $$\frac{x^3}{x+2}=x^3\cdot\frac{1}{2+x}=\frac{x^3}{2}\cdot\frac{1}{1+\frac{x}{2}}\quad (a = 1,\ r = -\frac{x}{2})$$ $$\frac{x^3}{x+2}=\frac{x^3}{2}\sum_{n=0}^\infty (-\frac{x}{2})^n=\sum_{n=0}^\infty (-1)^n\frac{x^{n+3}}{2^{n+1}}$$ Since this is a geometric series, we know that it converges when $|r| < 1$ or $|-\frac{x}{2}|< 1 \rightarrow x < 2$. Thus the radius of convergence is $R=2$ and the interval of convergence is $(-2, 2)$.

#### Differentiation and Integration of Power Series

:
If the power series $\sum c_n(x-a)^n$ has radius of convergence $R>0$ then the function defined by: $$f(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots=\sum c_n(x-a)^n$$ is differentiable and continuous on the interval $(a-R,a+R)$ and
1. $$f'(x)=c_1+2c_2(x-a)+3c_3(x-a)^2+\cdots=\sum nc_n(x-a)^{n-1}\quad\text{or}$$ $$\frac{d}{dx}\bigg[\sum c_n(x-a)^n\bigg]=\sum \frac{d}{dx}[c_n(x-a)^n]$$
2. $$\int f(x)dx=C+c_0(x-a)+c_1\frac{(x-a)^2}{2}+c_2\frac{(x-a)^3}{3}+\cdots=C+\sum c_n\frac{(x-a)^{n+1}}{n+1}\quad\text{or}$$ $$\int\bigg[\sum c_n(x-a)^n\bigg]dx=\sum \int[c_n(x-a)^n]dx$$

# 10 Taylor & Maclaurin Series

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If $f$ has a power series representation at $a$ such that $$f(x)=\sum c_n(x-a)^n\quad |x-a|< R$$ then its coefficients are given by $$c_n=\frac{f^{(n)}(a)}{n!}$$ Thus, $$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots$$
Definition: The Maclaurin Series is the Taylor Series centred at $a=0$. $$f(x)=f(0)+f'(0)(x-a)+\frac{f''(0)}{2!}(x-a)^2+\frac{f'''(0)}{3!}(x-a)^3+\cdots$$
In general, $f(x)$ is the sum of a Taylor series if $f(x)=\lim_{n\to\infty}T_n(x)$. We define $T_n(x)$ as the $n$th-degree Taylor polynomial of $f$ at $a$ and $R_n(x)$ as the remainder of the Taylor series such that $R_n(x)=f(x)-T_n(x)$.

:
If $\lim_{n\to\infty}R_n(x)=0$ for $|x-a| < R$, then $f$ is equal to the sum of the Taylor series on the interal $|x-a|< R$.

#### Taylor's Inequality

If $|f^{(n+1)}(x)|\leq M$ for $|x-a|\leq d$, then the remainder $R_n(x)$ of the Taylor series satisfies $$|R_n(x)|\leq \frac{M}{(n+1)!}|x-a|^{n+1}\quad\text{for }|x-a|\leq d$$ When using Taylor's Inequality, it is useful to note that $$\lim_{n\to\infty}\frac{x^n}{n!}=0\quad x\in\mathbb{R}$$

#### Multiplication and Division of Power Series

Power series can be added or subtracted, they can also be multiplied and divided using polynomial multiplication/division.

### Applications of Taylor Polynomials

#### 1. Obtaining non-elementary antiderivatives

e.g. Find $\int e^{-x^2}$
We use the Maclaurin series expansion for $e^{t}=\sum_{n=0}^\infty \frac{t^n}{n!}$, where $t=-x^2$. \begin{align*} e^{-x^2}&=\sum_{n=0}^\infty \frac{(-x^2)^n}{n!}\\ &=\sum_{n=0}^\infty (-1)^n\frac{x^2n}{n!}\\ &=1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\cdots \end{align*} To get $\int e^{-x^2}$, we antidifferentiate term-wise. $$\int e^{-x^2}=C+x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\cdots$$ We can use this series to obtain estimates of definite integrals of $e^{-x^2}$. $$\int_0^1e^{-x^2}=\bigg[x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\cdots\bigg]_0^1=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\cdots$$ Furthermore, we can estimate the error in the approximation for any given $n$. For example, if we use $T_5(x)$, we can estimate the error by noting that this is an Alternating Series, so we use the Alternating Series Estimation Theorem, $$R_5\leq b_{6}=\frac{1}{11\cdot 5!}=\frac{1}{1320}$$