Created the Real Analysis Theorems and Definitions packet
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\section{Approximate Integration}
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\textbf{Equal Partitions}
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If $f:[a,b] \to \R$ is continuous, we know that its Riemann integral exists. To find an approximate value for this integral with the minimum amount of calculation, it is convenient to consider partitions $\mathcal{P}_n$ of $[a,b]$ into \textit{n equal} subintervals having length $h_n:=(b-a)/n$. Hence $\mathcal{P}_n$ is the partition:
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\[a<a+h_n<a+2h_n<\dots<a+nh_n=b\]
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If we pick our tag points to be the \textit{left endpoints} and the \textit{right endpoints} of the subintervals, we obtain the \textbf{$n$th left approximation} given by
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\[L_n(f):=h_n\sum\limits_{k=1}^{n-1}f(a+kh_n)\]
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and the \textbf{$n$th right approximation} given by
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\[R_n(f):=h_n\sum\limits_{k=1}^{n}f(a+kh_n)\]
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\begin{theorem}
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If $f:[a,b] \to \R$ is monotone and if $T_n(f)$ is given by
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\[T_n(f):=h_n\left(\frac{1}{2}f(a)+\sum\limits_{k=1}^{n-1}f(a+kh_n)+\frac{1}{2}f(b)\right)\]
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then
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\[\left|\displaystyle\int_{a}^{b}f-T_n(f)\right|\leq |f(b)-f(a)|\cdot \frac{(b-a)}{2n}\]
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\end{theorem}
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Note that the function $T_n(f)$ defined above is called the \textbf{$n$th Trapezoidal Approximation} of $f$
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\begin{theorem}
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Let $f,f',$ and $f''$ be continuous on $[a,b]$ and let $T_n(f)$ be the $n$th Trapezoidal Approximation. Then there exists $c \in [a,b]$ such that
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\[T_n(f)-\displaystyle\int_{a}^{b}f=\frac{(b-a)h^2_n}{12}\cdot f''(c)\]
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\end{theorem}
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\begin{corollary}
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Let $f,f',$ and $f''$ be continuous, and let $|f''(x)|\leq B_2$ for all $x \in [a,b]$. Then
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\[\left|T_n(f)-\displaystyle\int_{a}^{b}f\right|\leq \frac{(b-a)h_n^2}{12} \cdot B_2 = \frac{(b-a)^3}{12n^2}\cdot B_2\]
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\end{corollary}
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If $\mathcal{P}_n$ is the equally spaced partition given before, the \textbf{Midpoint Approximation} of $f$ is given by
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\[M_n(f):=h_n\left(f\left(a+\frac{1}{2}h_n\right)+f\left(a+\frac{3}{2}h_n\right)+\dots+f\left(a\left(n-\frac{1}{2}\right)h_n\right)\right)=h_n\sum\limits_{k=1}^{n}f\left(a+\left(k-\frac{1}{2}\right)h_n\right)\]
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\begin{theorem}
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Let $f,f',$ and $f''$ be continuous on $[a,b]$ and let $M_n(f)$ be the $n$th Midpoint Approximation. Then there exists $\gamma \in [a,b]$ such that
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\[\displaystyle\int_{a}^{b}f-M_n(f)=\frac{(b-a)h_n^2}{24}\cdot f''(\gamma)\]
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\end{theorem}
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\begin{corollary}
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Let $f,f',$ and $f''$ be continuous, and let $|f''(x)| \leq B_2$ for all $x \in [a,b]$. Then
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\[\left|M_n(f)-\displaystyle\int_{a}^{b}f\right|\leq \frac{(b-a)h_n^2}{24}\cdot B_2=\frac{(b-a)^3}{24n^2}\cdot B_2\]
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\end{corollary}
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The \textbf{$n$th Simpson Approximation} is defined by
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\[S_n(f):=\frac{1}{3}h_n(f(a)+4f(a+h_n)+2f(a+2h_n)+4f(a+3h_n)+2f(a+4h_n)+\dots+2f(b-2h_n)+4f(b-h_n)+f(b))\]
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\begin{theorem}
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Let $f,f',f'',f^{(3)}$, and $f^{(4)}$ be continuous on $[a,b]$ and let $n \in \N$ be even. If $S_n(f)$ is the $n$th Simpson Approximation, then there exists $c \in [a,b]$ such that
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\[S_n(f)-\displaystyle\int_{a}^{b}f=\frac{(b-a)h_n^4}{180}\cdot f^{(4)}(c)\]
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\end{theorem}
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\begin{corollary}
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Let $f,f',f'',f^{(3)},$ and $f^{(4)}$ be continuous on $[a,b]$ and let $|f^{(4)}| \leq B_4$ for all $x \in [a,b]$. Then
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\[\left|S_n(f)-\displaystyle\int_{a}^{b}f\right|\leq \frac{(b-a)h_n^4}{180}\cdot B_4 = \frac{(b-a)^5}{180n^4}\cdot B_4\]
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\end{corollary}
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\begin{remark}
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The $n$th Midpoint Approximation $M_n(f)$ can be used to "step up" to the $(2n)$th Trapezoidal and Simpson Approximations by using the formulas
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\[T_{2n}(f)=\frac{1}{2}M_n(f)+\frac{1}{2}T_n(f)\ \ \ \ \text{and}\ \ \ \ S_{2n}(f)=\frac{2}{3}M_n(f)+\frac{1}{3}T_n(f)\]
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that are given in the exercises. Thus once the initial Trapezoidal Approximation $T_1=T_1(f)$ has been calculated, only the Midpoint Approximation $M_n=M_n(f)$ need be found. That is, we employ the following sequence of calculations:
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\[T_1=\frac{1}{2}(b-a)(f(a)+f(b));\]
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\begin{align*}
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M_1=(b-a)f(\frac{1}{2}(a+b)),\ \ \ \ & T_2=\frac{1}{2}M_1+\frac{1}{2}T_1, & S_2=\frac{2}{3}M_1+\frac{1}{3}T_1; \\
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M_2,\ \ \ \ & T_4=\frac{1}{2}M_2+\frac{1}{2}T_2, & S_4=\frac{2}{3}M_2+\frac{1}{3}T_2; \\
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M_4,\ \ \ \ & T_8=\frac{1}{2}M_4+\frac{1}{2}T_4, & S_8=\frac{2}{3}M_4+\frac{1}{3}T_4; \\
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\dots,\ \ \ \ & \dots, & \dots
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\end{align*}
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\end{remark}
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\chapter{The Riemann Integral}
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\subimport{./}{riemann-integral.tex}
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\subimport{./}{riemann-integrable-functions.tex}
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\subimport{./}{the-fundamental-theorem.tex}
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\subimport{./}{the-darboux-integral.tex}
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\subimport{./}{approximate-integration.tex}
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\section{Riemann Integrable Functions}
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\begin{theorem}[\textbf{Cauchy Criterion}]
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A function: $[a,b] \to \R$ belongs to $\mathcal{R}[a,b]$ if and only if for every $\varepsilon >0$ there exists $\eta_\varepsilon > 0$ such that if $\dot{\mathcal{P}}$ and $\dot{\mathcal{Q}}$ are any tagged partitions of $[a,b]$ with $||\dot{\mathcal{P}}||<\eta_\varepsilon$ and $||\dot{\mathcal{Q}}||<\eta_\varepsilon$, then
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\[|S(f;\dot{\mathcal{P}})-S(f;\dot{\mathcal{Q}})|<\varepsilon\]
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\end{theorem}
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\begin{theorem}[\textbf{Squeeze Theorem}]
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Let $f:[a,b] \to \R$. Then $f \in \mathcal{R}[a,b]$ if and only if for every $\varepsilon>0$ there exist functions $\alpha_\varepsilon$ and $\omega_\varepsilon$ in $\mathcal{R}[a,b]$ with
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\[\alpha_\varepsilon(x) \leq f(x) \leq \omega_\varepsilon(x)\ \forall\ x \in [a,b]\]
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and such that
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\[\displaystyle\int_{a}^{b}(\omega_\varepsilon-\alpha_\varepsilon)<\varepsilon\]
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\end{theorem}
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\begin{lemma}
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If $J$ is a subinterval of $[a,b]$ having endpoints $c < d$ and if $\varphi_J(x):=1$ for $x \in J$ and $\varphi_J(x):=0$ elsewhere in $[a,b]$, then $\varphi_J \in \mathcal{R}[a,b]$ and $\displaystyle\int_{a}^{b}\varphi_J=d-c$.
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\end{lemma}
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\begin{theorem}
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If $\varphi:[a,b] \to \R$ is a step function, then $\varphi \in \mathcal{R}[a,b]$.
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\end{theorem}
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\begin{theorem}
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If $f:[a,b] \to \R$ is continuous on $[a,b]$, then $f \in \mathcal{R}[a,b]$.
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\end{theorem}
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\begin{theorem}
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If $f:[a,b]\to\R$ is monotone on $[a,b]$, then $f \in \mathcal{R}[a,b]$.
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\end{theorem}
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\begin{theorem}[\textbf{Additivity Theorem}]
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Let $f:=[a,b] \to \R$ and let $c \in (a,b)$. Then $f \in \mathcal{R}[a,b]$ if and only if its restrictions to $[a,c]$ and $[c,b]$ are both Riemann integrable. In this case
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\[\displaystyle\int_{a}^{b}f=\displaystyle\int_{a}^{c}f+\displaystyle\int_{c}^{b}f\]
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\end{theorem}
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\begin{corollary}
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If $f \in \mathcal{R}[a,b]$, and if $[c,d]\subseteq [a,b]$, then the restriction of $f$ to $[c,d]$ is in $\mathcal{R}[c,d]$.
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\end{corollary}
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\begin{corollary}
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If $f \in \mathcal{R}[a,b]$ and if $a=c_0<c_1<\dots<c_m=b$, then the restrictions of $f$ to each of the subintervals $[c_{i-1},c_i]$ are Riemann integrable and
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\[\displaystyle\int_{a}^{b}f=\sum\limits_{i=1}^{m}\displaystyle\int_{c_{i-1}}^{c_i}f\]
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\end{corollary}
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\begin{definition}
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If $f \in \mathcal{R}[a,b]$ and if $\alpha, \beta \in [a,b]$ with $\alpha < \beta$, we define
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\[\displaystyle\int_{\beta}^{\alpha}f:=-\displaystyle\int_{\alpha}^{\beta}f\ \text{ and }\ \displaystyle\int_{\alpha}^{\alpha}f:=0\]
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\end{definition}
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\begin{theorem}
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If $f \in \mathcal{R}[a,b]$ and if $\alpha,\beta,\gamma$ are any numbers in $[a,b]$, then
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\[\displaystyle\int_{\alpha}^{\beta}f=\displaystyle\int_{\alpha}^{\gamma}f+\displaystyle\int_{\gamma}^{\beta}f\]
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in the sense that the existence of any two of these integrals implies the existence of the third integral and the equality.
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\end{theorem}
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\section{Riemann Integral}
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\begin{definition}
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If $I:=[a,b]$ is a closed bounded interval in $\R$, then a \textbf{partition} of $I$ is a finite, ordered set $\mathcal{P}:=(x_0, x_1, \dots, x_{n-1}, x_n)$ of point in $I$ such that
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\[a = x_0 < x_1 < \dots < x_{n-1} < x_n = b\]
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Often we will denote the partition $\mathcal{P}$ by the notation $\mathcal{P}=\{[x_{i-1},x_i]\}_{i=1}^n$. We define the \textbf{norm} (or \textbf{mesh}) of $\mathcal{P}$ to be the number
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\[||\mathcal{P}||:=\max\{x_1-x_0, x_2-x_1, \dots, x_n-x_{n-1}\}\]
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Thus the norm {of a partition is merely the length of the largest subinterval into which the partition divides $[a,b]$. Clearly, many partitions have the same norm, so the partition is \textit{not}} a function of the norm.
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\\If a point $t_i$ has been selected from each subinterval $I_i=[x_{i-1},x_i]$, for $i=1,2,\dots,n$, then the points are called \textbf{tags} of the subintervals of $I_i$. A set of ordered pairs
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\[\dot{\mathcal{P}}:=\{([x_{i-1},x_i],t_i)\}_{i=1}^{n}\]
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of subintervals and corresponding tags is called a \textbf{tagged partition} of $I$.
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\\If $\dot{\mathcal{P}}$ is the tagged partition given above, we define the \textbf{Riemann sum} of a function $f:[a,b] \to \R$ corresponding to $\dot{\mathcal{P}}$ to be the number
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\[S(f;\dot{\mathcal{P}}):=\sum\limits_{i=1}^{n} f(t_i)(x_i-x_{i-1})\]
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We will also use this notation when $\dot{\mathcal{P}}$ denotes a \textit{subset} of a partition, and not the entire partition.
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\end{definition}
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\begin{definition}
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A function $f:[a,b] \to \R$ is said to be \textbf{Riemann integrable} on $[a,b]$ if there exists a number $L \in \R$ such that for every $\varepsilon >0$ there exists $\delta_\varepsilon >0$ such that if $\dot{\mathcal{P}}$ is any tagged partition of $[a,b]$ with $||\dot{\mathcal{P}}||<\delta_\varepsilon$, then
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\[|S(f;\dot{\mathcal{P}})-L|<\varepsilon\]
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The set of all Riemann integrable functions on $[a,b]$ will be denoted by $\mathcal{R}[a,b]$.
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\end{definition}
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\begin{remark}
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It is sometimes said that the integral $L$ is "the limit" of the Riemann sums $S(f:\dot{\mathcal{P}})$ as the norm $||\dot{\mathcal{P}}|| \to 0$. However, since $S(f;\dot{\mathcal{P}})$ is not a function of $||\dot{\mathcal{P}}||$, this \textbf{limit} is not of the type that we have studied before.
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\\\\First we will show that if $f \in \mathcal{R}[a,b]$, then the number $L$ is uniquely determined. It will be called the \textbf{Riemann integral of $f$} over $[a,b]$. Instead of $L$, we will usually write
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\[L=\int_{a}^{b}f\ \text{ or }\ \int_{a}^{b}f(x)dx\]
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\end{remark}
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\begin{theorem}
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If $f \in \mathcal{R}[a,b]$, then the value of the integral is uniquely determined.
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\end{theorem}
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\begin{theorem}
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If $g$ is Riemann integrable on $[a,b]$ and if $f(x)=g(x)$ except for a finite number of points in $[a,b]$, then $f$ is Riemann integrable and $\displaystyle\int_{a}^{b}f=\displaystyle\int_{b}^{a}g$.
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\end{theorem}
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\begin{theorem}
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Suppose that $f$ and $g$ are in $\mathcal{R}[a,b]$. Then:
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\begin{enumerate}
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\item If $k \in \R$, the function $kf$ is in $\mathcal{R}[a,b]$ and
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\[\displaystyle\int_{a}^{b}kf=k\displaystyle\int_{a}^{b}f\]
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\item The function $f+g$ is in $\mathcal{R}[a,b]$ and
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\[\displaystyle\int_{a}^{b}(f+g)=\displaystyle\int_{a}^{b}f+\displaystyle\int_{a}^{b}g\]
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\item If $f(x) \leq g(x)$ for all $x \in [a,b]$, then
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\[\displaystyle\int_{a}^{b}f \leq \displaystyle\int_{a}^{b}g\]
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\end{enumerate}
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\end{theorem}
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\begin{theorem}
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If $f \in \mathcal{R}[a,b]$, then $f$ is bounded on $[a,b]$.
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\end{theorem}
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\section{The Darboux Integral}
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\begin{definition}[\textbf{Upper and Lower Sums}]
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Let $f:I \to \R$ be a bounded function on $I=[a,b]$ and let $\mathcal{P}=(x_0,x_1,\dots,x_n)$ be a partition of $I$. for $k=1,2,\dots,n$ we let
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\[m_k:=\inf \{f(x):x \in [x_{k-1},x_k]\},\ \ \ \ M_k:=\sup \{f(x):x \in [x_{k-1},x_k]\}\]
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The \textbf{lower sum} of $f$ corresponding to the partition $\mathcal{P}$ is defined to be
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\[L(f;\mathcal{P}):= \sum\limits_{k=1}^{n}m_k(x_k-x_{k-1})\]
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and the \textbf{upper sum} of $f$ corresponding to $\mathcal{P}$ is defined to be
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\[U(f;\mathcal{P}):=\sum\limits_{k=1}^{n}M_k(x_k-x_{k-1})\]
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\end{definition}
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\begin{lemma}
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If $f:=I\to\R$ is bounded and $\mathcal{P}$ is any partition of $I$, then $L(f;\mathcal{P})\leq U(f;\mathcal{P})$.
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\end{lemma}
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\begin{definition}
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If $\mathcal{P}:=(x_0,x_1,\dots,x_n)$ and $\mathcal{Q}:=(y_0,y_1,\dots,y_m)$ are partitions of $I$, we say that $\mathcal{Q}$ is a \textbf{refinement of } $\mathcal{P}$ if each partition point $x_k \in \mathcal{P}$ also belongs to $\mathcal{Q}$ (that is, if $\mathcal{P} \subseteq \mathcal{Q}$). A refinement $\mathcal{Q}$ of a partition $\mathcal{P}$ can be obtained by adjoining a finite number of points to $\mathcal{P}$.
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\end{definition}
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\begin{lemma}
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If $f:I\to\R$ is bounded, if $\mathcal{P}$ is a partition of $I$, and if $Q$ is a refinement of $\mathcal{P}$, then
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\[L(f;\mathcal{P})\leq L(f;\mathcal{Q})\ \text{ and }\ U(f;\mathcal{Q})\leq U(f;\mathcal{P})\]
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\end{lemma}
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\begin{lemma}
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Let $f:I\to\R$ be bounded. If $\mathcal{P}_1,\mathcal{P}_2$ are any two partitions of $I$, then $L(f;\mathcal{P}_1)\leq U(f;\mathcal{P}_2)$.
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\end{lemma}
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\begin{definition}
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We shall denote the collection of all partitions of the interval $I$ by $\mathscr{P}(I)$. Let $I:=[a,b]$ and let $f:I \to \R$ be a bounded function. The \textbf{lower integral of $f$ on $I$} is the number
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\[L(f):=\sup\{L(f;\mathcal{P}):\mathcal{P} \in \mathscr{P}(I)\}\]
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and the \textbf{upper integral of $f$ on $I$} is the number
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\[U(f):=\inf\{U(f;\mathcal{P}):\mathcal{P} \in \mathscr{P}(I)\}\]
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\end{definition}
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\begin{theorem}
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Let $I=[a,b]$ and let $f:I \to \R$ be a bounded function. Then the lower integral $L(f)$ and the upper integral $U(f)$ of $f$ on $I$ exist. Moreover,
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\[L(f)\leq U(f)\]
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\end{theorem}
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\begin{definition}
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Let $I=[a,b]$ and let $f:I \to \R$ be a bounded function. Then $f$ is said to be \textbf{Darboux integrable on $I$} if $L(f)=U(f)$. In this case the \textbf{Darboux Integral of $f$ over $I$} is defined to be the value $L(f)=U(f)$.
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\end{definition}
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\begin{theorem}[\textbf{Integrability Criterion}]
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Let $I:=[a,b]$ and let $f:I \to \R$ be a bounded function on $I$. Then $f$ is Darboux integrable on $I$ if and only if for each $\varepsilon > 0$ there is a partition $\mathcal{P}_\varepsilon$ of $I$ such that
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\[U(f;\mathcal{P}_\varepsilon)-L(f;\mathcal{P}_\varepsilon)<\varepsilon\]
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\end{theorem}
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\begin{corollary}
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Let $I=[a,b]$ and let $f:I \to \R$ be a bounded function. If $\{P_n:n \in \N\}$ is a sequence of partitions on $I$ such that
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\[\lim\limits_{n}(U(f;P_n)-L(f;P_n))=0,\]
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then $f$ is integrable and $\lim\limits_{n}L(f;P_n)=\displaystyle\int_{a}^{b}f=\lim\limits_{n}U(f;P_n)$.
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\end{corollary}
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\begin{theorem}
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If the function $f$ on the interval $I=[a,b]$ is either continuous or monotone on $I$, then $f$ is Darboux integrable on $I$.
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\end{theorem}
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\begin{theorem}[\textbf{Equivalence Theorem}]
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A function $f$ on $I=[a,b]$ is Darboux integrable if and only if it is Riemann integrable.
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\end{theorem}
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\section{The Fundamental Theorem}
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\begin{theorem}[\textbf{Fundamental Theorem of Calculus (First Form)}]
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Suppose there is a \textbf{finite} set $E$ in $[a,b]$ and functions $f,F:=[a,b] \to \R$ such that
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\begin{enumerate}
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\item $F$ is continuous on $[a,b]$,
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\item $F'(x)=f(x)$ for all $x \in [a,b]\setminus E$,
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\item $f$ belongs to $\mathcal{R}[a,b]$.
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\end{enumerate}
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Then we have
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\[\displaystyle\int_{a}^{b}f=F(b)-F(a)\]
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\end{theorem}
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\begin{remark}
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If the function $F$ is differentiable at every point of $[a,b]$, then (by \textit{Theorem 6.1.2}) hypothesis (a) is automatically satisfied. If $f$ is not defined for some point $c \in E$, we take $f(c):=0$. Even if $F$ is differentiable at every point of $[a,b]$, condition (c) is \textit{not automatically satisfied}, since there exists functions $F$ such that $F'$ is not Riemann integrable.
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\end{remark}
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\begin{definition}
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If $f \in \mathcal{R}[a,b]$, then the function defined by
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\[F(z):=\displaystyle\int_{a}^{z}f\ \text{for}\ z \in [a,b]\]
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is called the \textbf{indefinite integral} of $f$ with \textbf{basepoint} $a$. (sometimes a point other than $a$ is used as a basepoint)
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\end{definition}
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\begin{theorem}
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The indefinite integral $F$ defined by the above definition is continuous on $[a,b]$. In fact, if $|f(x)|\leq M$ for all $ x \in [a,b]$, then $|F(z)-F(w)|\leq M|z-w|$ for all $z,w \in [a,b]$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}[\textbf{Fundamental Theorem of Calculus (Second Form)}]
|
||||
Let $f \in \mathcal{R}[a,b]$ and let $f$ be continuous at a point $c \in [a,b]$. Then the indefinite integral, defined by \textit{Definition 7.3.1}, is differentiable at $c$ and $F'(c)=f(c)$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}
|
||||
If $f$ is continuous on $[a,b]$, then the indefinite integral $F$, defined by \textit{Definition 7.3.1}, is differentiable on $[a,b]$, and $F'(x)=f(x)$ for all $x \in [a,b]$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}[\textbf{Substitution Theorem}]
|
||||
Let $J:=[\alpha, \beta]$ and let $\varphi:J\to\R$ have a continuous derivative on $J$. If $f:I\to\R$ is continuous on an interval $I$ containing $\varphi(J)$, then
|
||||
\[\displaystyle\int_{\alpha}^{\beta}f(\varphi(t))\cdot\varphi'(t)dt=\displaystyle\int_{\varphi(\alpha)}^{\varphi(\beta)}f(x)dx\]
|
||||
\end{theorem}
|
||||
|
||||
\begin{definition}
|
||||
|
||||
\begin{enumerate}
|
||||
\item A set $Z \subset \R$ is said to be a \textbf{null set} if for every $\varepsilon>0$ there exists a countable collection $\{(a_k,b_k)\}_{k=1}^\infty$ of open intervals such that
|
||||
\[Z \subseteq \bigcup_{k=1}^{\infty}(a_k,b_k)\ \text{and}\ \sum\limits_{k=1}^{\infty}(b_k-a_k)\leq \varepsilon\]
|
||||
|
||||
\item If $Q(x)$ is a statement about the point $x \in I$, we say that $Q(x)$ holds \textbf{almost everywhere} on $I$ (or for \textbf{almost every} $x \in I$), if there exists a null set $Z \subset I$ such that $Q(x)$ holds for all $x \in I\setminus Z$. In this case, we may write
|
||||
\[Q(x)\ \text{for a.e. (almost everywhere)}\ x \in I\]
|
||||
\end{enumerate}
|
||||
\end{definition}
|
||||
|
||||
\begin{theorem}[\textbf{Lebesgue's Integrability Criterion}]
|
||||
A bounded function $f:[a,b] \to \R$ is Riemann integrable if and only if it is continuous almost everywhere on $[a,b]$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}[\textbf{Composition Theorem}]
|
||||
Let $f \in \mathcal{R}[a,b]$ with $f([a,b])\subseteq [c,d]$ and let $\varphi:[c,d] \to \R$ be continuous. Then the composition $\varphi \circ f$ belongs to $\mathcal{R}[a,b]$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{corollary}
|
||||
Suppose that $f \in \mathcal{R}[a,b]$. Then its absolute value $|f|$ is in $\mathcal{R}[a,b]$, and
|
||||
\[\left|\displaystyle\int_{a}^{b}f\right|\leq\displaystyle\int_{a}^{b}|f|\leq M(b-a),\]
|
||||
where $|f(x)|\leq M$ for all $x \in [a,b]$.
|
||||
\end{corollary}
|
||||
|
||||
\begin{theorem}[\textbf{The Product Theorem}]
|
||||
If $f$ and $g$ belong to $\mathcal{R}[a,b]$, then the product $fg$ belongs to $\mathcal{R}[a,b]$.
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}[\textbf{Integration by Parts}]
|
||||
Let $F,G$ be differentiable on $[a,b]$ and let $f:=F'$ and $g:=G'$ belong to $\mathcal{R}[a,b]$. Then
|
||||
\[\left.\displaystyle\int_{a}^{b}fG=FG\ \right|_a^b-\displaystyle\int_{a}^{b}Fg\]
|
||||
\end{theorem}
|
||||
|
||||
\begin{theorem}[\textbf{Taylor's Theorem with the Remainder}]
|
||||
Suppose that $f',\dots,f^{(n)},f^{(n+1)}$ exist on $[a,b]$ and that $f^{(n+1)} \in \mathcal{R}[a,b]$. Then we have
|
||||
\[f(b)=f(a)+\frac{f'(a)}{1!}(b-a)+\dots+\frac{f^{(n)}(a)}{n!}(b-a)^n+R_n,\]
|
||||
where the remainder is given by
|
||||
\[R_n=\frac{1}{n!}\displaystyle\int_{a}^{b}f^{(n+1)}(t)\cdot(b-t)^n dt\]
|
||||
\end{theorem}
|
||||
Reference in New Issue
Block a user