%!TEX root = main.tex
\documentclass[main.tex]{subfiles}

\begin{document}
	\section{Integrals}

	\subsection{Introduction}

	We seek to come up with a formal definition for the area between the $x$-axis and a positive function. For the concept of area to make sense, we will assume that all functions discussed in this chapter are \textit{bounded}, unless stated otherwise.

	\begin{figure}[H]
		\centering
		\begin{asy}import graph;
			size(8cm);
			real f(real x){ return sin(0.5*x)-(0.1*x)+5; }

			draw(graph(f,0,12));

			draw((0,0)--(12,0));
			label("$y=f(x)$", (12,f(12)),E);
			draw((0,0)--(0,f(0)), dashed);
			draw((12,0)--(12,f(12)), dashed);
			label("$a$",(0,0),S);
			label("$b$",(12,0),S);
		\end{asy}
	\end{figure}

	First, we introduce new definitions to formalize the concept of the area under the curve.

	\begin{definition}
		Given an interval $[a,b]$, a \vocab{partition} $P$ is a set of points
		\[a=x_0 < x_1 < x_2 < \ldots < x_n = b.\]
	\end{definition}

	By dividing the interval into smaller subintervals, we are essentially chopping up the region beneath $y=f(x)$ into \textit{rectangles} that either leave remaining space in the intended region or cover the entire region plus some extra space. Here is an illustration of the former case:

	\begin{figure}[H]
		\centering
		\begin{asy}import graph;
		size(8cm);
		real f(real x){ return sin(0.5*x)-(0.1*x)+5; }

		draw(graph(f,0,12));

		draw((0,0)--(12,0));
		label("$y=f(x)$", (12,f(12)),E);
		draw((0,0)--(0,f(0)));
		draw((0,f(0))--(1.75,f(0)));
		draw((1.74,0)--(1.74,f(4.2)));
		draw((1.74,f(4.2))--(4.2,f(4.2)));
		draw((4.2,0)--(4.2,f(4.2)));
		draw((4.2,f(7.77))--(7.77,f(7.77)));
		draw((7.77,0)--(7.77,f(7.77)));
		draw((7.77,f(9.827))--(11,f(9.827)));
		draw((11,0)--(11,f(9.827)));
		draw((11,f(9.827))--(11,f(11)));
		draw((11,f(11))--(12,f(11)));
		draw((12,0)--(12,f(11)));
		label("$a$",(0,0),S);
		label("$x_1$",(1.74,0),S);
		label("$x_2$",(4.2,0),S);
		label("$x_3$",(7.77,0),S);
		label("$x_4$",(11,0),S);
		label("$b$",(12,0),S);
		\end{asy}
	\end{figure}

	How would we know where to extend the rectangles to? In this case, for a subinterval $[x_{i-1},x_i]$, the height of the corresponding rectangle would be the minimum value of $f(x)$ on $[x_{i-1},x_i]$.

	We have assumed $f(x)$ to be bounded, but it is not necessarily continuous (although we have assumed it to be continuous in the diagram above). Therefore, it would not be appropriate to use $\min$ or $\max$ to define the height of the rectangles, but rather, $\inf$ and $\sup$.

	Thus, the height of a rectangle on $[x_{i-1},x_i]$ in this case would be $\inf\left( \{f(x)\mid x_{i-1} \leq x \leq x_i \} \right)$. Then, the area of an individual rectangle would be $(x_i-x_{i-1})\cdot \inf\left( \{f(x)\mid x_{i-1} \leq x \leq x_i \} \right)$. This is certainly an unwieldly expression, so we introduce convenient notation for brevity.

	\begin{definition}
		On a subinterval $[x_{i-1},x_i]$, let $m_i = \inf\left( \{f(x)\mid x_{i-1} \leq x \leq x_i \} \right)$.
	\end{definition}

	Then the area of a particular rectangle in this diagram would be $m_i (x_i-x_{i-1})$. Note that all of these rectangles are below the graph, leading to our expression for the sum of all the areas of the rectangles:

	\begin{definition}
		For a function $f$ and partition $P$ of $n$ points, we define the \vocab{lower sum}, or the sum of the areas of all rectangles below the graph, as \[L(f,P) = \sum\limits^n_{i=1} m_i (x_i-x_{i-1}).\]
	\end{definition}

	Now consider the other case, where the rectangles cover the entire region in addition to some extra space:

	\begin{figure}[H]
		\centering
		\begin{asy}import graph;
		size(8cm);
		real f(real x){ return sin(0.5*x)-(0.1*x)+5; }

		draw(graph(f,0,12));

		draw((0,0)--(12,0));

		label("$y=f(x)$", (12,f(12)),E);

		draw((0,0)--(0,f(1.74)));
		draw((0,f(1.74))--(1.74,f(1.74)));
		draw((1.74,0)--(1.74,f(2.739)));
		draw((1.74,f(2.739))--(4.2,f(2.739)));
		draw((4.2,0)--(4.2,f(2.739)));
		draw((4.2,f(4.2))--(7.77,f(4.2)));
		draw((7.77,0)--(7.77,f(4.2)));
		draw((7.77,f(7.77))--(11,f(7.77)));
		draw((11,0)--(11,f(7.77)));
		draw((11,f(7.77))--(11,f(11)));
		draw((11,f(12))--(12,f(12)));
		draw((12,0)--(12,f(12)));
		label("$a$",(0,0),S);
		label("$x_1$",(1.74,0),S);
		label("$x_2$",(4.2,0),S);
		label("$x_3$",(7.77,0),S);
		label("$x_4$",(11,0),S);
		label("$b$",(12,0),S);
		\end{asy}
	\end{figure}

	Now, the height of a rectangle on $[x_{i-1},x_i]$ is the maximum value of $f(x)$ on that subinterval. Thus, we establish analogous definitions.

	\begin{definition}
		On a subinterval $[x_{i-1},x_i]$, let $M_i = \sup\left( \{f(x)\mid x_{i-1} \leq x \leq x_i \} \right)$.
	\end{definition}

	Likewise, we consider the sum of all areas of the rectangles, which are above the graph in this case.

	\begin{definition}
		For a function $f$ and partition $P$ of $n$ points, we define the \vocab{upper sum}, or the sum of the areas of all rectangles above the graph, as \[U(f,P) = \sum\limits^n_{i=1} M_i (x_i-x_{i-1}).\]
	\end{definition}

	Before moving on to discussing the purpose of defining these sums, we first state and prove some expected results related to these.

	The following lemma is a symbolic representation of an intuitive idea: if we chop up the region into more rectangles, the lower sum is greater than before (so it is approaching the actual area from below) and the upper sum is lesser than before (so it is approaching the actual area from above).

	\begin{lemma}
		\label{lem:chop}
		Suppose $P \subseteq P'$, where $P$ and $P'$ are partitions. Then $L(f,P) \leq L(f,P')$ and $U(f,P) \geq U(f,P')$.
	\end{lemma}
	\begin{proof}
		First, let $[a,b]$ be the interval for the partitions $P$ and $P'$. We will prove $L(f,P) \leq L(f,P')$ by induction, and the proof will be analogous for upper sums.

		To start, we will show that the inequality holds for $P'$ containing exactly one more point than $P$.

		Consider $x^\star \in [a,b]$, where $x^\star \notin P$ and let $P' = P \cup \{x^\star\}$. Note that we represent the partition $P$ as
		\[a = x_0 < x_1 < x_2 < \ldots < x_n = b.\]
		Suppose $x_{k-1} < x^\star < x_k$, such that $P'$ can be represented as
		\[a = x_0 < \ldots < x_{k-1} < x^\star < x_k < \ldots < x_n = b.\]
		Then we have
		\begin{align*}
		L(f,P) &= \sum\limits^n_{i=1} m_i (x_i-x_{i-1}),\\
		L(f,P') &= \sum\limits^{k-1}_{i=1} m_i (x_i-x_{i-1}) + m_k^\star (x^\star - x_{k-1}) + m_k^{\star \star} (x_k - x^\star) + \sum\limits^n_{i=k+1} m_i (x_i-x_{i-1}),
		\end{align*}
		where we define
		\begin{align*}
		m_k^\star &= \inf\left( \{f(x) \mid x_{k-1} \leq x \leq x^\star \} \right),\\
		m_k^{\star\star} &= \inf\left( \{f(x) \mid x^\star \leq x \leq x_k \} \right).
		\end{align*}

		It follows that $L(f,P')$ will share the same summands as $L(f,P)$, except that $m_k(x_k-x_{k-1})$ is replaced by $m_k^\star (x^\star - x_{k-1}) + m_k^{\star \star} (x_k - x^\star)$. Thus, to prove $L(f,P) \leq L(f,P')$, it suffices to prove that
		\[m_k(x_k-x_{k-1}) \leq m_k^\star (x^\star - x_{k-1}) + m_k^{\star \star} (x_k - x^\star),\]
		which can be rewritten as
		\[m_k (x^\star - x_{k-1}) + m_k (x_k - x^\star) \leq m_k^\star (x^\star - x_{k-1}) + m_k^{\star \star} (x_k - x^\star).\]
		But note that $[x_{k-1},x^\star] \subset [x_{k-1},x_k]$ and $[x^\star,x_k] \subset [x_{k-1},x_k]$, so we conclude that $m_k \leq m_k^\star$ and $m_k \leq m_k^{\star\star}$, and the inequality is clearly true as a result.

		We have shown that $L(f,P) \leq L(f,P')$ holds for $P'$ containing exactly one more point than $P$. However, for a general $P' \supseteq P$, we know that $P$ and $P'$ are finite collections of points, so we can simply inductively add points to $P$ until the partition becomes $P'$ (since the inequality will always hold in this manner).
	\end{proof}

	\begin{theorem}
		For any partitions $P$ and $Q$, $L(f,P) \leq U(f,Q)$.
	\end{theorem}
	\begin{proof}
		Note that $P \subset P\cup Q$, so by \Cref{lem:chop}, $L(f,P) \leq L(f,P\cup Q)$. Obviously, $L(f,P \cup Q) \leq U(f,P\cup Q)$ since the supremum of $f(x)$ on a subinterval is always greater than or equal to the infimum of $f(x)$ on that same subinterval. Lastly, we apply \Cref{lem:chop} again to get $U(f,P\cup Q) \leq U(f,Q)$. We can now string these inequalities together to conclude that $L(f,P) \leq U(f,Q)$.
	\end{proof}

	This theorem establishes that the set of all lower sums is bounded above by any upper sum, and vice-versa. Thus, by \Cref{def:derrick}, $\sup\left( \{L(f,P) \mid P \text{ is a partition of } [a,b]\}\right)$ and $\inf\left( \{U(f,P) \mid P \text{ is a partition of } [a,b]\}\right)$ exist.

	If the area, $A$, of the region between $y=f(x)$ and the $x$-axis is to be meaningful, we should have \[\forall \text{ partitions } P,\ L(f,P) \leq A \leq U(f,P).\] Furthermore, there should be \textit{only one value} of $A$ for which this is true (it would be unreasonable to have multiple possible values represent the area!).

	Note that \[\forall \text{ partitions } P,\ L(f,P) \leq \sup\left( \{L(f,P)\}\right) \leq \inf\left( \{U(f,P)\}\right) \leq U(f,P),\] so it would only be reasonable to have $A = \sup\left( \{L(f,P)\}\right) = \inf\left( \{U(f,P)\}\right)$. This leads us into our much-awaited definition:

	\begin{definition}
		The function $f$ is \vocab{integrable} on $[a,b]$ if \[\sup\left( \{L(f,P) \mid P \text{ is a partition of } [a,b]\}\right)=\inf\left( \{U(f,P) \mid P \text{ is a partition of } [a,b]\}\right).\]

		If $f$ is integrable, then we define the \vocab{integral} of $f$ on $[a,b]$ to be equal to this common value, and denote it as
		\[\int\limits_{a}^{b} f.\]
	\end{definition}

	\begin{problem}
		Define $f(x) = 1$. Is $f$ integrable on $[0,1]$? What about \[f(x) = \left\{\begin{aligned}&1 & x&\in\QQ\\ &0 & x&\not\in\QQ\end{aligned}\right.?\]
	\end{problem}
	\begin{soln}
		For the first function, no matter how we choose the partition of points, we will always have $L(f,P)=U(f,P) = 1$. Thus, we can say that the area of the region between $f$ and the $x$-axis is $1$, i.e. \[\int\limits_{0}^{1} f = 1.\] This is consistent with our previous knowledge on the area of the unit square, which is $1$.

		Now consider the second function. For some arbitrary partition $P$, consider the subinterval $[x_{i-1},x_i]$. By previous results, we know there exists an irrational number in $[x_{i-1},x_i]$, so $f(x)$ will be $0$ somewhere on that interval. Thus, $m_i$ is automatically $0$, since the function can only either be $0$ or $1$.

		Likewise, we know that there is a rational number in $[x_{i-1},x_i]$, so $M_i$ is automatically $1$. Thus, $L(f,P)=0$ and $U(f,P)=1$. However, there is no \textit{unique} number between $0$ or $1$, and it follows that $f$ is not integrable.
	\end{soln}

	Now recall the result of \Cref{thm:supeqinf}. This immediately leads to another condition for integrability:
	\begin{theorem}
		\label{thm:intweak}
		$f$ is integrable on $[a,b]$ $\lliff$ \[\forall \eps > 0\ \exists P,Q \text{ s.t. } U(f,Q) - L(f,P) < \eps.\]
	\end{theorem}

	We can now use this result to prove a slightly stronger theorem:

	\begin{theorem}
		\label{thm:intstrong}
		$f$ is integrable on $[a,b]$ $\lliff$ \[\forall \eps > 0\ \exists P \text{ s.t. } U(f,P) - L(f,P) < \eps.\]
	\end{theorem}
	\begin{proof}
		The left direction is simple, by \Cref{thm:intweak} (let $Q= P$).
		Now we prove the right direction. Given $\eps >0$, by \Cref{thm:intweak}, there exist $R$ and $S$ such that $U(f,S) - L(f,R) < \eps$. Let $P=R\cup S$. We have
		\begin{align*}
		U(f,P) \leq U(f,S),\\
		L(f,P) \geq L(f,R).
		\end{align*}
		Thus, $U(f,P) - L(f,P) \leq U(f,S) - L(f,R) < \eps$, and we are done.
	\end{proof}
    We will also prove a theorem that will allow us to calculate integrals more easily.
    \begin{theorem}
        \label{thm:subsetpart}
        If $A,B$ are sets with $A\leq B$ and $\widetilde{A}\subseteq A$ and $\widetilde{B}\subseteq B$ satisfy $\text{sup}\widetilde{A}=\text{inf}\widetilde{B},$ $\text{sup}A,\text{inf}B$ are both equal to this common value.
    \end{theorem}
    \begin{proof}
        Clearly $\text{sup} A\geq \text{sup} \widetilde{A}$ and $\text{inf} B\leq \text{inf} \widetilde{B}.$ But as $A\leq B,$ we have $\text{sup}A\leq \text{inf}B.$ Thus
        \[\text{sup}\widetilde{A}\leq \text{sup}A\leq \text{inf}B\leq \text{inf}\widetilde{B},\]
        but as the extremes are equal, it follows that everything in between must equal the common value.
    \end{proof}
    We apply these theorems to calculate the area under a parabola.
    \begin{problem}
        For $a>0,$ show $\int_0^a x^2=\tfrac{a^3}{3}.$
    \end{problem}
    \begin{proof}
        In order to approximate the area with lower and upper sums, it will be convenient to consider evenly dividing partitions. Define $P_n=\{0,\tfrac{a}{n},\tfrac{2a}{n},\cdots, a\}.$ We also note that as $x^2$ is increasing, $\sup\{f(x)\mid x_{i-1}\leq x\leq x_i\}=x_i^2,\inf\{f(x)\mid x_{i-1}\leq x\leq x_i\}=x_{i-1}^2.$ \\ \\
        Then by definition,
        \begin{align*}
            U(x^2,P_n)&=\sum_{i=1}^nx_i^2(x_i-x_{i-1})\\
            &= \sum_{i=1}^n\left(\frac{ai}{n}\right)^2\cdot \frac{a}{n} \\
            &=\frac{a^3}{n^3}\sum_{i=1}^n i^2\\
            &=\frac{a^3}{n^3}\cdot \frac{n(n+1)(2n+1)}{6}\\
            &=\frac{a^3}{6}\cdot \left(2+\frac{3}{n}+\frac{1}{n^2}\right).
        \end{align*}
    This shows that $\inf U(x^2,P_n)=\tfrac{a^3}{3}$ (the sequence is decreasing and the limit is clearly $\tfrac{a^3}{3}$). Similarly, we can show $\sup L(x^2,P_n)=\tfrac{a^3}{3},$ so by \Cref{thm:subsetpart} the integral exists and is equal to $\tfrac{a^3}{3}.$
    \end{proof}
    \subsection{Conditions for Integrability}
    In this section, we will show that nondecreasing/nonincreasing functions and continuous functions are always integrable. First, we will introduce a useful definition:
    \begin{definition}
    Define the \vocab{span} of $f$ on $[a,b]$ to be $\sup \{f(x)\mid x\in [a,b]\}-\inf\{f(x)\mid x\in [a,b]\}.$
    \end{definition}
    This definition is useful in proving integrability because it accounts for the differences between the upper and lower sums in the subintervals. This gives
    \[U(f,P)-L(f,P)=\sum_{i=1}^n(x_i-x_{i-1})\cdot \text{span}(f\text{ on }[x_{i-1},x_i])\]
    \begin{theorem}
        \label{thm:nondecint}
        If $f$ is nondecreasing on $[a,b],$ it is integrable on $[a,b].$
    \end{theorem}
    \begin{proof}
    To prove this, we first prove the Small-span lemma:
    \begin{lemma}[Small-span]
        \label{lemma:smallspan}
        If $f$ is continuous on $[a,b],$ for all $\eps>0$ there exists a partition $P=(x_0,x_1,\cdots, x_n)$ such that the span of $f$ on each $[x_{i-1},x_i]$ is less than $\eps.$
    \end{lemma}
    \begin{proof}
        Fix $\eps>0,$ and call a partition \textit{suitable} if the span of $f$ on each subinterval is less than $\eps.$
        Let $S$ be the set $\{a\leq x\leq b\mid [a,x]\text{ has a suitable partition}\}.$ By continuity at $a,$ there is some $a+\delta\in S.$ Moreover $S$ is bounded above by $b,$ so $c=\sup S$ exists. \\ \\
        But by continuity at $c,$ we can choose $\delta$ such that the span of $f$ on $[c-\delta,c+\delta]$ is less than $\eps.$ Moreover, as $S$ contains every number in $[a,c),$ $c-\delta\in S$ and there exists a suitable partition from $[a,c-\delta].$ Gluing this partition with $[c-\delta,c+\delta]$ shows that $[a,c]$ has a suitable partition, contradicting the maximality of $c.$ This shows that $c=b$ is necessary. Now using continuity at $b$ shows that $b\in S,$ so $[a,b]$ has a suitable partition, as desired.
    \end{proof}
    Now the theorem is easy. Pick $P$ such that the span of $f$ on each subinterval is less than $\tfrac{\eps}{b-a}.$ Then
    \begin{align*}
        U(f,P)-L(f,P)&=\sum_{i=1}^n(x_i-x_{i-1})\cdot \text{span}(f\text{ on }[x_{i-1},x_i])\\
        &<\sum_{i=1}^n(x_i-x_{i-1})\cdot \frac{\eps}{b-a}\\
        &=(b-a)\cdot\frac{\eps}{b-a}=\eps,
    \end{align*}
    proving that $f$ is integrable.
    \end{proof}
    \begin{theorem}
        \label{thm:nondecint}
        If $f$ is nondecreasing on $[a,b],$ $f$ is integrable on $[a,b]/$
    \end{theorem}
	\begin{proof}
    	We again consider equal length subintervals and note that $sup\{f(x)\mid x_{i-1}\leq x\leq x_i\}=f(x_i),$ $\inf\{f(x)\mid x_{i-1}\leq x\leq x_i\}=f(x_{i-1}).$ Then
    	\begin{align*}
            U(f,P)-L(f,P)&=\sum_{i=1}^n(x_i-x_{i-1})\cdot \text{span}(f\text{ on }[x_{i-1},x_i])\\
            &= \sum_{i=1}^n\frac{1}{n}(f(x_i)-f(x_{i-1}))
            &=\frac{f(b)-f(a)}{n},
        \end{align*}
        and as this quantity can be made arbitarily small by taking $n$ large, we conclude that $f$ is integrable.
	\end{proof}
	\begin{problem}
	Given $f(x)=\begin{cases}1,\ x\neq 1 \\ 2,\ x=1\end{cases},$ find $\int_0^2f.$
	\end{problem}
	\subsection{Integral Properties}
	Using our intuitive notion of area, we will prove some useful facts about the integral. We first introduce some new notation:
	\begin{definition}
	Define $\int_a^bf(x)dx=\int_a^bf.$
	\end{definition}
	The $dx$ here cannot exist independently. Its meaning will become clear as we prove more results about integrals.
	\begin{theorem}
	    \label{lrshiftint}
	    $\int_a^bf(x)dx=\int_{a+c}^{b+c}f(x-c)dx.$
	\end{theorem}
	\begin{proof}
	This theorem says that we can shift integrals horizontally without changing the area: \\
	%insert diagram
	For any partition $P=(x_0,\cdots, x_n),$ let $P'=(x_0+c,\cdots, x_n+c).$ We claim that $L(f(x),P)=L(f(x-c),P').$ Indeed, each of the sets $\{f(x)\mid x_{i-1}\leq x\leq x_i\}$ are equal to the corresponding sets $\{f(x-c)\mid x_{i-1}+c\leq x\leq x_i+c\},$ so the $\sup,\inf$ are equal as well. The lengths of the subintervals are also unchanged, so both the lower and upper sums are equal. It follows that the integrals must also be equal, as desired.
	\end{proof}
	\begin{theorem}
	    \label{intaddint}
	    For $a<b<c,$ $\int_a^bf+\int_b^cf=\int_a^cf.$
	\end{theorem}



\end{document}