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diff --git a/usth/MATH1.4/homework/limits.tex b/usth/MATH1.4/homework/limits.tex new file mode 100644 index 0000000..c4d31e4 --- /dev/null +++ b/usth/MATH1.4/homework/limits.tex @@ -0,0 +1,1727 @@ +\documentclass[a4paper,12pt]{article} +\usepackage[utf8]{inputenc} +\usepackage[english,vietnamese]{babel} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{enumerate} +\usepackage{mathtools} +\usepackage{pgfplots} +\usepackage{siunitx} +\title{Calculus Homework} +\author{Nguyễn Gia Phong} +\date{Winter 2018} +\newcommand{\ud}{\,\mathrm{d}} +\newcommand{\leibniz}[3][]{\frac{\mathrm{d} #1 #2}{\mathrm{d} #3 #1}} +\DeclareMathOperator{\erf}{erf} + +\begin{document} +\maketitle +\setcounter{section}{1} +\section{Limits} + +\setcounter{subsection}{2} +\subsection{Limit Laws} +Evaluate the limit: +\[\lim_{x \to 2}\sqrt{\frac{2x^2 + 1}{3x - 2}} += \sqrt{\lim_{x \to 2}\frac{2x^2 + 1}{3x - 2}} += \sqrt{\frac{2 \cdot 2^2 + 1}{3 \cdot 2 - 2}} += \frac{3}{2} \tag{9}\] + +\[\lim_{x \to 4}\frac{x^2 - 4x}{x^2 - 3x - 4} += \lim_{x \to 4}\frac{x}{x + 1} += \frac{4}{5} \tag{12}\] + +\begin{align*} + \lim_{t \to 0}\frac{\sqrt{1 + t} - \sqrt{1 - t}}{t} +&= \lim_{t \to 0}\frac{2t}{t(\sqrt{1 + t} + \sqrt{1 - t})} \\ +&= \lim_{t \to 0}\frac{2}{(\sqrt{1 + t} + \sqrt{1 - t})} \\ +&= \frac{2}{\sqrt{1} + \sqrt{1}} \\ +&= 1 \tag{25} +\end{align*} + +\noindent\textbf{40. }Prove that +$\lim_{x \to 0^+}\sqrt{x}e^{\sin\frac{\pi}{x}} = 0$. + +Given $\varepsilon > 0$, let $\delta = \frac{1}{9}\varepsilon^2$. +If $0 < x < 0 + \delta$ then \[0 < \sqrt{x}e^{\sin\frac{\pi}{x}} +\leq e\sqrt{\delta} < 3\sqrt{\frac{\varepsilon^2}{9}} +\Longrightarrow |\sqrt{x}e^{\sin\frac{\pi}{x}} - 0 | < \varepsilon\] + +Thus, by the definition of right-hand limit, +\[\lim_{x \to 0^+}\sqrt{x}e^{\sin\frac{\pi}{x}} = 0\] + +\noindent\textbf{59. }Prove that $\lim_{x \to 0}f(x) = 0$ if +\[f(x) = \begin{cases} + x^2\text{ if } x\text{ is rational} \\ + 0\text{ if } x\text{ is irrational} + \end{cases}\] + +Given $\varepsilon > 0$, let $\delta = \sqrt{\varepsilon}$. +If $0 < |x - 0| < \delta$, then $0 < x^2 < \varepsilon$ +or $|f(x) - 0| < \varepsilon$. Thus, by the definition of a limit, +\[\lim_{x \to 0}f(x) = 0\] + +\noindent\textbf{61. }If $f(x) = \begin{cases} + 1\text{ if } x \geq 0 \\ + 0\text{ if } x < 0 + \end{cases}$ +and $g(x) = \begin{cases} + 0\text{ if } x \geq 0 \\ + 1\text{ if } x < 0 + \end{cases}$ +then $f(x)g(x) = 0$. Thus $\lim_{x \to 0}f(x)g(x) = 0$ though neither +$\lim_{x \to 0}f(x)$ nor $\lim_{x \to 0}g(x)$ exists. + +\subsection{The precise definition of a limit} +\textbf{3. }Given $f(x) = \sqrt{x}$, if $|x - 4| < 1.44$ then +$|\sqrt{x} - 2| < 0.4$. + +\noindent\textbf{21. }Prove that $\lim_{x \to 2}\frac{x^2 + x - 6}{x - 2} = 5$. + +Given $\varepsilon > 0$, let $\delta = \varepsilon$. +If $0 < |x - 2| < \delta$, then \[|x + 3 - 5| < \varepsilon +\iff \left|\frac{x^2 + x - 6}{x - 2} - 5\right| < \varepsilon\] + +Thus, by the definition of a limit, +$\lim_{x \to 2}\frac{x^2 + x - 6}{x - 2} = 5$. + +\noindent\textbf{39. }Prove that $\lim_{x \to 0}f(x)$ does not exist if +\[f(x) = \begin{cases} + 0\text{ if } x\text{ is rational} \\ + 1\text{ if } x\text{ is irrational} + \end{cases}\]. + +Suppose $\lim_{x \to 0}f(x) = L$, hence by the definition of limit, +for every $\varepsilon > 0$, there exists $\delta > 0$ that +\[0 < |x - 0| < \delta \Rightarrow |f(x) - L| < \varepsilon \tag{$*$}\] + +For $L = 0$, consider $\varepsilon = |L - 1|$. For every $\delta$, there is +at least one irrational $x \in (0, \delta)$, which turns $(*)$ into a false +statement: \[0 < |x| < \delta \Rightarrow |1 - L| < |L - 1|\] + +For $L \neq 0$, consider $\varepsilon = |L|$. For every $\delta$, there is +at least one rational $x \in (0, \delta)$, which turns $(*)$ into a false +statement: \[0 < |x| < \delta \Rightarrow |L| < |L|\] + +Conclusion: The assumption is incorrect; in other words, $\lim_{x \to 0}f(x)$ +does not exist. + +\subsection{Continuity} +\textbf{22. }Explain why the function $f$ is discontinuous at the given number +$a = 3$. +\begin{align*} +f(x) &= \begin{cases} + \frac{2x^2 - 5x - 3}{x - 3}\text{ if } x \neq 3 \\ + 6\text{ if } x = 3 + \end{cases}\\ + &= \begin{cases} + 2x + 1\text{ if } x \neq 3 \\ + 6\text{ if } x = 3 + \end{cases} +\end{align*} + +Since $\lim_{x \to 3}f(x) = \lim_{x \to 3}(2x + 1) = 7 \neq 6 = f(3)$, +$f$ is discontinuous at $3$. + +\begin{tikzpicture} + \begin{axis}[ + axis x line=middle, axis y line=middle, + xlabel={$x$}, ylabel={$f(x)$}, + xlabel style={at=(current axis.right of origin), anchor=west}, + ylabel style={at=(current axis.above origin), anchor=south}] + \addplot[domain=-1:4,blue]{2*x + 1}; + % This part is a bit hacky but it works XD + \addplot[white, mark=*, only marks] coordinates {(3,7)}; + \addplot[blue, mark=o, only marks] coordinates {(3,7)}; + \addplot[blue, mark=*, only marks] coordinates {(3,6)}; + \end{axis} +\end{tikzpicture} + +\noindent\textbf{26. }$G(x) = \frac{x^2 + 1}{2x^2 - x - 1}$ is a rational +function so it is continuous at every number in its domain. + +\noindent\textbf{38. }Since $arctan$ is an inverse trigonometric function and +thus continuous at every number in its domain and +$\lim_{x \to 2}\frac{x^2 - 4}{3x^2 - 6x} = \lim_{x \to 2}\frac{x + 2}{3x} = +\frac{2}{3}$, \[\lim_{x \to 2}\arctan\frac{x^2 - 4}{3x^2 - 6x} = +\arctan\frac{2}{3}\] + +\subsection{To Infinity and Beyond!} +Find the limit: +\[\lim_{x \to -\infty}\frac{\sqrt{9x^6 - x}}{x^3 + 1} += \lim_{x \to -\infty}\frac{\sqrt{9 - \frac{1}{x^5}}}{-1 - \frac{1}{x^3}} += -3 \tag{24}\] + +\subsection{Derivatives} +\textbf{24. }If $g(x) = x^4 - 2$, +\begin{align*} +g'(1) &= \lim_{h \to 0}\frac{g(1 + h) - g(1)}{h}\\ + &= \lim_{h \to 0}\frac{(1 + h)^4 - 2 - (1^4 - 2)}{h}\\ + &= \lim_{h \to 0}\frac{h^4 + 4h^3 + 6h^2 + 4h + 1 - 1}{h}\\ + &= \lim_{h \to 0}(h^3 + 4h^2 + 6h + 4)\\ +\end{align*} + +An equation of the tangent line to $g$ at $(1, -1)$: +\[y - g(1) = g'(1)(x - 1) \iff y = 4x - 5\] + +\noindent Determine whether $f'(0)$ exists. +\[f(x) = \begin{cases} + x\sin\frac{1}{x}\text{ if } x \neq 0 \\ + 0\text{ if } x = 0 + \end{cases}\tag{53}\] +\begin{align*} +f'(0) &= \lim_{h \to 0}\frac{f(0 + h) - f(0)}{h}\\ + &= \lim_{h \to 0}\frac{h\sin\frac{1}{h}}{h}\\ + &= \lim_{h \to 0}\sin\frac{1}{h}\tag{does not exist}\\ +\end{align*} + +\[f(x) = \begin{cases} + x^2\sin\frac{1}{x}\text{ if } x \neq 0 \\ + 0\text{ if } x = 0 + \end{cases}\tag{54}\] +\begin{align*} +f'(0) &= \lim_{h \to 0}\frac{f(0 + h) - f(0)}{h}\\ + &= \lim_{h \to 0}\frac{h^2\sin\frac{1}{h}}{h}\\ + &= \lim_{h \to 0}h\sin\frac{1}{h} +\end{align*} + +Since $\forall h \neq 0, -|h| \leq h\sin\frac{1}{h} \leq |h|$ +and $\lim_{h \to 0}(-|h|) = \lim_{h \to 0}|h| = 0$, +according to the Squeeze Theorem, $f'(0) = 0$. + +\section{Differentiation} +\setcounter{subsection}{3} +\subsection{The chain rule} +Find the derivative of the function. + +\[y = \cos\sqrt{\sin(\tan{\pi x})}\tag{45}\] +\begin{align*} +\dot{y} &= -\sqrt{\sin(\tan{\pi x})}' \cdot \sin\sqrt{\sin(\tan{\pi x})}\\ + &= \frac{\sin'(\tan{\pi x}) \cdot \sin\sqrt{\sin(\tan{\pi x})}} + {2\sqrt{\sin(\tan{\pi x})}}\\ + &= \frac{\tan'{\pi x} \cdot \cos(\tan{\pi x}) + \cdot \sin\sqrt{\sin(\tan{\pi x})}} + {2\sqrt{\sin(\tan{\pi x})}}\\ + &= \frac{\pi\sec^2{\pi x} \cdot \cos(\tan{\pi x}) + \cdot \sin\sqrt{\sin(\tan{\pi x})}} + {2\sqrt{\sin(\tan{\pi x})}} +\end{align*} + +\[y = [x + (x + \sin^2{x})^3]^4 \tag{46}\] +\begin{align*} +\dot{y} &= 4[x + (x + \sin^2{x})^3]' [x + (x + \sin^2{x})^3]^3\\ + &= 4[1 + 3(x + \sin^2{x})' (x + \sin^2{x})^2] + [x + (x + \sin^2{x})^3]^3\\ + &= 4[1 + 3(1 + \sin{2x})(x + \sin^2{x})^2] + [x + (x + \sin^2{x})^3]^3 +\end{align*} + +\setcounter{subsection}{6} +\subsection{Applications in Sciences} +\textbf{9. }A rock is thrown vertically upward from the surface of Mars, its +height after $t$ seconds is $h = 15t - 1.86t^2$. +\[\frac{\ud h}{\ud t}(2) += (t \mapsto 15 - 3.72t)(2) += 7.56\text{ (m/s)}\tag{a}\] +\[h = 25 \iff 15t - 1.86t^2 = 25 + \iff t = \frac{375 \mp 25\sqrt{39}}{93}\tag{b}\] + +So at $t \approx 2.35$ s or $t \approx 5.71$ the Rock's height is 25 m. Its +velocity at this point is +\[v = (t \mapsto 15 - 3.72t)\left(\frac{375 \mp 25\sqrt{39}}{93}\right) + = \pm 6.24\text{ (m/s)}\] + +\pagebreak\noindent\textbf{10. }A particle moves with position function +\[s = t^4 - 4t^3 - 20t^2 + 20t \qquad t \geq 0\] +\[v = 20 \iff \dot{s} = 20 \iff 4t^3 - 12t^2 - 40t + 20 = 20 \tag{a}\] + +Since $t$ is nonnegative, the particle has a velocity of 20 m/s at $t = 0$ and +$t = 5$ s. +\[a = 0 \iff \dot{v} = 0 \iff 12t^2 - 24t - 40 = 0 \tag{b}\] + +Since $t$ is nonnegative, the acceleration is 0 at $t = \sqrt\frac{13}{3} - 1$ +s. This is when the instantaneous speed of the particle ($|v|$) reaches its +critical value. + +\noindent\textbf{21. }The force $F$ acting on a body with velocity $v$ and mass +$m = m_0 \Big/ \sqrt{1 - \frac{v^2}{c^2}}$ (where $m_0$ is the mass of the particle +at rest and $c$ is the speed of light) is the rate of change of momentum: +\begin{align*} +F &= \frac{\ud(mv)}{\ud t}\\ + &= \frac{\ud}{\ud t}\left(\frac{m_0v}{\sqrt{1 - \frac{v^2}{c^2}}}\right)\\ + &= m_0\frac{\ud v}{\ud t}\cdot + \frac{\ud}{\ud v}\left(\frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}\right)\\ + &= m_0a\frac{\ud}{\ud v}\left(\frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}\right)\\ + &= m_0a\frac{\sqrt{1 - \frac{v^2}{c^2}} + - v\frac{\ud\sqrt{1 - v^2/c^2}}{\ud v}} + {1 - \frac{v^2}{c^2}}\\ + &= m_0a\frac{\sqrt{1 - \frac{v^2}{c^2}} + - \frac{v}{2}\cdot\frac{-2v}{c^2\sqrt{1 - v^2/c^2}}} + {1 - \frac{v^2}{c^2}}\\ + &= m_0a\frac{\sqrt{1 - \frac{v^2}{c^2}} + \frac{v^2}{c^2\sqrt{1 - v^2/c^2}}} + {1 - \frac{v^2}{c^2}}\\ + &= m_0a\frac{1 - \frac{v^2}{c^2} + \frac{v^2}{c^2}} + {\left(1 - \frac{v^2}{c^2}\right)^\frac{3}{2}}\\ + &= \frac{m_0a}{\left(1 - \frac{v^2}{c^2}\right)^\frac{3}{2}} +\end{align*} + +\noindent\textbf{30. }The frequency of vibrations a vibrating violin string is +given by \[f = \frac{1}{2L}\sqrt{\frac{T}{\rho}} \qquad T \geq 0, \rho > 0\] +\begin{enumerate}[(a)] + \item The rate of change of the frequency with respect to + \begin{enumerate}[(i)] + \item The length: + $\frac{\ud f}{\ud L} = \frac{-1}{L^2}\sqrt{\frac{T}{\rho}}$. + \item The tension: $\frac{\ud f}{\ud T} = \frac{1}{4L\sqrt{T\rho}}$. + \item The density: + $\frac{\ud f}{\ud L} = \frac{-1}{L2}\sqrt{\frac{T}{\rho^3}}$. + \end{enumerate} + \item The pitch of a note gets higher when the string is shorter and lower + when the tension or density is increased. +\end{enumerate} + +\noindent\textbf{35. }Applying the gas law +\[PV = nRT \iff T = \frac{PV}{nR}\] + +The rate of change of temperature can be easily calculated via differentiation: +\begin{align*} + \frac{\ud T}{\ud t} +&= \frac{\ud}{\ud t}\left(\frac{PV}{nR}\right)\\ +&= \frac{1}{nR}\left(P\frac{\ud V}{\ud t} + V\frac{\ud P}{\ud t}\right)\\ +&= \frac{8.0 \cdot 0.15 + 10 \cdot 0.10}{10 \cdot 0.0821}\\ +&= \frac{1.2 + 1}{10 \cdot 0.0821}\\ +&= \frac{2}{10 \cdot 0.0821}\\ +&= 2\text{ (K/s)} +\end{align*} + +(In the calculation above, significant figures are taken into consideration.) + +\pagebreak\subsection{Exponential Growth and Decay} +\textbf{4. }Let $P(t)$ be the bacteria count after $t$ hours. As the bacteria +culture grows with constant relative growth rate, +\[\frac{\ud P}{\ud t} = kP \Longrightarrow P(t) = P(0)e^{kt}\] + +Since $P(2) = 400$ and $P(6) = 25600$, +\begin{align*} + \begin{cases} + P(0)e^{2k} = 400\\ + P(0)e^{6k} = 25600 + \end{cases} + &\iff + \begin{cases} + P(0)e^{2k} = 400\\ + e^{4k} = 64 + \end{cases}\\ + &\iff + \begin{cases} + P(0)e^{2k} = 400\\ + e^{2k} = 8 + \end{cases}\\ + &\iff + \begin{cases} + P(0) = 50\\ + k = \frac{\ln 8}{2} \approx 104\% + \end{cases} +\end{align*} + +Thus (a) the relative growth rate is 104\%, (b) the initial size of the culture +is 50 and (c) the number of bacteria after $t$ hours is $50\sqrt{8^t}$. + +The number of cells after 4.5 hours: +\[P(4.5) = 50\sqrt{8^{4.5}} \approx 5382\tag{d}\] + +The rate of growth after 4.5 hours: +\begin{align*} + \frac{\ud P}{\ud t}(4.5) +&= 50\frac{\ud \sqrt{8}^t}{\ud t}(4.5)\\ +&= 50\left(t \mapsto \sqrt{8^t}\ln\sqrt{8}\right)(4.5)\\ +&= 25\cdot8^{2.25}\ln 8\\ +&\approx 5596\text{ (bacteria per minute)}\tag{e} +\end{align*} + +The population reach 50000 when +\[50\sqrt{8^t} = 50000 \iff 8^t = 10^6 \iff t = \log_2{100} +\approx 6.64\text{ (days)}\tag{f}\] + +\noindent\textbf{8. }Given 50 mg of $^{90}$Sr which has a half-life of 28 days. +\begin{enumerate}[(a)] + \item Formula of the mass remaining after $t$ days: $m(t) = 50\cdot2^{-t/28}$. + \item The mass remaining after 40 days: + $m(40) = 50\cdot\frac{1}{2}^{10/7} \approx 19\text{ (mg)}$. + \item To decay to a mass of 2 mg, it takes + $-28\log_2\frac{2}{50} \approx 130\text{ (days)}$. + \item The graph of the mass function:\\ + \begin{tikzpicture} + \begin{axis}[ + axis x line=middle, axis y line=middle, + xlabel={$t$}, ylabel={$m$}, + xlabel style={at=(current axis.right of origin), anchor=west}, + ylabel style={at=(current axis.above origin), anchor=south}, + enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}] + \addplot[domain=0:130,magenta]{50*2^(-x/28)}; + \end{axis} + \end{tikzpicture} +\end{enumerate} + +\noindent\textbf{16. }Let $T(t)$ be the temperature of the coffee after $t$ +minutes. The surrounding temperature is \ang{20}C, so Newton's Law of Cooling +states that \[\frac{\ud T}{\ud t} = k(T - 20)\] + +If we let $y = T - 20$, then $y(0) = T(0) - 20 = 95 - 20 = 75$, so $y$ +satisfies \[\frac{\ud y}{\ud t} = ky \iff y(t) = 75e^{kt}\] + +When the temperature of the coffee is \ang{70}C, its cooling rate is \ang{1}C +per minute, i.e. +\begin{align*} + \begin{cases} + y(t) + 20 = 70\\ + ky(t) = -1 + \end{cases} + &\iff\begin{cases} + y(t) = 50\\ + k = \frac{-1}{50} + \end{cases}\\ + \Longrightarrow 75e^{-t/50} = 50 + &\iff t = 50\ln1.5 \approx 20\text{ (minutes)} +\end{align*} + +\subsection{Related rates} +\textbf{10. }A particle is moving along a hyperbola $xy = 8$ +\begin{align*} + \Longrightarrow \frac{\ud (xy)}{\ud t} = \frac{\ud 8}{\ud t} + &\iff y\frac{\ud x}{\ud t} + x\frac{\ud y}{\ud t} = 0\\ + &\iff 2 \cdot \frac{\ud x}{\ud t} + 4 \cdot 3 = 0\\ + &\iff \frac{\ud x}{\ud t} = -6\text{ (cm/s)} +\end{align*} + +\noindent\textbf{12. }Let $D(t)$ (cm) be the diameter of the ball at minute +$t$, its surface area is $A(D) = \pi D^2$ (cm$^2$). +\[\frac{\ud A}{\ud t} = 1 \iff \frac{\ud A}{\ud D} \cdot \frac{\ud D}{\ud t} = 1 + \iff 2\pi D \frac{\ud D}{\ud t} = 1 \iff \frac{\ud D}{\ud t} + = \frac{1}{2\pi D}\] + +Thus the decreasing rate of the diameter when it is 10 cm: +\[\frac{\ud D}{\ud t}(10) = \frac{1}{20\pi}\text{ (cm/s)}\] + +\noindent\textbf{14. } + +\begin{tikzpicture} + \begin{axis}[nodes near coords align=right, xlabel={W -- E}, ylabel={S -- N}] + \node[label=A, shape=circle, fill, inner sep=1.5pt] at (0,0) {}; + \addplot[->] plot coordinates {(0,0) (35,0)}; + \node[label={180:B}, shape=circle, fill, inner sep=1.5pt] at (150,0) {}; + \addplot[->] plot coordinates {(150,0) (150,25)}; + \end{axis} +\end{tikzpicture} +\[\begin{cases} + \Delta x(t) = x_B(t) - x_A(t)\\ + \Delta y(t) = y_B(t) - y_A(t) + \end{cases} +\iff\begin{cases} + \Delta x(t) = 150 - 35t\\ + \Delta y(t) = 25t + \end{cases}\] +\[\Longrightarrow\Delta s(t) = \sqrt{1850t^2 - 10500t + 22500} + \Longrightarrow\frac{\ud s}{\ud t} = \frac{1850t - 5250} + {\sqrt{1850t^2 - 10500t + 22500}}\] +\[\Longrightarrow\frac{\ud s}{\ud t}(4) + = \frac{1850\cdot4 - 5250} + {\sqrt{1850\cdot16 - 10500\cdot4 + 22500}} + = \frac{2150}{\sqrt{10100}} + = \frac{215\sqrt{101}}{101} + \approx 21\text{ (km/h)}\] + +\noindent\textbf{27. }Let $h(t)$ (ft) be the height of the cone at minute $t$. +Volume of the cone is +\begin{align*} + V(h) = \frac{\pi h^2}{12} + &\Longrightarrow\frac{\ud V}{\ud t} = \frac{\pi h}{6}\cdot\frac{\ud h}{\ud t} + = 30 + \iff\frac{\ud h}{\ud t} = \frac{180}{\pi h}\\ + &\Longrightarrow\frac{\ud h}{\ud t}(10) = \frac{180}{10\pi} = \frac{18}{\pi} + \approx 5.7\text{ (ft/s)} +\end{align*} + +\section{Applications of derivative} +\subsection{Max and Min} +Find the absolute min and max values of $f$. + +\[f(x) = 3x^4 - 4x^3 - 12x^2 + 1, \qquad [-2, 3]\tag{51}\] + +Since $f'(x) = 12x^3 - 12x^2 - 24x$, we have $f'(x) = 0$ when +$x \in \{-1, 0, 2\}$. The values of $f$ at these critical numbers are +\begin{align*} + f(-1) &= 3 + 4 - 12 + 1 = -4\\ + f(0) &= 1\\ + f(2) &= 48 - 32 - 48 + 1 = -31 +\end{align*} + +The values of $f$ at endpoints are +\begin{align*} + f(-2) &= 48 + 32 - 48 + 1 = 33\\ + f(3) &= 243 - 108 - 108 + 1 = 28 +\end{align*} + +Comparing these five numbers and using the Closed Interval Method, we see that +the absolute minimum value is $f(2) = -31$ and the absolute maximum value is +$f(-2) = 33$. + +\[f(t) = t\sqrt{4 - t^2}, \qquad [-1, 2]\tag{55}\] + +Since $f'(t) = \frac{4 - 2t^2}{\sqrt{4 - t^2}}$, with $t \in [-1, 2]$, we have +$f'(t) = 0$ when $t = \sqrt{2}$. The value of $f$ at this critical number is +$f(\sqrt{2}) = 2$. The value of $f$ at endpoints are $f(-1) = -\sqrt{3}$ and +$f(2) = 0$. Comparing these 3 numbers and using the Closed Interval Method, we +see that the absolute minimum value is $f(-1) = -\sqrt{3}$ and the absolute +maximum value is $f(\sqrt{2}) = 2$. + +\[f(x) = xe^{-x^2/8}, \qquad [-1, 4]\tag{59}\] + +With $x \in [-1, 4]$, we have $f'(x) = \left(1-\frac{x^2}{4}\right)e^{-x^2/8}$ +when $x = 2$. Comparing values of $f$ at this critical number and endpoints, +the minimum value is $f(-1) = -e^{-1/8}$ and +the maximum value is $f(2) = 2e^{-1/2}$. + +\subsection{The Mean Theorem} +\textbf{26. }Let $h$ be the function that $h(x) = f(x) - g(x)$. Since both $f$ +and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$, $h$ inherits +the same properties. By applying the Mean Value Theorem to $h$ on the interval +$[a, b]$, we get a number $c \in (a, b)$ such that +\begin{align*} + &h(b) - h(a) = (b - a)h'(c)\\ +\iff&f(b) - g(b) - f(a) + g(a) = (b - a)(f'(c) - g'(c))\\ +\iff&f(b) - g(b) = (b - a)(f'(c) - g'(c)) +\end{align*} +$b - a > 0$ and $f'(c) - g'(c) < 0$ so $f(b) - g(b) < 0$ or $f(b) < g(b)$. + +\begin{align*} +x > 0 &\iff x + 1 > 1\\ + &\iff \sqrt{x + 1} > 1\\ + &\iff \sqrt{x + 1} - 1 > 0\\ + &\Longrightarrow \left(\sqrt{x + 1} - 1\right)^2 > 0\\ + &\iff x + 1 - 2\sqrt{x + 1} + 1 > 0\\ + &\iff x + 2 > 2\sqrt{x + 1}\\ + &\iff \sqrt{1 + x} < 1 + \frac{1}{2}x\tag{27} +\end{align*} + +\noindent\textbf{33. }Prove the identity +\[\arcsin\frac{x - 1}{x + 1} = 2\arctan\sqrt{x} - \frac{\pi}{2}\] + +Let $\frac{-\pi}{2} \leq y = \arcsin\frac{x - 1}{x + 1} \leq \frac{\pi}{2}$ and +$z = \arctan\sqrt{x}$, then +\begin{align*} +\begin{cases} + \sin{y} = \frac{x - 1}{x + 1}\\ + \tan{z} = \sqrt{x} +\end{cases} +\Longrightarrow& +\begin{cases} + \frac{\ud\sin{y}}{\ud x} = \frac{\ud}{\ud x}\left(\frac{x - 1}{x + 1}\right)\\ + \frac{\ud\tan{z}}{\ud x} = \frac{\ud\sqrt{x}}{\ud x} +\end{cases}\\ +\iff& +\begin{cases} + \cos{y}\cdot\frac{\ud y}{\ud x} = \frac{2}{(x + 1)^2}\\ + \left(\tan^2{z} + 1\right)\frac{\ud z}{\ud x} = \frac{1}{2\sqrt{x}} +\end{cases}\\ +\iff& +\begin{cases} + \sqrt{1 - \sin^2{y}}\cdot\frac{\ud y}{\ud x} = \frac{2}{(x + 1)^2}\\ + \left(\sqrt{x}^2 + 1\right)\frac{\ud z}{\ud x} = \frac{1}{2\sqrt{x}} +\end{cases}\\ +\iff& +\begin{cases} + \sqrt{\frac{4x}{(x + 1)^2}}\cdot\frac{\ud y}{\ud x} = \frac{2}{(x + 1)^2}\\ + (x + 1)\frac{\ud z}{\ud x} = \frac{1}{2\sqrt{x}} +\end{cases}\\ +\iff& +\begin{cases} + \frac{\ud y}{\ud x} = \frac{1}{|x + 1|\sqrt{x}}\\ + \frac{\ud z}{\ud x} = \frac{1}{2(x + 1)\sqrt{x}} +\end{cases}\\ +\end{align*} + +For all $x \geq 0$ or $x + 1 \geq 1 > 0$ +\begin{align*} + \frac{\ud}{\ud x}\left(2\arctan\sqrt{x} - \arcsin\frac{x - 1}{x + 1}\right) +&= 2\frac{\ud z}{\ud x} - \frac{\ud y}{\ud x}\\ +&= \frac{2}{2(x + 1)\sqrt{x}} - \frac{1}{(x + 1)\sqrt{x}}\\ +&= 0 +\end{align*} + +Thus the function +$x \mapsto 2\arctan\sqrt{x} - \arcsin\frac{x - 1}{x + 1}$ is constant on its +domain $[0, \infty)$. Consequently, in $[0, \infty)$ +\begin{align*} + 2\arctan\sqrt{x} - \arcsin\frac{x - 1}{x + 1} +&= \left(x \mapsto 2\arctan\sqrt{x} - \arcsin\frac{x - 1}{x + 1}\right)(0)\\ +&= 2\arctan\sqrt{0} - \arcsin\frac{-1}{1}\\ +&= 0 - \frac{-\pi}{2}\\ +&= \frac{\pi}{2}\\ +\iff \arcsin\frac{x - 1}{x + 1} &= 2\arctan\sqrt{x} - \frac{\pi}{2} +\end{align*} + +\subsection{Shape of a graph} +\textbf{75. }Given two funtions $f$ and $g$ which are positive and concave +upward on $I$, i.e. for all $x$ in $I$ +\[\begin{cases} + f(x) > 0\\ + f''(x) > 0\\ + g(x) > 0\\ + g''(x) > 0 + \end{cases}\] + +Second derivative of the product function $fg$: +\[(f(x)g(x))'' = (f'(x)g(x) + f(x)g'(x))' + = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x)\] + +If $f$ and $g$ are both increasing or decreasing, then $f'(x)g'(x) > 0$, which +means $\forall x \in I, (f(x)g(x))'' > 0$, or $fg$ is concave upward on $I$. +Otherwise, $f$ is increasing and $g$ is decreasing for instance, $fg$ may be +either concave upward, concave downward or linear: + +\begin{center} + \begin{tikzpicture}[domain=0:8] + \begin{axis}[ + axis x line = middle, axis y line = middle, + xlabel={$x$}, ylabel={$y$}, ymin=-8, + xlabel style={at=(current axis.right of origin), anchor=west}, + ylabel style={at=(current axis.above origin), anchor=south}, + enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}, + legend pos=south east] + \addplot[color=blue, samples=100, smooth]{-1/x}; + \addplot[color=cyan, samples=100, smooth]{-x^(3/2)}; + \addplot[color=red, samples=100, smooth]{sqrt(x)}; + \legend{$f(x) = \frac{-1}{x}$, $g(x) = -x\sqrt{x}$, + $f(x)g(x) = \sqrt{x}$} + \end{axis} + \end{tikzpicture} + + \begin{tikzpicture}[domain=0:4] + \begin{axis}[ + axis x line = middle, axis y line = middle, + xlabel={$x$}, ylabel={$y$}, ymin=-8, + xlabel style={at=(current axis.right of origin), anchor=west}, + ylabel style={at=(current axis.above origin), anchor=south}, + enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}] + \addplot[color=blue, samples=100, smooth]{-1/x}; + \addplot[color=cyan, samples=100, smooth]{-x^3}; + \addplot[color=red, samples=100, smooth]{x^2}; + \legend{$f(x) = \frac{-1}{x}$, $g(x) = -x^3$, $f(x)g(x) = x^2$} + \end{axis} + \end{tikzpicture} + + \begin{tikzpicture}[domain=0:4] + \begin{axis}[ + axis x line = middle, axis y line = middle, + xlabel={$x$}, ylabel={$y$}, ymin=-8, + xlabel style={at=(current axis.right of origin), anchor=west}, + ylabel style={at=(current axis.above origin), anchor=south}, + enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}, + legend pos=south east] + \addplot[color=blue, samples=100, smooth]{-1/x}; + \addplot[color=cyan, samples=100, smooth]{-x^2}; + \addplot[color=red, samples=100, smooth]{x}; + \legend{$f(x) = \frac{-1}{x}$, $g(x) = -x^2$, $f(x)g(x) = x$} + \end{axis} + \end{tikzpicture} +\end{center} + +\pagebreak\noindent\textbf{76. }In order for $h = f(g(x))$ to be concave +upward on $\mathbb R$ +\begin{align*} + h'' > 0 &\iff (f \circ g)'' > 0\\ + &\iff ((f' \circ g) \cdot g')' > 0\\ + &\iff (f' \circ g)' \cdot g' + (f' \circ g) \cdot g'' > 0\\ + &\iff (f'' \circ g) \cdot (g')^2 + (f' \circ g) \cdot g'' > 0 +\end{align*} + +Because $f$ and $g$ are given to be concave upward on $\mathbb R$, i.e. +$f'' > 0$ and $g'' > 0$, and $\forall x \in \mathbb R, g^2(x) \geq 0$, so if +$f' > 0$ or $f$ is an increasing function, $h$ will be concave upward. + +\noindent\textbf{77. }Show that $\tan{x} > x$ for $0 < x < \frac{\pi}{2}$. + +Let $f$ be the function that $f(x) = \tan{x} - x$. On $(0, \frac{\pi}{2})$, +$\sin{x}\cos{x} \neq 0$ thus $\tan{x}$ exists and is nonzero. Therefore, +$f'(x) = \tan^2(x) > 0$ or $f$ is increasing on $[0, \frac{\pi}{2}]$, which +means for all x in $(0, \frac{\pi}{2})$, +\[f(x) > f(0) \iff \tan{x} - x > \tan0 - 0 \iff \tan{x} > x\] + +\noindent\textbf{78. } Use mathematical induction to prove that for all +positive integer $n$, +\[\forall x \geq 0, \qquad e^x \geq 1 + \sum_{i=1}^n\frac{x^i}{i!}\tag{$*$}\] + +Let $f$ be the function of domain $[0, \infty)$ that $f(x) = e^x - x$, then +for all $x \geq 0$, $f'(x) = e^x - 1 > f'(0) = 0$ (since it is obvious that $f'$ +is an increasing function). Hence +$\forall x \geq 0, e^x - x > e^0 - 0 = 1 \iff \forall x \geq 0, e^x > 1 + x$, +i.e. $(*)$ is true for $n = 1$. + +Suppose that $(*)$ is also true for $n = k$ ($k \in \mathbb N^*$). For all +nonnegative $x$, +\[e^x \geq 1 + \sum_{i=1}^k\frac{x^i}{i!} + \iff e^x - 1 - \sum_{i=1}^k\frac{x^i}{i!} \geq 0\] + +Let $g$ be the function of domain $[0, \infty)$ that +$g(x) = e^x - \sum_{i=1}^{k+1}\frac{x^{i}}{i!}$, then for all positive $x$ +\[g'(x) = e^x - \sum_{i=1}^{k+1}\frac{ix^{i-1}}{i!} + = e^x - 1 - \sum_{i=2}^{k+1}\frac{x^{i-1}}{(i - 1)!} + = e^x - 1 - \sum_{i=1}^{k}\frac{x^{i}}{i!} \geq 0\] + +This means $g$ in a non-decreasing function on $[0, \infty)$ +\[e^x - \sum_{i=1}^{k+1}\frac{x^{i}}{i!} + \geq e^0 - \sum_{i=1}^{k+1}\frac{0^{i}}{i!} = 1\] + +This expression shows that $(*)$ is true for $n = k + 1$. Therefore, by +mathematical induction, it is true for all positive integers $n$. + +\subsection{Rule of the Hospital} +Find the limit + +\noindent\textbf{54. }Since $\lim_{x \to 0^+}\ln x = -\infty$ +and $\lim_{x \to 0^+}\frac{1}{\sqrt x} = \infty$ +\[\lim_{x \to 0^+}x^{\sqrt x} += \lim_{x \to 0^+}e^{\sqrt{x}\cdot\ln{x}} += e^{\lim_{x \to 0^+}\frac{\ln x}{1/\sqrt x}} += e^{\lim_{x \to 0^+}\frac{-2x\sqrt x}{x}} += e^{-2\lim_{x \to 0^+}\sqrt x} += e^0 += 1\] + +\noindent\textbf{60. }Since +$\lim_{x \to \infty}\ln{2}\ln x = \lim_{x \to \infty}(1 + \ln x) = \infty$ +\[\lim_{x \to \infty}x^\frac{\ln 2}{1 + \ln x} += e^{\lim_{x \to \infty}\frac{\ln{2}\ln x}{1 + \ln x}} += e^{\lim_{x \to \infty}\ln{2}} += e^{\ln 2} += 2\] + +\setcounter{subsection}{6} +\subsection{Optimization Problems} +\textbf{44. }Given $E(v) = \frac{aLv^3}{v - u}$ +\[\frac{\ud E}{\ud v} += aL\frac{3v^2(v - u) - v^3}{(v - u)^2} += aL\frac{2v^3 - 3uv^2}{(v - u)^2}\] + +Since $v > u > 0$, $E$ has only one absolute extreme value, at the only +critical number $v = 1.5u$. Applying the First Derivative Test for Absolute +Extreme Values, $v = 1.5u$ is shown to be the value of $v$ that minimizes $E$. + +\noindent\textbf{45. }Given +$S = 6sh + \frac{3}{2}s^2(\sqrt{3}\cdot\csc\theta - \cot\theta)$. +\[\frac{\ud S}{\ud\theta} += \frac{3}{2}s^2\frac{\ud}{\ud\theta}(\sqrt{3}\cdot\csc\theta - \cot\theta) += \frac{3s^2(1 - {\sqrt{3}\cdot\cos\theta})}{2\sin^2\theta}\] + +We have $\frac{\ud S}{\ud\theta} = 0$ when $\theta = \arccos\frac{\sqrt 3}{3}$. +Applying the First Derivative Test for Absolute Extreme Values, this value of +$\theta$ is shown to minimize $S$ to $6sh + \frac{3s^2}{\sqrt 2}$. + +\noindent\textbf{76. }Using Poiseuille's Law, we have the total resistance of +the blood along the path ABC is +\begin{multline*} +R = R_{AB} + R_{BC} + = C\frac{a - b\cot\theta}{r_1^4} + C\frac{b}{r_2^4 \sin\theta} + = C\left(\frac{a - b\cot\theta}{r_1^4} + \frac{b\csc\theta}{r_2^4}\right)\\ +\Longrightarrow\frac{\ud R}{\ud\theta} += \frac{Cb}{r_1^4 \sin^2\theta} - \frac{Cb\cos\theta}{r_2^4\sin^2\theta} += \frac{Cb}{sin^2\theta}\left(\frac{1}{r_1^4} - \frac{\cos\theta}{r_2^4}\right) +\end{multline*} + +We have $\frac{\ud R}{\ud\theta} = 0$ when $\cos\theta = (r_2/r_1)^4$. At this +angle, the resistance is minimized (can be shown using the First Derivative +Test for Absolute Extreme Values, but like in the two previous exercises, I'm +too lazy to evaluate it). When $\frac{r_2}{r_1} = \frac{2}{3}$, the optimal +branching angle is $\theta \approx \ang{79}$. + +\section{Integral} +\subsection{Areas} +\textbf{4. }Estimate the area under the graph of $f(x) = \sqrt{x}$ from $x = 0$ +to $x = 4$. +\[\lim_{n \to \infty}R_n += \lim_{n \to \infty}\frac{4 - 0}{n}\sum_{i = 1}^n\sqrt\frac{4i}{n} += \lim_{n \to \infty}\frac{8}{n}\sum_{i = 1}^n\sqrt\frac{i}{n}\] +\[\lim_{n \to \infty}L_n += \lim_{n \to \infty}\frac{4 - 0}{n}\sum_{i = 0}^{n - 1}\sqrt\frac{4i}{n} += \lim_{n \to \infty}\frac{8}{n}\sum_{i = 0}^{n - 1}\sqrt\frac{i}{n}\] + +For estimation, consider $n \to 4$: +\[\lim_{n \to 4}R_n += \frac{8}{4}\sum_{i = 1}^4\sqrt\frac{i}{4} += \sum_{i = 1}^4\sqrt i += 1 + \sqrt 2 + \sqrt 3 + 2 +\approx 6.1463\] +\[\lim_{n \to 4}L_n += \frac{8}{4}\sum_{i = 0}^3\sqrt\frac{i}{4} += \sum_{i = 0}^3\sqrt i += 0 + 1 + \sqrt 2 + \sqrt 3 +\approx 4.1463\] + +\noindent\textbf{5. }Estimate the area under the graph of $f(x) = 1 + x^2$ from +$x = -1$ to $x = 2$. +\begin{align*} + \lim_{n \to \infty}R_n +=&\lim_{n \to \infty}\frac{2 + 1}{n}\sum_{i = 1}^n + f\left(-1 + i\frac{2 + 1}{n}\right)\\ +=&\lim_{n \to \infty}\frac{3}{n}\sum_{i = 1}^n + \left(1 + \left(\frac{3i}{n} - 1\right)^2\right) +\end{align*} + +Similarly, +\begin{align*} + \lim_{n \to \infty}L_n +=&\lim_{n \to \infty}\frac{3}{n}\sum_{i = 0}^{n - 1} + \left(1 + \left(\frac{3i}{n} - 1\right)^2\right)\\ + \lim_{n \to \infty}M_n +=&\lim_{n \to \infty}\frac{3}{n}\sum_{i = 1}^n + \left(1 + \left(\frac{3i - 3/2}{n} - 1\right)^2\right) +\end{align*} + +For $n \to 3$, +\begin{align*} + \lim_{n \to 3}R_n +=&\sum_{i = 1}^3\left(1 + (i - 1)^2\right) += 1 + 2 + 5 += 8\\ + \lim_{n \to 3}M_n +=&\sum_{i = 1}^3\left(1 + \left(i - \frac{3}{2}\right)^2\right) += \frac{5}{4} + \frac{5}{4} + \frac{13}{4} += 5.75\\ + \lim_{n \to 3}L_n +=&\sum_{i = 0}^2\left(1 + (i - 1)^2\right) += 2 + 1 + 2 += 5 +\end{align*} + +For $n \to 6$, +\begin{align*} + \lim_{n \to 6}R_n +=&\frac{1}{2}\sum_{i = 1}^6\left(1 + \left(\frac{i}{2} - 1\right)^2\right) += \frac{1}{2}\left(\frac{5}{4} + 1 + \frac{5}{4} + 2 + \frac{13}{4} + 5\right) += 6.875\\ + \lim_{n \to 6}M_n +=&\frac{1}{2}\sum_{i = 1}^6\left(1 + \left(\frac{2i - 5}{4}\right)^2\right) += \frac{1}{2}\left(\frac{25}{16} + \frac{17}{16} + \frac{17}{16} + + \frac{25}{16} + \frac{41}{16} + \frac{65}{16}\right) += 5.9375\\ + \lim_{n \to 6}L_n +=&\frac{1}{2}\sum_{i = 0}^5\left(1 + \left(\frac{i}{2} - 1\right)^2\right) += \frac{1}{2}\left(2 + \frac{5}{4} + 1 + \frac{5}{4} + 2 + \frac{13}{4}\right) += 5.375 +\end{align*} + +\noindent\textbf{16. }The height (in feet) above the earth's surface of the +\textit{Endeavour}, 62 seconds after liftoff, can be estimated with the assist +of Python (which, coincidentally, has been utilized by NASA recently): +\begin{verbatim} +>>> time = 0, 10, 15, 20, 32, 59, 62, 125 +>>> velocity = 0, 185, 319, 447, 742, 1325, 1445, 4151 +>>> sum(map(int.__mul__, velocity, +... map(int.__sub__, time[1:], time[:-1]))) +122928 +\end{verbatim} + +\subsection{The Definite Integral} +Evaluate the integral. +\begin{align*} + \int_2^5(4 - 2x)\ud x &= \left.4x\right]_2^5 - \left.x^2\right]_2^5 += 12 - 21 = -9\tag{21}\\ + \int_0^2(2x - x^3)\ud x &= \left.x^3\right]_0^2 - \left.\frac{x^4}{4}\right]_0^2 += 9 - 16 = -7\tag{24} +\end{align*} + +\noindent\textbf{33. }Evaluate integral by interpreting it in terms of areas. +\begin{align*} + \int_0^2 f(x)\ud x &= 4\tag{a}\\ + \int_0^5 f(x)\ud x &= 10\tag{b}\\ + \int_5^7 f(x)\ud x &= -3\tag{c}\\ + \int_0^9 f(x)\ud x &= \int_0^5 f(x)\ud x + \int_5^9 f(x)\ud x + = 10 - 8 = 2\tag{d} +\end{align*} + +\noindent\textbf{50. }Given $f(x) = \begin{cases} + 3\text{ for } x < 3\\ + x\text{ for } x \geq 3 + \end{cases}$ +\[\int_0^5 f(x)\ud x = \int_0^3 3\ud x + \int_3^5 x\ud x += \left.3x\right]_0^3 + \left.\frac{x^2}{2}\right]_3^5 += 9 + 8 = 17\] + +\subsection{The Fundamental Theorem of Calculus} +\textbf{3. }Let $g(x) = \int_0^x f(t)\ud t$. +\begin{enumerate}[(a)] +\item By interpreting the above integral in terms of areas, we get $g(0) = 0$, + $g(1) = 2$, $g(2) = 5$, $g(3) = 7$ and $g(6) = 3$. +\item $g$ is increasing on $(0, 3)$. +\item $g$ has a maximum value of 7 at $x = 3$. +\item \begin{tikzpicture} + \begin{axis}[ + axis x line = middle, axis y line = middle, + xlabel={$x$}, ylabel={$y$}, + xlabel style={at=(current axis.right of origin), anchor=west}, + ylabel style={at=(current axis.above origin), anchor=south}, + enlarge y limits={rel=0.1}, enlarge x limits={rel=0.1}] + \addplot[color=magenta, domain=0:1]{2}; + \addplot[color=blue, domain=0:1]{2*x}; + \addplot[color=magenta, domain=1:2]{2*x}; + \addplot[color=blue, domain=2:3]{-2*x^2 + 12*x -11}; + \addplot[color=magenta, domain=2:3]{-4*x + 12}; + \addplot[color=blue, domain=1:2]{x^2 + 1}; + \addplot[color=magenta, domain=3:5]{-x + 3}; + \addplot[color=blue, domain=3:5]{-x^2/2 + 3*x + 2.5}; + \addplot[color=magenta, domain=5:6]{-2}; + \addplot[color=blue, domain=5:6]{-2*x + 15}; + \addplot[color=magenta, domain=6:7]{2*x - 14}; + \addplot[color=blue, domain=6:7]{x^2 - 14*x + 51}; + \legend{$f(x)$, $g(x)$} + \end{axis} + \end{tikzpicture} +\end{enumerate} + +\noindent Find the derivative of the function. +\begin{align*} + \frac{\ud}{\ud x}\int_1^{\sqrt x}\frac{z^2}{z^4 + 1}\ud z +&= \frac{\ud}{\ud\sqrt x}\left(\int_1^{\sqrt x}\frac{z^2}{z^4 + 1}dz\right) + \frac{\ud\sqrt x}{\ud x}\\ +&= \frac{x}{x^2 + 1} \cdot \frac{1}{2\sqrt x}\\ +&= \frac{\sqrt x}{2x^2 + 2}\tag{14} +\end{align*} + +\begin{align*} + \frac{\ud}{\ud x}\int_0^{\tan x}\sqrt{t + \sqrt t}\ud t +&= \frac{\ud}{\ud\tan x}\left(\int_0^{\tan x}\sqrt{t + \sqrt t}\ud t\right) + \frac{\ud\tan x}{\ud x}\\ +&= \frac{\sqrt{\tan x + \sqrt{\tan x}}}{\cos^2 x}\tag{15} +\end{align*} + +\noindent\textbf{64. }Given the \textbf{error function} +\[\erf(x) = \frac{2}{\sqrt\pi}\int_0^x e^{-t^2}\ud t + \Longrightarrow \erf'(x) = \frac{2e^{-x^2}}{\sqrt\pi}\] + +\[\int_a^b e^{-t^2}\ud t = \frac{\sqrt\pi}{2}\int_a^b \erf'(t)\ud t + = \frac{\sqrt\pi}{2}[\erf(b) - \erf(a)]\tag{a}\] + +With $y = e^{x^2}\erf(x)$ +\begin{align*} +y' &= \left(e^{x^2}\right)'\erf(x) + e^{x^2}\erf'(x)\\ + &= 2xe^{x^2}\erf(x) + e^{x^2}\frac{2e^{-x^2}}{\sqrt\pi}\\ + &= 2xe^{x^2}\erf(x) + \frac{2}{\sqrt\pi}\\ + &= 2xy + \frac{2}{\sqrt\pi}\tag{b} +\end{align*} + +\subsection{Infinite Integral} +\textbf{56. }Let $y(x)$ be the vertical postion at a distance of $x$ miles from +the start of the trail, then $y'(x) = f(x)$. +Thus, $\int_3^5 f(x)\ud x = y(5) - y(3)$, which is the vertical displacement from +3 to 5 miles. + +\noindent\textbf{63. }Total mass of the rod: +\[\int_0^4 (9 + 2\sqrt x)\ud x += \left.9x\right]_0^4 + \left.\frac{2\sqrt{x^3}}{3}\right]_0^4 += 36 + \frac{16}{3} = 41\frac{1}{3}\text{ (kg)}\] + +\noindent\textbf{64. }Amount of water flowing from the tank during the first 10 +minutes: +\[\int_0^{10} (200 - 4t)\ud t += \left.200t\right]_0^{10} - \left.2t^2\right]_0^{10} += 2000 - 200 = 1800\text{ (l)}\] + +\subsection{The Substitution Rule} +\textbf{74. }Given $f(x) = \sin\sqrt[3]x$. + +Since $f(-x) = \sin\sqrt[3]{-x} = \sin-\sqrt[3]x = -\sin\sqrt[3]x = -f(x)$, +$f$ is an odd function. Hence $\int_{-2}^3 f(x)\ud x = \int_2^3 f(x)\ud x$. + +For $2 \leq x \leq 3$, $0 \leq \sqrt[3]2 \leq \sqrt[3]x \leq \sqrt[3]3 \pi$, +thus $\sin\sqrt[3]x \geq 0$ and $\int_2^3 f(x)\ud x \geq 0$. Futhermore, +$\sin\sqrt[3]x \leq 1$ so $\int_2^3 f(x)\ud x \leq \int_2^3 \ud x = 1$. + +\noindent Evaluate the integral. +\begin{align*} + \int_{-2}^2(x + 3)\sqrt{4 - x^2}\ud x +&= \int_{-2}^2 x\sqrt{4 - x^2}\ud x + 3\int_{-2}^2\sqrt{4 - x^2}\ud x\\ +&= 0 + 3 \cdot 2\pi\\ +&= 6\pi\tag{77} +\end{align*} + +\begin{align*} + \int_0^{24}\left(85 - 0.18\cos\frac{\pi t}{12}\right)\ud t +&= \left.85t\right]_0^{24} + - \frac{54}{25\pi}\int_0^{24}\frac{\pi t}{12}'\cos\frac{\pi t}{12}\ud t\\ +&= 2040 - \frac{54}{25\pi}\int_0^{2\pi}\cos x \ud x\\ +&= 2040 - \left.\frac{54\sin x}{25\pi}\right]_0^{2\pi}\\ +&= 2040\tag{80} +\end{align*} + +\begin{align*} + 400 + \int_0^3 450.268e^{1.12567t}\ud t +&= 400 + 400\int_0^3 1.12567e^{1.12567t}\ud t\\ +&= 400 + \left.400e^{1.12567t}\right]_0^3\\ +&= 400e^{1.12567 \cdot 3}\\ +&\approx 11713\tag{82} +\end{align*} + +\section{Applications of Integration} +\subsection{Areas Between Curves} +Evaluate the integral + +\begin{align*} + \int_{-1}^1\left|e^x - x^2 + 1\right|\ud x +&=\int_{-1}^1\left(e^x - x^2 + 1\right)\ud x\\ +&=\left[e^x - \frac{x^3}{3} + x\right]_{-1}^1\\ +&=e - \frac{1}{3} + 1 - \frac{1}{e} - \frac{1}{3} + 1\\ +&=e - \frac{1}{e} + \frac{4}{3}\tag{5} +\end{align*} + +\begin{align*} + \int_1^4\left|x^2 - 3x + 4\right|\ud x +&=\int_1^4\left(x^2 - 3x + 4\right)\ud x\\ +&=\left[\frac{x^3}{3} - \frac{3x^2}{2} + 4x\right]_1^4\\ +&=\frac{64 - 1}{3} - \frac{48 - 3}{2} + 16 - 4\\ +&=\frac{21}{2}\tag{7} +\end{align*} + +\begin{align*} + \int_1^2\left|\frac{1}{x} - \frac{1}{x^2}\right|\ud x +&=\int_1^2\left(\frac{1}{x} - \frac{1}{x^2}\right)\ud x\\ +&=\left[\frac{1}{3x^3} - \frac{1}{2x^2}\right]_1^2\\ +&=\frac{1}{24} - \frac{1}{8} - \frac{1}{3} + \frac{1}{2}\\ +&=\frac{1}{12}\tag{9} +\end{align*} + +\noindent\textbf{53. }Find the values of $c$ such that +\begin{align*} + \int_{-|c|}^{|c|}\left|x^2 - c^2 - c^2 + x^2\right|\ud x = 576 +&\iff \int_{-|c|}^{|c|}\left(c^2 - x^2\right)\ud x = 288\\ +&\iff \left[c^2 x - \frac{x^3}{3}\right]_{-|c|}^{|c|} = 288\\ +&\iff \frac{4\left|c^3\right|}{3} = 288\\ +&\iff |c| = 6\\ +&\iff c = \pm 6 +\end{align*} + +\noindent\textbf{54. }Find the area of the region enclosed by the line $y = mx$ +and the curve $y = \frac{x}{x^2 + 1}$. + +Those two curves enclose a region if and only if the following equations has +two unique solutions, i.e. $m \in (0, 1)$ +\[mx = \frac{x}{x^2 + 1} \iff mx^3 + (m - 1)x = 0 + \iff x \in \left\{0, \pm\frac{1 - m}{m}\right\}\] + +The area of the region would then be +\begin{align*} +A&=\int_{\frac{m-1}{m}}^{\frac{1-m}{m}}\left|mx - \frac{x}{x^2 + 1}\right|\ud x\\ + &=\int_{\frac{m-1}{m}}^0\left(\frac{x}{x^2 + 1} - mx\right)\ud x + + \int_0^{\frac{1-m}{m}}\left(mx - \frac{x}{x^2 + 1}\right)\ud x\\ + &=\int_{\frac{m-1}{m}}^0\left(\frac{x}{x^2 + 1} - mx\right)\ud x + + \int_0^{\frac{1-m}{m}}\left(mx - \frac{x}{x^2 + 1}\right)\ud x\\ + &=\int_{\frac{m-1}{m}}^0\frac{1}{x^2 + 1}\cdot\frac{\ud x^2}{\ud x}\ud x - + \left.\frac{mx^2}{2}\right]_{\frac{m-1}{m}}^0 + + \left.\frac{mx^2}{2}\right]_0^{\frac{1-m}{m}} - + \int_0^{\frac{1-m}{m}}\frac{1}{x^2 + 1}\cdot\frac{\ud x^2}{\ud x}\ud x\\ + &=\frac{(m - 1)^2}{m} + + 2\int_{\left(\frac{m-1}{m}\right)^2}^0\frac{1}{x + 1}\ud x\\ + &=\frac{m^2 - 2m + 1}{m} + + 2\left.\ln(|x + 1|)\right]_{\frac{m^2 - 2m + 1}{m^2}}^0\\ + &=\frac{m^2 - 2m + 1}{m} - + 2\ln\left(\frac{2m^2 - 2m + 1}{m^2}\right) +\end{align*} + +\subsection{Volumes} +Evaluate the integral + +\begin{align*} + \int_1^2\pi\left(2 - \frac{x}{2}\right)^2 \ud x +&=\pi\int_1^2\left(4 - x + \frac{x^2}{4}\right)\ud x\\ +&=\pi\left[4x - x^2 + \frac{x^3}{12}\right]_1^2\\ +&=\pi\left(4 - 3 + \frac{7}{12}\right)\\ +&=\frac{19\pi}{12}\tag{1} +\end{align*} + +\[\int_1^5\pi(x - 1)\ud x += \pi\left[\frac{x^2}{2} - x\right]_1^5 += \pi(12 - 4) += 8\pi\tag{3}\] + +\[\int_0^9 4\pi y \ud y = \left.2\pi y^2\right]_0^9 = 162\pi\tag{5}\] + +\[\int_0^1\pi\left|x^2 - x^6\right|\ud x += \pi\left[\frac{x^3}{3} - \frac{x^7}{7}\right]_0^1 += \frac{4\pi}{21}\tag{7}\] + +\begin{align*} + \int_{-2}^2\pi\left|\frac{x^4}{16} - 25 + 10x^2 - x^4\right|\ud x +&= 2\pi\int_0^2\left(\frac{15x^4}{16} - 10x^2 + 25\right)\ud x\\ +&= 2\pi\left[-\frac{10x^3}{3} + 25x + \frac{3x^5}{16}\right]_0^2\\ +&= \frac{88\pi}{3}\tag{8} +\end{align*} + +\begin{align*} + \int_0^1\pi\left|\left(\sqrt x - 1\right)^2 - \left(x^2 - 1\right)^2\right|\ud x +&= \int_0^1\pi\left|x - 2\sqrt x - x^4 + 2x^2\right|\ud x\\ +&= \int_0^1\pi\left(x^4 - 2x^2 - x + 2\sqrt x\right)\ud x\\ +&= \pi\left[\frac{x^5}{5} - \frac{2x^3}{3} + - \frac{x^2}{2} + \frac{4\sqrt x^3}{3}\right]_0^1\\ +&= \pi\left(\frac{1}{5} - \frac{2}{3} - \frac{1}{2} + \frac{4}{3}\right)\\ +&= \frac{11\pi}{30}\tag{11} +\end{align*} + +\begin{align*} + \int_0^h\left(a + \frac{x}{h}(b-a)\right)^2\ud x +&= \int_0^h\left(a^2 - \frac{ax}{h}(b-a) + \frac{x^2}{h^2}(b-a)^2\right)\ud x\\ +&= \left[a^2 x - \frac{ax^2}{2h}(b-a) + \frac{x^3}{3h^2}(b-a)^2\right]_0^h\\ +&= ha^2 - \frac{ha}{2}(b-a) + \frac{h}{3}(b-a)^2\\ +&= ha^2 - \frac{hab}{2} + \frac{ha^2}{2} + + \frac{ha^2}{3} - \frac{2hab}{3} + \frac{hb^2}{3}\\ +&= \frac{11ha^2 - 7hab + 2hb^2}{6}\tag{50} +\end{align*} + +\begin{align*} +A&=\int_{-r}^r\left(\pi\left(R + \sqrt{r^2 - x^2}\right)^2 - + \pi\left(R - \sqrt{r^2 - x^2}\right)^2\right)\ud x\\ + &=\int_{-r}^r 4\pi R\sqrt{r^2 - x^2}\ud x\\ + &=2\pi R\int_{-r}^r 2\sqrt{r^2 - x^2}\ud x\\ + &=2\pi R \cdot \pi r^2\\ + &=2\pi^2 R r^2\tag{61} +\end{align*} + +\setcounter{subsection}{3} +\subsection{Work} +\textbf{7. }Spring constant: $k = F(4) / 4 = 10g / 4 = 2.5g$ (lbf/in) + +Work done by stretching the spring from its natural length to 6 in: +\[\int_0^6 kx\ud x = \left.k\frac{x^2}{2}\right]_0^6 = 18k = 45g\text{ (lbf.in)}\] + +\noindent\textbf{9. }Suppose that 2 J of work is needed to stretch a spring from its +natural length of 30 cm to a length of 42 cm. + +\begin{align*} + \int_0^{0.12}kx\ud x = 2 +&\iff \left.\frac{kx^2}{2}\right]_0^{0.12} = 2\\ +&\iff 0.0072k = 2\\ +&\iff k = \frac{2500}{9}\text{ (N/m)} +\end{align*} + +\begin{align*} + \int_{0.05}^{0.1}kx\ud x +&=\int_{0.05}^{0.1}\frac{2500x}{9}\ud x\\ +&=\left.\frac{1250x^2}{9}\right]_{0.05}^{0.1}\\ +&=\frac{1250(0.01 - 0.0025)}{9}\\ +&=\frac{25}{24}\text{ (J)}\tag{a} +\end{align*} + +\[x = \frac{F}{k} = \frac{30 \cdot 9}{2500} = \frac{27}{250}\text{ (m)}\tag{b}\] + +\noindent Evaluate the integral +\[\int_0^{50}mgx\ud x = \left.\frac{25gx^2}{2}\right]_0^{50} = 31250g\text{ (ft.lbf)}\tag{13}\] + +\begin{align*} +W&=\lim_{n \to \infty}\sum_{i = 1}^n F_i\frac{3i}{n}\\ + &=\lim_{n \to \infty}\sum_{i = 1}^n m_i g\frac{3i}{n}\\ + &=\lim_{n \to \infty}\sum_{i = 1}^n A_i\frac{3}{n}\rho g\frac{3i}{n}\\ + &=\lim_{n \to \infty}\sum_{i = 1}^n + 3\left(1 - \frac{i}{n}\right)8\left(1 - \frac{i}{n}\right) + 1000g\frac{3i}{n}\cdot\frac{3}{n}\\ + &=\lim_{n \to \infty}\sum_{i = 1}^n + \frac{80000}{3}\left(3 - \frac{3i}{n}\right)^2\frac{3i}{n}\cdot\frac{3}{n}\\ + &=\int_0^3 \frac{80000}{3}\left(9x - 6x^2 + x^3\right)\ud x\\ + &=\frac{80000}{3}\left[\frac{9x^2}{2} - 2x^3 + \frac{x^4}{4}\right]_0^3\\ + &=180000\text{ (J)}\tag{21} +\end{align*} + +\begin{align*} +W&=\lim_{n \to \infty}\sum_{i = 1}^n m_i g\frac{8i}{n}\\ + &=\lim_{n \to \infty}\sum_{i = 1}^n A_i\frac{8}{n}\rho g\frac{8i}{n}\\ + &=\lim_{n \to \infty}\sum_{i = 1}^n + \pi\left(6 - 3\frac{i}{n}\right)^2 62.5g\frac{8i}{n}\cdot\frac{8}{n}\\ + &=\lim_{n \to \infty}\sum_{i = 1}^n + \frac{1125\pi g}{128}\left(16 - \frac{8i}{n}\right)^2\frac{8i}{n}\cdot\frac{8}{n}\\ + &=\frac{1125\pi g}{128}\int_0^8\left(16 - x\right)^2 x\ud x\\ + &=\frac{1125\pi g}{128}\left[128x^2 - \frac{32x^3}{3} + \frac{x^4}{4}\right]_0^8\\ + &=33000\pi g\text{ (ft.lbf)}\tag{23} +\end{align*} + +\subsection{Average Value of a Function} +\textbf{9. }Given the function $f(x) = (x - 3)^2$ on $[2, 5]$. +\[f_{ave} += \frac{1}{5 - 2}\int_2^5 (x - 3)^2 \ud x += \frac{1}{3}\left[\frac{x^3}{3} - 3x^2 + 9x\right]_2^5 += 1\tag{a}\] + +Since $x \in [2, 5]$, +\[f(c) = f_{ave} \iff (c - 3)^2 = 1 \iff c = 4\tag{b}\] + +\begin{tikzpicture} + \begin{axis}[ + axis x line=middle, axis y line=middle, + xlabel={$x$}, ylabel={$y$}, area style, + xlabel style={at=(current axis.right of origin), anchor=west}, + ylabel style={at=(current axis.above origin), anchor=south}] + \addplot[domain=2:5,blue]{x^2 - 6*x + 9}; + \addplot[domain=2:5]{1}; + \end{axis} +\end{tikzpicture} + +\noindent\textbf{12. }Given $f(x) = 2\sin x - \sin 2x$ on $[0, \pi]$. +\begin{align*} +f_{ave} &= \frac{1}{\pi}\int_0^\pi\left(2\sin x - \sin 2x\right)\ud x\\ + &= \frac{1}{\pi}\left[\frac{\cos 2x}{2} - 2\cos x\right]_0^\pi\\ + &= \frac{4}{\pi} +\end{align*} + +$f(c) = f_{ave} \iff 2\sin x - \sin 2x = \frac{4}{\pi}$, i.e. +$x \approx 1.24$ or $x \approx 2.81$ on $[0, \pi]$. + +\noindent\textbf{13. }Since $f$ is continuous, apply Mean Value Theorem on +$[1, 3]$, + +\[\exists c \in [1, 3], f(c) = \frac{1}{3 - 1}\int_1^3 f(x)\ud x += \frac{8}{2} = 4\] + +\section{Techniques of Integration} +\subsection{Integration by Parts} +Evaluate the integral +\begin{align*} + \int x \cos 5x \ud x +&= \int\frac{x}{5} \ud\sin 5x\\ +&= \frac{x\sin 5x}{5} - \int\sin 5x \ud\frac{x}{5}\\ +&= \frac{x\sin 5x}{5} + \frac{x\cos 5x}{25} + C\tag{3} +\end{align*} + +\begin{align*} + \int e^{2\theta}\sin 3\theta \ud\theta +&= \int\frac{\sin 3\theta}{2} \ud e^{2\theta}\\ +&= \frac{e^{2\theta}\sin 3\theta}{2} + - \int\frac{3}{2}e^{2\theta}\cos 3\theta \ud\theta\\ +&= \frac{e^{2\theta}\sin 3\theta}{2} + - \frac{3}{2}\int\frac{\cos 3\theta}{2} \ud e^{2\theta}\\ +&= \frac{e^{2\theta}\sin 3\theta}{2} + - \frac{3e^{2\theta}\cos 3\theta}{4} + - \frac{9}{4}\int e^{2\theta}\sin 3\theta \ud\theta\\ +&= \frac{4}{13}e^{2\theta} + \left(\frac{\sin 3\theta}{2} - \frac{\cos 3\theta}{4}\right) + C\\ +&= \frac{e^{2\theta}}{13}(2\sin 3\theta - 3\cos 3\theta) + C\tag{17} +\end{align*} + +\setcounter{section}{8} +\section{Differential Equations} +\setcounter{subsection}{2} +\subsection{Separable Equations} +Solve the equation. +\begin{align*} + \leibniz{y}{x} = xy^2 + &\iff y^{-2}\ud y = x\ud x\\ + &\;\Longrightarrow \int\frac{\ud y}{y^2} = \int x\ud x + \qquad\text{(for $y\neq 0$)}\\ + &\iff C_y - \frac{1}{y} = C_x + \frac{x^2}{2}\\ + &\iff \frac{C - x^2}{2} = \frac{1}{y}\\ + &\iff y = \frac{2}{C - x^2}\tag{1} +\end{align*} +\begin{align*} + (y + \sin y)\leibniz{y}{x} = x + x^3 + &\iff \int(y + \sin y)\ud y = \int(x + x^3)\ud x\\ + &\iff \frac{y^2}{2} - \cos y = \frac{x^2}{2} + \frac{x^4}{4} + C\tag{5}\\ +\end{align*} +\begin{align*} + \leibniz{y}{t} = \frac{t}{y\exp(y + t^2)} + &\iff \int ye^y\ud y = \int te^{-t^2}\ud t\\ + &\iff (y - 1)e^y = C - \frac{1}{2e^{t^2}}\tag{7} +\end{align*} +\begin{align*} + \leibniz{u}{t} = \frac{2t + \sec^2 t}{2u} + &\iff \int 2u\ud u = \int(2t + \sec^2 t)\ud t\\ + &\iff u^2 = t^2 + \tan t + C\tag{13} +\end{align*} +Since $u(0) = -5$, $u = -\sqrt{t^2 + \tan t + 25}$. +\begin{align*} + \leibniz{y}{x} = xy + &\iff \int\frac{\ud y}{y} = \int x\ud x\qquad\text{(since $y\neq 0$)}\\ + &\iff \ln|y| = \frac{x^2}{2} + C\\ + &\iff |y| = \exp\left(\frac{x^2}{2} + C\right)\tag{19} +\end{align*} +Since $y(0) = 1$, $y = \exp(x^2/2)$.\pagebreak +\begin{align*} + y(x) = 2 + \int_2^x[t - ty(t)]\ud t + &\;\Longrightarrow y - 2 = \int(x - xy)\ud x\\ + &\iff \leibniz{(y - 2)}{x} = x - xy\\ + &\iff \int\frac{\ud y}{1 - y} = \int x\ud x\\ + &\iff C - \frac{x^2}{2} = \ln|1 - y|\\ + &\iff y = 1 \pm \exp\left(C - \frac{x^2}{2}\right)\tag{33} +\end{align*} +Since $y(2) = 2$ (which can be trivially obtained from the original condition), +$y = 1 + \exp(2 - x^2/2)$. + +\subsection{Models for Population Growth} +\textbf{3.} The Pacific halibut fishery has been modeled +by the differential equation +\begin{align*} + \leibniz{y}{t} = ky\left(1 - \frac{y}{M}\right) + &\;\Longrightarrow \int\left(\frac{1}{y} + \frac{1}{M-y}\right)\ud y + = \int k\ud t\\ + &\iff \ln|y| - \ln|M - y| = kt + C\\ + &\iff \left|\frac{M}{y} - 1\right| = e^{-kt-C}\\ + &\iff \frac{M}{y} = 1 \pm e^{-kt-C}\\ + &\iff y = \frac{M}{1 \pm e^{-kt-C}} + \tag{$*$} +\end{align*} + +As $M = 8\times 10^7$, $k = 0.71$ and $y(0) = 2\times 10^7$, +from $(*)$ we get $\pm e^{-C} = 3$ and thus +\[y = \frac{M}{1 + 3e^{-kt}}\] + +For $t = 1$, $y \approx 3.2\times 10^7$. +For $y = 4\times 10^7$, $t = (\ln 3)/0.71$. + +\noindent\textbf{5.} Suppose a population grows according to a logistic model +\[\leibniz{P}{t} = kP\left(1 - \frac{P}{M}\right) +\iff P(t) = \frac{M}{1 \pm e^{-kt-C}}\] +with initial population $P(0) = 1000$ and carrying capacity $M = 10000$. + +Suppose $P(1) = 2500$, +\[\begin{dcases} + \frac{10000}{1\pm e^{-C}} &= 1000\\ + \frac{10000}{1\pm e^{-k-C}} &= 2500 +\end{dcases} +\iff\begin{cases} + \pm e^{-C} &= 9\\ + \pm e^{-k-C} &= 3 +\end{cases} +\iff\begin{cases} + \pm := +\\ + C = -\ln 9\\ + k = \ln 3 +\end{cases}\] + +After another 3 years, the population will be +\[P(3) = \left(t\mapsto\frac{10000}{1+3^{2-t}}\right)(1+3) = 9000\] + +\subsection{Linear Equations} +Solve the differential equation. +\begin{align*} + \leibniz{y}{x} + y = x + &\iff e^x\leibniz{y}{x} + y\leibniz{e^x}{x} = xe^x\\ + &\iff \int\ud ye^x = \int x\ud e^x\\ + &\iff ye^x = e^x(x - 1) + C\\ + &\iff y = x - 1 + Ce^{-x}\tag{7} +\end{align*} +\begin{align*} + x\leibniz{y}{x} + y = \sqrt x + &\iff \int\ud xy = \int\sqrt x\ud x\\ + &\iff xy = \frac{2x\sqrt x}{3} + C\\ + &\iff y = \frac{2\sqrt x}{3} + \frac{C}{x}\tag{9} +\end{align*} +\begin{align*} + x^2\leibniz{y}{x} + 2xy = \ln x + &\iff \int\ud yx^2 = \int\ln x\ud x\\ + &\iff yx^2 = x(\ln x - 1) + C\\ + &\iff y = \frac{\ln x - 1}{x} + \frac{C}{x^2} +\end{align*} +Since $y(1) = 2$, $C = 3$. +\begin{align*} + L\leibniz{I}{t} + RI = \mathcal E + &\iff e^{Rt/L}\left(\leibniz{I}{t} + \frac{R}{L}I\right) + = \frac{\mathcal E}{L}e^{Rt/L}\\ + &\iff \int\ud Ie^{Rt/L} = \frac{\mathcal E}{L}\int e^{Rt/L}\ud t\\ + &\iff Ie^{Rt/L} = \frac{\mathcal E}{R}e^{Rt/L} + C\\ + &\iff I = \frac{\mathcal E}{R} + \frac{C}{\exp(Rt/L)} +\end{align*} +Since $\mathcal E = 40$ V, $L = 2$ H, $R = 10\,\Omega$ and $I(0) = 0$, +$I(t) = 4 - 4/\exp 5t$ and $I(0.1) = 4 - 4/\sqrt e$. + +\allowdisplaybreaks +\setcounter{section}{10} +\section{Lazy Evaluation} +\setcounter{subsection}{2} +\subsection{The Integral Test and Estimates of Sums} +\textbf{34. }Using Leonhard Euler's calculation of the exact sum of +the $p$-series with $p = 2$: +\[\zeta(2) = \sum_{n=1}^\infty\frac{1}{n^2} += \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i^2} = \frac{\pi^2}{6}\] +\[\sum_{n=2}^\infty\frac{1}{n^2} = \lim_{n\to\infty}\sum_{i=2}^n\frac{1}{i^2} += \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i^2} - \frac{1}{1^2} += \frac{\pi^2}{6} - 1\tag{a}\] +\[\sum_{n=3}^\infty\frac{1}{(n + 1)^2} += \lim_{n\to\infty}\sum_{i=4}^{n+1}\frac{1}{i^2} += \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i^2} - \sum_{i=1}^3\frac{1}{i^2} += \frac{\pi^2}{6} - \frac{49}{36}\tag{b}\] +\[\sum_{n=1}^\infty\frac{1}{(2n)^2} += \lim_{n\to\infty}\sum_{i=1}^n\frac{1}{4i^2} += \frac{1}{4}\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i^2} += \frac{\pi^2}{24}\tag{c}\] + +\noindent Determine if the series is convergent or divergent using +the Integral Test. + +\[\sum_{n=2}^\infty \frac{1}{n(\ln n)^2}\tag{22}\] + +\begin{align*} + \int_2^\infty\frac{1}{x(\ln x)^2} +&= \lim_{t\to\infty}\int_2^t\frac{1}{x(\ln x)^2}\ud x\\ +&= \lim_{t\to\infty}\int_2^t\frac{1}{(\ln x)^2}\ud\ln x\\ +&= \lim_{t\to\infty}\int_{\ln 2}^{\ln t}\frac{1}{x^2}\ud x\\ +&= \lim_{t\to\infty}\left.\frac{-1}{x}\right]_{\ln 2}^{\ln t}\\ +&= \lim_{t\to\infty}\left(\frac{1}{\ln{2}} - \frac{1}{\ln t}\right)\\ +&= \frac{1}{\ln 2} +\end{align*} + +Thus by the Integral Test, the given series is convergent. + +\[\sum_{n=3}^\infty\frac{n^2}{e^n}\tag{24}\] + +\begin{align*} + \int_3^\infty\frac{x^2}{e^x} +&= \lim_{t\to\infty}\int_3^t\frac{x^2}{e^x}\ud x\\ +&= \lim_{t\to\infty}\int_3^t -x^2 \ud e^{-x}\\ +&= \lim_{t\to\infty}\left(\int_3^t e^{-x}\ud x^2 + - \left.x^2 e^{-x}\right]_3^t\right)\\ +&= \lim_{t\to\infty}\left(-\int_3^t 2x\ud e^{-x} + + \left.\frac{x^2}{e^x}\right]_t^3\right)\\ +&= \lim_{t\to\infty}\left(\int_3^t e^{-x}\ud 2x + + \left[\frac{2x}{e^x} + \frac{x^2}{e^x}\right]_t^3\right)\\ +&= \lim_{t\to\infty}\left(2\int_3^t e^{-x}\ud x + + \left.\frac{2x + x^2}{e^x}\right]_t^3\right)\\ +&= \lim_{t\to\infty}\left.\frac{2 + 2x + x^2}{e^x}\right]_t^3\\ +&= \lim_{t\to\infty}\left(\frac{17}{e^3} - \frac{2 + 2t + t^2}{e^t}\right)\\ +&= \frac{17}{e^3} +\end{align*} + +Thus by the Integral Test, the given series is convergent.\pagebreak + +\[\sum_{n=1}^\infty\frac{\cos(\pi n)}{\sqrt n}\tag{27}\] + +Since $x \mapsto \cos(\pi x)/\sqrt n$ is neither positive +(e.g. $\cos 3\pi/\sqrt 3 = -1$) nor ultimately decreasing, the Integral Test +cannot be used to determine whether the series is convergent. + +\subsection{The Comparison Test} +Determine whether the series is convergent or divergent. +\[\sum_{n=1}^\infty\frac{\sqrt{n^4 + 1}}{n^3 + n^2}\tag{25}\] + +We use the Limit Comparison Test with +\[a_n = \frac{\sqrt{n^4 + 1}}{n^3 + n^2} \qquad b_n = \frac{1}{n}\] +and obtain +\[\lim_{n\to\infty}\frac{a_n}{b_n} += \lim_{n\to\infty}\frac{\sqrt{n^4 + 1}}{n^2 + n} += \lim_{n\to\infty}\frac{\sqrt{1 + \frac{1}{n^4}}}{1 + \frac{1}{n}} += 1 > 0\] + +Since this limit exists and $\sum\frac{1}{n}$ is divergent ($p$-series with +$p = 1$), the given series diverges by the Limit Comparison Test. + +\[\sum_{n=1}^\infty\frac{1}{n!}\tag{29}\] +\[\lim_{n\to\infty}\frac{1/(n+1)!}{1/n!} += \lim_{n\to\infty}\frac{1}{n + 1} = 0 < 1\] + +Thus by the Ratio Test, the given series is absolutely convergent. + +\[\sum_{n=1}^\infty\frac{n!}{n^n}\tag{30}\] +\[\frac{n!}{n^n} = \frac{2}{n^2}\cdot\frac{n!}{2n^{n-2}} \leq \frac{2}{n^2}\] + +Since both $\sum n!/n^n$ and $\sum 2/n^2$ are series with positive terms and +$\sum 2/n^2$ converges because it is a constant time of $p$-series with +$p = 2$, by the Comparison Test, $\sum n!/n^n$ is convergent. + +\[\sum_{n=1}^\infty\frac{1}{n\sqrt[n]n}\tag{32}\] + +We use the Limit Comparison Test with +\[a_n = \frac{1}{n\sqrt[n]n} \qquad b_n = \frac{1}{n}\] +and obtain +\[\lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{1}{\sqrt[n]n}\] + +Since $\frac{1}{\sqrt[n]n}' = \frac{1 - \ln n}{n^2\sqrt[n]n}$ is negative on +$(e, \infty)$, $n \mapsto \frac{1}{\sqrt[n]n}$ is ultimately decreasing. +Additionally, $\frac{1}{\sqrt[n]n} \geq \frac{1}{\sqrt[n]1} = 1$ on this +interval, thus +\[\lim_{n\to\infty}\frac{1}{\sqrt[n]n} += \inf\left\{\frac{1}{\sqrt[n]n}: n \in \mathbb N_3\right\} = 1 > 0\] + +Therefore, the given series diverges by the Limit Comparison Test, +as $\sum\frac{1}{n}$ is divergent ($p$-series with $p = 1$). + +\subsection{Alternating Series} +Test the series for convergence or divergence. + +\[\sum_{n=1}^\infty (-1)^n\frac{n^n}{n!}\tag{19}\] + +Since $\frac{(n + 1)^{n + 1}}{(n + 1)!} > \frac{n^n}{n!}$, the given +alternating series diverges. + +\[\sum_{n=1}^\infty (-1)^n\left(\sqrt{n + 1} - \sqrt n\right)\tag{20}\] + +For all $n$, +\begin{align*} + n^2 + 2n < n^2 + 2n + 1 +&\iff \sqrt{n(n + 2)} < n + 1\\ +&\iff n + \sqrt{n(n + 2)} + n + 2 < 4n + 4\\ +&\iff \sqrt{n + 2} + \sqrt n < 2\sqrt{n + 1}\\ +&\iff \sqrt{n + 2} - \sqrt{n + 1} < \sqrt{n + 1} - \sqrt n\tag{i} +\end{align*} +\[\lim_{n\to\infty}\left(\sqrt{n + 1} - \sqrt n\right) += \lim_{n\to\infty}\frac{1}{\sqrt{n + 1} + \sqrt n} += \lim_{n\to\infty}\frac{1/\sqrt n}{\sqrt{1 + 1/\sqrt{n}} + 1} += 0\tag{ii}\] + +Thus, by the Alternating Series Test, the given series is convergent. + +\subsection{Absolute Convergence} +Determine whether the series is absolutely convergent, +conditionally convergent, or divergent. + +\[\sum_{n=2}^\infty\left(\frac{-2n}{n + 1}\right)^{5n}\tag{22}\] +\[\lim_{n\to\infty}\sqrt[n]{\left(\frac{2n}{n + 1}\right)^{5n}} += \lim_{n\to\infty}\left(\frac{2n}{n + 1}\right)^5\\ += \left(\lim_{n\to\infty}\frac{2}{1 + 1/n}\right)^5\\ += 32 > 1\] + +Thus the given series diverges by the Root Test. + +\[\sum_{n=1}^\infty\prod_{i=1}^n\frac{2i}{3i + 2}\tag{30}\] +\[\lim_{n\to\infty}\frac{\prod_{i=1}^{n+1}\frac{2i}{3i + 2}} + {\prod_{i=1}^n\frac{2i}{3i + 2}} += \lim_{n\to\infty}\frac{2n + 2}{3n + 5} += \lim_{n\to\infty}\frac{2 + 2/n}{3 + 5/n} += \frac{2}{3} < 1\] + +Thus by the Ratio Test, the given series is absolutely convergent. + +\setcounter{subsection}{7} +\subsection{Power Series} +Find the radius of convergence and the interval of convergence of the series. + +\[\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n + 1)!}\tag{14}\] + +Let $a_n = (-1)^n x^{2n+1}/(2n + 1)!$, +\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| += \lim_{n\to\infty}\left|\frac{x^2}{4n^2 + 10n + 6}\right| += 0 < 1\] + +Thus by the Ratio Test, the series is convergent for all $x$ and the radius of +convergence is $R = \infty$. + +\[\sum_{n=1}^\infty\frac{3^n (x+4)^n}{\sqrt n}\tag{17}\] + +Let $a_n = 3^n (x+4)^n / \sqrt n$, +\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| += \lim_{n\to\infty}\left|\frac{3(x + 4)\sqrt n}{\sqrt{n + 1}}\right| += 3|x + 4|\] + +Using the Ratio Test, we see that the series converges if $|x + 4| < 1/3$ and +it diverges if $|x + 4| > 1/3$, thus the radius of convergence is $R = 1/3$. + +When $|x + 4| = 1/3$, the series is either $\sum (-3)^n / \sqrt n$ or +$\sum 3^n / \sqrt n$, both of which diverge by the Test for Divergence. +Therefore the interval of convergence is $(-13/3, -11/3)$. + +\[\sum_{n=1}^\infty\frac{b^n}{\ln n}(x - a)^n,\qquad b > 0\tag{22}\] + +Let $a_n = b^n (x - a)^n / \ln n$, +\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| += \lim_{n\to\infty}\left|\frac{b(x - a)\ln n}{\ln(n + 1)}\right| += b|x - a|\] + +Using the Ratio Test, we see that the series converges if $|x - a| < b^{-1}$ +and it diverges if $|x - a| > b^{-1}$, thus the radius of convergence is +$R = b^{-1}$. + +When $|x - a| = b^{-1}$, the series is $\sum (\pm b)^n / \ln n$, which diverges +by the Test for Divergence. Therefore the interval of convergence is +$(a - b^{-1}, a + b^{-1})$. + +\[\sum_{n=2}^\infty\frac{x^{2n}}{n(\ln n)^2}\tag{26}\] + +Let $a_n = x^{2n} / n / (\ln n)^2$, +\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| += \lim_{n\to\infty}\left|\frac{x^2 n \ln^2 n}{(n + 1)\ln^2(n + 1)}\right| += x^2\] + +Using the Ratio Test, we see that the series converges if $|x| < 1$ and it +diverges if $|x| > 1$, therefore the radius of convergence is $R = 1$. + +When $x = \pm 1$, $a_n = n^{-1}/(\ln n)^2$, which is defined by a continuous, +positive and decreasing function $x \mapsto x^{-1}/(\ln x)^2$ on $[2, \infty)$. +\begin{align*} + \int_2^\infty\frac{1}{x(\ln x)^2}\ud x +&= \lim_{t\to\infty}\int_2^t\frac{1}{(\ln x)^2}\ud\ln x\\ +&= \lim_{t\to\infty}\int_{\ln 2}^{\ln t}\frac{1}{x^2}\ud x\\ +&= \lim_{t\to\infty}\left.\frac{1}{3x^3}\right]_{\ln t}^{\ln 2}\\ +&= \frac{1}{3(\ln 2)^3} +\end{align*} + +By the Integral Test, $\sum_{n=2}^\infty n^{-1}/(\ln n)^2$ converges, and thus +the interval if of convergence of the given power series is $[-1, 1]$. + +\subsection{Representations of Functions as Power Series} +Find a power series representation for the function and determine the interval +of convergence. + +\[f(x) = \frac{5}{1 - 4x^2} + = 5\sum_{n=0}^\infty\left(4x^2\right)^n + = \sum_{n=0}^\infty 5 \cdot 2^{2n} x^{2n}\tag{4}\] + +Interval of convergence is $(-1, 1)$. + +\[f(x) = \frac{x^2}{a^3 - x^3} + = \frac{x^2}{a^3}\sum_{n=0}^\infty\left(\frac{x}{a}\right)^{3n} + = \sum_{n=0}^\infty\frac{x^{3n + 2}}{a^{3n + 3}}\tag{10}\] + +Interval of convergence is $(-a, a)$. + +\begin{align*} +f(x) &= \frac{x + 2}{2x^2 - x - 1}\\ + &= \frac{1}{x - 1} - \frac{1}{2x + 1}\\ + &= -\sum_{n=0}^\infty x^n - \sum_{n=0}^\infty (-2x)^n\\ + &= \sum_{n=0}^\infty (-1 - (-2)^n)x^n\tag{12} +\end{align*} + +Interval of convergence is $(-1, 1) \cap (-1/2, 1/2) = (-1/2, 1/2)$. + +\noindent\textbf{40. }Find the sum of the series when $|x| < 1$. +\begin{align*} + \sum_{n=1}^\infty nx^{n - 1} +&= \sum_{n=1}^\infty x^{n - 1} + \sum_{n=1}^\infty (n - 1)x^{n - 1}\\ +&= \sum_{n=0}^\infty x^n + \sum_{n=0}^\infty nx^n\\ +&= \frac{1}{1 - x} + x\sum_{n=1}^\infty nx^{n - 1}\\ +&= \frac{1}{(1 - x)^2}\tag{a} +\end{align*} + +\[\sum_{n=1}^\infty nx^n += x\sum_{n=1}^\infty nx^{n - 1} += \frac{x}{(1 - x)^2}\tag{b.i}\] + +\[\sum_{n=1}^\infty \frac{n}{2^n} += \left(x \mapsto \frac{x}{(1 - x)^2}\right)\left(\frac{1}{2}\right) += 2\tag{b.ii}\] + +\begin{align*} + \sum_{n=2}^\infty n(n - 1)x^n +&= \sum_{n=2}^\infty 2(n - 1)x^n + \sum_{n=2}^\infty (n - 1)(n - 2)x^n\\ +&= 2\sum_{n=1}^\infty (n - 1)x^n + x\sum_{n=1}^\infty n(n - 1)x^n\\ +&= 2\left(\sum_{n=1}^\infty nx^n + 1 - \sum_{n=0}^\infty x^n\right) : (1 - x)\\ +&= 2\left(\frac{x}{(1 - x)^2} + 1 - \frac{1}{1 - x}\right) : (1 - x)\\ +&= \frac{2x^2}{(1 - x)^3}\tag{c.i} +\end{align*} + +\[\sum_{n=2}^\infty \frac{n^2 - n}{2^n} += \left(x \mapsto \frac{2x^2}{(1 - x)^3}\right)\left(\frac{1}{2}\right) += 4\tag{c.ii}\] + +\[\sum_{n=1}^\infty \frac{n^2}{2^n} += \sum_{n=2}^\infty \frac{n^2 - n}{2^n} + \sum_{n=1}^\infty nx^n += 4 + 2 = 6\tag{c.iii}\] + +\subsection{Taylor and Maclaurin Series} +Find the Taylor series for $f$ centered at the given value of $a$ and the +associative radius of convergence. + +\[f(x) = \ln x,\qquad a = 2\tag{15}\] +\begin{align*} + f(x) +&= \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n\\ +&= \ln 2 + \sum_{n=1}^\infty\left(x\mapsto\binom{-1}{n-1}\frac{1}{nx^n}\right) + (2)\cdot(x - 2)^n\\ + &= \ln 2 + \sum_{n=1}^\infty(-1)^{n-1}\frac{(x - 2)^n}{n2^n} +\end{align*} + +Let $a_n = (-1)^{n-1}(x - 2)^n/(n2^n)$, +\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| += \lim_{n\to\infty}\frac{n}{2n + 2}|x - 2| += \frac{|x - 2|}{2}\] + +Using the Ratio Test, we see $f(x) = \ln 2 + \sum a_n$ converges if +$|x - 2| < 2$ and it diverges if $|x - 2| > 2$, therefore the associative +radius of convergence is $R = 2$. + +\[f(x) = \frac{1}{x},\qquad a = -3\tag{16}\] +\begin{align*} + f(x) +&= \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n\\ +&= \sum_{n=0}^\infty + \left(x\mapsto\binom{-1}{n}\frac{1}{x^{n+1}}\right)(-3)\cdot(x + 3)^n\\ +&= \sum_{n=0}^\infty(-1)^n\frac{(x + 3)^n}{(-3)^{n+1}}\\ +&= \sum_{n=0}^\infty\frac{-(x + 3)^n}{3^{n+1}} +\end{align*} + +Let $a_n = (x + 3)^n/3^{n+1}$, +\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| += \lim_{n\to\infty}\frac{|x + 3|}{3} += \frac{|x + 3|}{3}\] + +Using the Ratio Test, we see that the series converges if $|x + 3| < 3$ and it +diverges if $|x + 3| > 3$, therefore the associative radius of convergence is +$R = 3$. + +\[f(x) = \sin x,\qquad a = \frac{\pi}{2}\tag{18}\] +\begin{align*} +f(x) &= \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n\\ + &= \sum_{n=0}^\infty\frac{\cos\frac{n\pi}{2}}{n!} + \left(x - \frac{\pi}{2}\right)^n\\ + &= \sum_{n=0}^\infty\frac{(-1)^n}{(2n)!} + \left(x - \frac{\pi}{2}\right)^{2n} +\end{align*} + +Let $a_n = (-1)^n(x - \pi/2)^{2n}/(2n)!$, +\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| += \lim_{n\to\infty}\frac{(x - \pi/2)^2}{(2n + 2)(2n + 1)} += 0 < 1\] + +Using the Ratio Test, we see that the series converges for all $x$, thus the +associative radius of convergence is $R = \infty$. + +\[f(x) = \sqrt x,\qquad x = 16\tag{20}\] +\begin{align*} + f(x) +&= \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x - a)^n\\ +&= \sum_{n=0}^\infty\left(x\mapsto\binom{\frac{1}{2}}{n}x^{1/2-n}\right)(16) + \cdot (x - 16)^n\\ +&= \sum_{n=0}^\infty\binom{\frac{1}{2}}{n}\frac{4(x - 16)^n}{16^n} +\end{align*} + +Let $a_n = 4\binom{1/2}{n}(x - 16)^n/16^n$, +\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| += \lim_{n\to\infty}\left|\frac{1/2 - n}{n + 1}\right|\frac{|x - 16|}{16} += \frac{|x - 16|}{16}\] + +Using the Ratio Test, we see that the series converges if $|x - 16| < 16$ and +it diverges if $|x - 16| > 16$, therefore the associative radius of convergence +is $R = 16$. + +\noindent\textbf{55. }Use series to evaluate the limit. +\begin{align*} + \lim_{x \to 0}\frac{x - \ln(1 + x)}{x^2} +&= \lim_{x \to 0}\frac{x - \sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}}{x^2}\\ +&= \lim_{x \to 0}\frac{\sum_{n=2}^\infty (-1)^n\frac{x^n}{n}}{x^2}\\ +&= \lim_{x \to 0}\sum_{n=0}^\infty (-1)^n\frac{x^n}{n + 2}\\ +&= \lim_{x \to 0}\frac{1}{2} + + \lim_{x \to 0}\sum_{n=1}^\infty (-1)^n\frac{x^n}{n + 2}\\ +&= \frac{1}{2} +\end{align*} + +\newpage\noindent Find the sum of the series. +\begin{align*} + \sum_{n=0}^\infty\frac{(-\ln 2)^n}{n!} +&= \left(x \mapsto \sum_{n=0}^\infty\frac{x^n}{n!}\right)(-\ln 2)\\ +&= (x \mapsto e^x)\left(\ln\frac{1}{2}\right)\\ +&= \exp\left(\ln\frac{1}{2}\right)\\ +&= \frac{1}{2}\tag{68} +\end{align*} + +\begin{align*} +\sum_{n=0}^\infty\frac{(-1)^n}{(2n + 1)2^{2n + 1}} +&= \left(x \mapsto \sum_{n=0}^\infty(-1)^n\frac{2^{2n + 1}}{2n + 1}\right) + \left(\frac{1}{2}\right)\\ +&= \left(x \mapsto \tan^{-1}x\right)\left(\frac{1}{2}\right)\\ + &= \tan^{-1}\frac{1}{2}\tag{70} +\end{align*} + +\noindent\textbf{72. }If $f(x) = \left(1 + x^3\right)^{30}$, what is +$f^{(58)}(0)$? +\begin{align*} + f(x) = \sum_{n=0}^{30}\binom{30}{n}x^{3n} +&\Longrightarrow f'(x) = \sum_{n=0}^{30}\binom{30}{n}x^{3n - 1}3n\\ +&\Longrightarrow f''(x) = \sum_{n=1}^{30}\binom{30}{n}x^{3n - 2}3n(3n - 1)\\ +&\Longrightarrow f^{(58)}(x) = \sum_{n=20}^{30}\binom{30}{n}x^{3n - 58} + \prod_{i=0}^{57}(3n - i)\\ +&\Longrightarrow f^{(58)}(0) = \sum_{n=20}^{30}\binom{30}{n}0^{3n - 58} + \prod_{i=0}^{57}(3n - i) = 0 +\end{align*} +\end{document} |