From 6e54175d3aaeee9ad8543ba96804b654c005dbd8 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Tue, 16 May 2023 13:42:37 -0600 Subject: [PATCH] Finish Apostol 1.15. --- Bookshelf/Apostol.tex | 536 ++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 519 insertions(+), 17 deletions(-) diff --git a/Bookshelf/Apostol.tex b/Bookshelf/Apostol.tex index 17ce443..1abdff1 100644 --- a/Bookshelf/Apostol.tex +++ b/Bookshelf/Apostol.tex @@ -57,15 +57,17 @@ Such a number $B$ is also known as the \textbf{greatest lower bound}. Let $s$ be a \nameref{sec:def-step-function} defined on $[a, b]$, and let $P = \{x_0, x_1, \ldots, x_n\}$ be a \nameref{sec:def-partition} of $[a, b]$ such that $s$ is constant on the open subintervals of $P$. -Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval, - so that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k, - \quad k = 1, 2, \ldots, n.$$ +Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval + of $P$, so that + $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k, \quad k + = 1, 2, \ldots, n.$$ The \textbf{integral of $s$ from $a$ to $b$}, denoted by the symbol $\int_a^b s(x)\mathop{dx}$, is defined by the following formula: $$\int_a^b s(x) \mathop{dx} = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}).$$ -Furthermore, we define - $$\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx} \quad\text{and}\quad - \int_a^a s(x) \mathop{dx} = 0.$$ +If $a < b$, then we define + $$\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx}.$$ +If $a = b$, then we define + $$\int_a^b s(x) \mathop{dx} = 0.$$ \section{\defined{Partition}}% \label{sec:def-partition} @@ -1707,8 +1709,10 @@ Then $P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$. $s$ and $t$ remain constant on every open subinterval of $P$. - Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$. - Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $P$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P$. + Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of + $P$. By definition of the \nameref{sec:def-integral-step-function}, \begin{align*} \int_a^b \left[ s(x) + t(x) \right] \mathop{dx} @@ -1735,14 +1739,14 @@ For every real number $c$, we have By definition of a step function, there exists a \nameref{sec:def-partition} $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open subinterval of $P$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P$. Then $c \cdot s$ is a step function with step partition $P$. - By definition of the \nameref{sec:def-integral-step-function}, \begin{align*} \int_a^b c \cdot s(x) \mathop{dx} - & = \sum_{k=1}^n (c \cdot s(x)) \cdot (x_k - x_{k-1}) \\ - & = \sum_{k=1}^n c \cdot (s(x) \cdot (x_k - x_{k-1})) \\ - & = c \sum_{k=1}^n s(x) \cdot (x_k - x_{k-1}) \\ + & = \sum_{k=1}^n c \cdot s_k \cdot (x_k - x_{k-1}) \\ + & = c \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) \\ & = c \int_a^b s(x) \mathop{dx}. \end{align*} @@ -1791,8 +1795,10 @@ If $s(x) < t(x)$ for every $x$ in $[a, b]$, then By construction, $P$ is a step partition for both $s$ and $t$. Thus $s$ and $t$ remain constant on every open subinterval of $P$. - Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$. - Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $P$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P$. + Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of + $P$. By definition of the \nameref{sec:def-integral-step-function}, \begin{align*} \int_a^b s(x) \mathop{dx} @@ -1823,7 +1829,8 @@ Then as a subdivision point. Then $Q$ is a step partition of $s$ and there exists some $0 < i < n$ such that $x_i = c$. - Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $Q$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $Q$. By definition of the \nameref{sec:def-integral-step-function}, \begin{align*} \int_a^b s(x) \mathop{dx} @@ -1857,7 +1864,8 @@ Then By definition of a step function, there exists a \nameref{sec:def-partition} $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open subinterval of $P$. - Let $s_k$ denote the value of $s$ on the $k$th open subinterval of $P$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P$. Let $c$ be a real number. Then $t(x) = s(x - c)$ is a step function on closed interval $[a + c, b + c]$ @@ -1938,7 +1946,7 @@ Then Let $s$ be a step function on closed interval $[a, b]$. Then - $$\int_a^b s(x) \mathop{dx} = - \int_{-b}^{-a} s(-x) \mathop{dx}.$$ + $$\int_a^b s(x) \mathop{dx} = -\int_{-b}^{-a} s(-x) \mathop{dx}.$$ \begin{proof} @@ -1951,4 +1959,498 @@ Then \end{proof} +\chapter{Exercises 1.15}% +\label{chap:exercises-1.15} + +\section{Exercise 1.15.1}% +\label{sec:exercise-1.15.1} + +Compute the value of each of the following integrals. + +\subsection{\partial{Exercise 1.15.1a}}% +\label{sub:exercise-1.15.1a} + +$\int_{-1}^3 \floor{x} \mathop{dx}$. + +\begin{proof} + + Let $s(x) = \floor{x}$ with domain $[-1, 3]$. + By construction, $s$ is a step function with partition + $P = \{-1, 0, 1, 2, 3\} = \{x_0, x_1, x_2, x_3, x_4\}$. + Let $s_k$ denote the constant value $s$ takes on the $k$th open subinterval of + $P$. + By definition of the \nameref{sec:def-integral-step-function}, + \begin{align*} + \int_{-1}^3 \floor{x} \mathop{dx} + & = \sum_{k=1}^4 s_k \cdot (x_k - x_{k-1}) \\ + & = -1 + 0 + 1 + 2 \\ + & = 2. + \end{align*} + +\end{proof} + +\subsection{\partial{Exercise 1.15.1c}}% +\label{sub:exercise-1.15.1c} + +$\int_{-1}^3 \left(\floor{x} + \floor{x + \frac{1}{2}}\right) \mathop{dx}$. + +\begin{proof} + + Let $s(x) = \floor{x} + \floor{x + \frac{1}{2}}$ with domain $[-1, 3]$. + By construction, $s$ is a step function with partition + \begin{align*} + P + & = \{-1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, + 3\} \\ + & = \{x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8\}. + \end{align*} + Let $s_k$ denote the constant value $s$ takes on the $k$th open subinterval of + $P$. + By definition of the \nameref{sec:def-integral-step-function}, + \begin{align*} + \int_{-1}^3 \floor{x} + \floor{x + \frac{1}{2}} + & = \sum_{k=1}^8 s_k \cdot (x_k - x_{k-1}) \\ + & = \frac{1}{2} \sum_{k=1}^8 s_k \\ + & = \frac{1}{2} \left( -2 - 1 + 0 + 1 + 2 + 3 + 4 + 5 \right) \\ + & = 6. + \end{align*} + +\end{proof} + +\subsection{\partial{Exericse 1.15.1e}}% +\label{sub:exercise-1.15.1e} + +$\int_{-1}^3 \floor{2x} \mathop{dx}$. + +\begin{proof} + + Let $s(x) = \floor{2x}$. + By \nameref{sec:hermites-identity}, + $s(x) = \floor{x} + \floor{x + \frac{1}{2}}$. + Thus, by \nameref{sub:exercise-1.15.1c}, + $$\int_{-1}^3 \floor{2x} \mathop{dx} = 6.$$ + +\end{proof} + +\section{\partial{Exercise 1.15.3}}% +\label{sec:exercise-1.15.3} + +Show that + $\int_a^b \floor{x} \mathop{dx} + \int_a^b \floor{-x} \mathop{dx} = a - b$. + +\begin{proof} + + Let $s(x) = \floor{x}$ and $t(x) = \floor{-x}$, both with domain $[a, b]$. + Let $x_1$, $\ldots$, $x_{n-1}$ denote the integers found in interval $(a, b)$. + Then $P = \{x_0, x_1, \ldots, x_n\}$, $x_0 = a$ and $x_n = b$, is a step + \nameref{sec:def-partition} of both $s$ and $t$. + Let $s_k$ and $t_k$ denote the constant values $s$ and $t$ take on the $k$th + open subinterval of $P$ respectively. + By \nameref{sub:exercise-1.11.4b}, $\floor{-x} = -\floor{x} - 1$ for all $x$ + in every open subinterval of $P$. + That is, $s_k = -t_k - 1$. + By definition of the \nameref{sec:def-integral-step-function}, + \begin{align*} + \int_a^b \floor{x} \mathop{dx} + \int_a^b \floor{-x} \mathop{dx} + & = \sum_{k=1}^n s_k (x_k - x_{k-1}) + + \sum_{k=1}^n t_k (x_k - x_{k-1}) \\ + & = \sum_{k=1}^n (x_k - x_{k-1}) \cdot (s_k + t_k) \\ + & = \sum_{k=1}^n (x_k - x_{k-1}) \cdot (-t_k - 1 + t_k) \\ + & = \sum_{k=1}^n (x_{k-1} - x_k) \\ + & = x_0 - x_n \\ + & = a - b. + \end{align*} + +\end{proof} + +\section{Exercise 1.15.5}% +\label{sec:exercise-1.15.5} + +\subsection{\partial{Exercise 1.15.5a}}% +\label{sub:exercise-1.15.5a} + +Prove that $\int_0^2 \floor{t^2} \mathop{dt} = 5 - \sqrt{2} - \sqrt{3}$. + +\begin{proof} + + Let $s(t) = \floor{t^2}$ with domain $[0, 2]$. + Then $s$ is a \nameref{sec:def-step-function} with partition + $P = \{0, 1, \sqrt{2}, \sqrt{3}, 2\} = \{x_0, x_1, \ldots, x_4\}$. + Let $s_k$ denote the constant value that $s$ takes in the $k$th open + subinterval of $P$. + By the \nameref{sec:def-integral-step-function}, + \begin{align*} + \int_0^2 \floor{t^2} \mathop{dt} + & = \sum_{k=1}^4 s_k \cdot (x_k - x_{k-1}) \\ + & = 0 \cdot (1 - 0) + 1 \cdot (\sqrt{2} - 1) + + 2 \cdot (\sqrt{3} - \sqrt{2}) + 3 \cdot (2 - \sqrt{3}) \\ + & = 5 - \sqrt{2} - \sqrt{3}. + \end{align*} + +\end{proof} + +\subsection{\partial{Exercise 1.15.5b}}% +\label{sub:exercise-1.15.5b} + +Compute $\int_{-3}^3 \floor{t^2} \mathop{dt}$. + +\begin{proof} + + Let $s(t) = \floor{t^2}$ with domain $[0, 3]$. + Then $s$ is a \nameref{sec:def-step-function} with \nameref{sec:def-partition} + \begin{align*} + P + & = \{\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, + \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{9}\} \\ + & = \{x_0, x_1, \ldots, x_9\}. + \end{align*} + Let $s_k$ denote the constant value that $s$ takes in the $k$th open + subinterval of $P$. + By the \nameref{sec:def-integral-step-function}, + \begin{align} + \int_0^3 \floor{t^2} \mathop{dt} + & = \sum_{k=1}^9 s_k \cdot (x_k - x_{k-1}) + \nonumber \\ + & = \sum_{k=0}^8 k \cdot (\sqrt{k + 1} - \sqrt{k}). + \label{sub:exercise-1.15.5b-eq1} + \end{align} + We notice $\floor{t^2}$ is symmetric about the $y$-axis. + Thus + \begin{equation} + \label{sub:exercise-1.15.5b-eq2} + \int_{-3}^0 \floor{t^2} \mathop{dt} = \int_0^3 \floor{t^2}. + \end{equation} + By \nameref{sec:step-additivity-with-respect-interval-integration}, + \begin{align*} + \int_{-3}^3 \floor{t^2} \mathop{dt} + & = \int_{-3}^0 \floor{t^2} \mathop{dt} + + \int_0^3 \floor{t^2} \mathop{dt} \\ + & = 2\int_0^3 \floor{t^2} \mathop{dt} + & \eqref{sub:exercise-1.15.5b-eq2} \\ + & = 2 \left[\sum_{k=0}^8 k \cdot (\sqrt{k + 1} - \sqrt{k})\right]. + & \eqref{sub:exercise-1.15.5b-eq1} + \end{align*} + +\end{proof} + +\section{Exercise 1.15.7}% +\label{sec:exercise-1.15.7} + +\subsection{\partial{Exercise 1.15.7a}}% +\label{sub:exercise-1.15.7a} + +Compute $\int_0^9 \floor{\sqrt{t}} \mathop{dt}$. + +\begin{proof} + + Let $s(t) = \floor{\sqrt{t}}$ with domain $[0, 9]$. + Then $s$ is a \nameref{sec:def-step-function} with \nameref{sec:def-partition} + $P = \{0, 1, 4, 9\} = \{x_0, x_1, x_2, x_3\}$. + Let $s_k$ denote the constant value that $s$ takes in the $k$th open + subinterval of $P$. + By the \nameref{sec:def-integral-step-function}, + \begin{align*} + \int_0^9 \floor{\sqrt{t}} \mathop{dt} + & = \sum_{k=1}^3 s_k \cdot (x_k - x_{k-1}) \\ + & = 0 \cdot (1 - 0) + 1 \cdot (4 - 1) + 2 \cdot (9 - 4) \\ + & = 13. + \end{align*} + +\end{proof} + +\subsection{\partial{Exercise 1.15.7b}}% +\label{sub:exercise-1.15.7b} + +If $n$ is a positive integer, prove that + $$\int_0^{n^2} \floor{\sqrt{t}} \mathop{dt} = n(n - 1)(4n + 1) / 6.$$ + +\begin{proof} + + Define predicate $P(n)$ as + \begin{equation} + \label{sub:exercise-1.15.7b-eq1} + \int_0^{n^2} \floor{\sqrt{t}} \mathop{dt} = \frac{n(n - 1)(4n + 1)}{6}. + \end{equation} + We use induction to prove $P(n)$ holds for all integers satisfying $n > 0$. + + \paragraph{Base Case}% + + Let $n = 1$. + Define $s(t) = \floor{\sqrt{t}}$ with domain $[0, 1]$. + Then $s$ is a \nameref{sec:def-step-function} with + \nameref{sec:def-partition} $P = \{0, 1\} = \{x_0, x_1\}$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P$. + By definition of the \nameref{sec:def-integral-step-function}, the left-hand + side of \eqref{sub:exercise-1.15.7b-eq1} evaluates to + \begin{align*} + \int_0^{n^2} \floor{\sqrt{t}} \mathop{dt} + & = \int_0^1 \floor{\sqrt{t}} \mathop{dt} \\ + & = \sum_{k=1}^1 s_k \cdot (x_k - x_{k-1}) \\ + & = 0. + \end{align*} + The right-hand side of \eqref{sub:exercise-1.15.7b-eq1} likewise evaluates + to $0$. + Thus $P(1)$ holds. + + \paragraph{Induction Step}% + + Let $n > 0$ be a positive integer and suppose $P(n)$ is true. + Define $s(t) = \floor{\sqrt{t}}$ with domain $[0, (n + 1)^2]$. + Then $s$ is a \nameref{sec:def-step-function} with + \nameref{sec:def-partition} + \begin{align*} + P + & = \{0, 1, 4, \ldots, n^2, (n + 1)^2\} \\ + & = \{x_0, x_1, \ldots, x_n, x_{n + 1}\}. + \end{align*} + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P$. + By definition of the \nameref{sec:def-integral-step-function}, it follows + that + \begin{align*} + & \int_0^{(n + 1)^2} s(x) \mathop{dx} \\ + & = \sum_{k=1}^{n + 1} s_k \cdot (x_k - x_{k-1}) \\ + & = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}) + + \left[s_{n+1} \cdot (x_{n + 1} - x_n)\right] \\ + & = \int_0^{n^2} s(x) \mathop{dx} + + \left[s_{n+1} \cdot (x_{n + 1} - x_n)\right] \\ + & = \int_0^{n^2} s(x) \mathop{dx} + + \left[ n \cdot ((n + 1)^2 - n^2) \right] \\ + & = \int_0^{n^2} s(x) \mathop{dx} + \left[ 2n^2 + n \right] \\ + & = \frac{n(n - 1)(4n + 1)}{6} + 2n^2 + n + & \text{induction hypothesis} \\ + & = \frac{n(n - 1)(4n + 1) + 12n^2 + 6n}{6} \\ + & = \frac{4n^3 + 9n^2 + 5n}{6} \\ + & = \frac{(n^2 + n)(4n + 5)}{6} \\ + & = \frac{(n + 1)((n + 1) - 1)(4(n + 1) + 1)}{6}. + \end{align*} + Thus $P(n + 1)$ holds. + + \paragraph{Conclusion}% + + By mathematical induction, it follows for all positive integers $n$, $P(n)$ + is true. + +\end{proof} + +\section{\partial{Exercise 1.15.9}}% +\label{sec:exercise-1.15.9} + +Show that the following property is equivalent to + \nameref{sec:step-expansion-contraction-interval-integration}: + \begin{equation} + \label{sec:exercise-1.15.9-eq1} + \int_{ka}^{kb} f(x) \mathop{dx} = k \int_a^b f(kx) \mathop{dx}. + \end{equation} + +\begin{proof} + + Let $f$ be a step function on closed interval $[a, b]$ and $k \neq 0$. + Applying \nameref{sec:step-expansion-contraction-interval-integration} to the + right-hand side of \eqref{sec:exercise-1.15.9-eq1} yields + $$k\int_{ka}^{kb} f(kx / k) \mathop{dx} = + k\left[k\int_a^b f(kx) \mathop{dx}\right].$$ + Simplifying the left-hand side and dividing both sides by $k$ immediately + yields the desired result. + +\end{proof} + +\section{Exercise 1.15.11}% +\label{sec:exercise-1.15.11} + +If we instead defined the integral of step functions as + \begin{equation*} + \label{sec:exercise-1.15.11-eq1} + \int_a^b s(x) \mathop{dx} = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}), + \end{equation*} + a new and different theory of integration would result. +Which of the following properties would remain valid in this new theory? + +\subsection{\partial{Exercise 1.15.11a}}% +\label{sub:exercise-1.15.11a} + +$\int_a^b s + \int_b^c s = \int_a^c s$. + +\note{This property mirrors + \nameref{sec:step-additivity-with-respect-interval-integration}} + +\begin{proof} + + The above property is valid. + + \divider + + WLOG, suppose $a < b < c$. + Let $s$ be a step function defined on closed interval $[a, c]$. + By definition of a \nameref{sec:def-step-function}, there exists a + \nameref{sec:def-partition} such that $s$ is constant on each open + subinterval of $P$. + Let $Q = \{x_0, x_1, \ldots, x_n\}$ be a refinement of $P$ that includes $b$ + as a subdivision point. + Then $Q$ is a step partition of $s$ and there exists some $0 < i < n$ such + that $x_i = c$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $Q$. + By \eqref{sec:exercise-1.15.11-eq1}, + \begin{align*} + \int_a^c s + & = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\ + & = \sum_{k=1}^i s_k^3 \cdot (x_k - x_{k-1}) + + \sum_{k=i+1}^n s_k^3 \cdot (x_k - x_{k-1}) \\ + & = \int_a^b s + \int_b^c s. + \end{align*} + +\end{proof} + +\subsection{\partial{Exercise 1.15.11b}}% +\label{sub:exercise-1.15.11b} + +$\int_a^b (s + t) = \int_a^b s + \int_a^b t$. + +\note{This property mirrors the \nameref{sec:step-additive-property}.} + +\begin{proof} + + The above property is invalid. + + \divider + + Let $s$ and $t$ be step functions on closed interval $[a, b]$. + By definition of a step function, there exists a \nameref{sec:def-partition} + $P_s$ such that $s$ is constant on each open subinterval of $P_s$. + Likewise, there exists a partition $P_t$ such that $t$ is constant on each + open subinterval of $P_t$. + Therefore $s + t$ is a step function with step partition + $$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\},$$ + the common refinement of $P_s$ and $P_t$ with subdivision points + $x_0$, $x_1$, $\ldots$, $x_n$. + + $s$ and $t$ remain constant on every open subinterval of $P$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P_s$. + Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of + $P_t$. + By \eqref{sec:exercise-1.15.11-eq1}, + \begin{align*} + \int_a^b s + t + & = \sum_{k=1}^n (s_k + t_k)^3 \cdot (x_k - x_{k-1}) \\ + & = \sum_{k=1}^n + \left[ s_k^3 + 3s_k^2t_k + 3s_kt_k^2 + t_k^3 \right] \\ + & = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \;+ \\ + & \quad\qquad + \sum_{k=1}^n t_k^3 \cdot (x_k - x_{k-1}) \;+ \\ + & \quad\qquad + \sum_{k=1}^n (3s_k^2t_k + 3s_kt_k^2) \cdot (x_k - x_{k - 1}) \\ + & = \int_a^b s + \int_a^b t + + \sum_{k=1}^n (3s_k^2t_k + 3s_kt_k^2) \cdot (x_k - x_{k - 1}). + \end{align*} + Since this last addend does not necessarily equal $0$, the desired property is + invalid. + +\end{proof} + +\subsection{\partial{Exercise 1.15.11c}}% +\label{sub:exercise-1.15.11c} + +$\int_a^b c \cdot s = c \int_a^b s$. + +\note{This property mirrors the \nameref{sec:step-homogeneous-property}.} + +\begin{proof} + + The above property is invalid. + + \divider + + Let $s$ be a step function on closed interval $[a, b]$. + By definition of a step function, there exists a \nameref{sec:def-partition} + $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open + subinterval of $P$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P$. + Then $c \cdot s$ is a step function with step partition $P$. + By \eqref{sec:exercise-1.15.11-eq1}, + \begin{align*} + \int_a^b c \cdot s + & = \sum_{k=1}^n (c \cdot s_k)^3 \cdot (x_k - x_{k-1}) \\ + & = \sum_{k=1}^n c^3 \cdot s_k^3 \cdot (x_k - x_{k-1}) \\ + & = c^3 \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\ + & = c^3 \int_a^b s. + \end{align*} + Since $c^3$ does not necessarily equal $c$, the desired property is invalid. + +\end{proof} + +\subsection{\partial{Exercise 1.15.11d}}% +\label{sub:exercise-1.15.11d} + +$\int_{a+c}^{b+c} s(x) \mathop{dx} = \int_a^b s(x + c) \mathop{dx}$. + +\note{This property mirrors \nameref{sec:step-invariance-under-translation}.} + +\begin{proof} + + The above property is valid. + + \divider + + Let $s$ be a step function on closed interval $[a + c, b + c]$. + By definition of a \nameref{sec:def-step-function}, there exists a \nameref{sec:def-partition} $P = \{x_0, x_1, \ldots, x_n\}$ such that $s$ is constant on each open subinterval of $P$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of $P$. + + Let $c$ be a real number. + Then $t(x) = s(x + c)$ is a step function on closed interval $[a, b]$ with partition $Q = \{x_0 - c, x_1 - c, \ldots, x_n - c\}$. + Furthermore, $t$ is constant on each open subinterval of $Q$. + Let $t_k$ denote the value of $t$ on the $k$th open subinterval of $Q$. + By construction, $t_k = s_k$. + By \eqref{sec:exercise-1.15.11-eq1}, + \begin{align*} + \int_{a+c}^{b+c} s(x) \mathop{dx} + & = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\ + & = \sum_{k=1}^n s_k^3 \cdot ((x_k - c) - (x_{k-1} - c)) \\ + & = \sum_{k=1}^n t_k^3 \cdot ((x_k - c) - (x_{k-1} - c)) \\ + & = \int_a^b t(x) \mathop{dx} \\ + & = \int_a^b s(x + c) \mathop{dx}. + \end{align*} + +\end{proof} + +\subsection{\partial{Exercise 1.15.11e}}% +\label{sub:exercise-1.15.11e} + +If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. + +\note{This property mirrors the \nameref{sec:step-comparison-theorem}.} + +\begin{proof} + + The above property is valid. + + \divider + + Let $s$ and $t$ be step functions on closed interval $[a, b]$. + By definition of a \nameref{sec:def-step-function}, there exists a \nameref{sec:def-partition} + $P_s$ such that $s$ is constant on each open subinterval of $P_s$. + Likewise, there exists a partition $P_t$ such that $t$ is constant on each + open subinterval of $P_t$. + Let $$P = P_s \cup P_t = \{x_0, x_1, \ldots, x_n\}$$ be the common refinement + of $P_s$ and $P_t$ with subdivision points $x_0$, $x_1$, $\ldots$, $x_n$. + + By construction, $P$ is a step partition for both $s$ and $t$. + Thus $s$ and $t$ remain constant on every open subinterval of $P$. + Let $s_k$ denote the constant value of $s$ on the $k$th open subinterval of + $P$. + Let $t_k$ denote the constant value of $t$ on the $k$th open subinterval of + $P$. + By \eqref{sec:exercise-1.15.11-eq1}, + \begin{align*} + \int_a^b s + & = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}) \\ + & < \sum_{k=1}^n t_k^3 \cdot (x_k - x_{k-1}) \\ + & = \int_a^b t. + \end{align*} + +\end{proof} + \end{document}