Finish Apostol 1.13.

finite-set-exercises
Joshua Potter 2023-05-16 08:56:12 -06:00
parent 8d6a569fa7
commit 0e6a4f810d
1 changed files with 71 additions and 24 deletions

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@ -63,6 +63,9 @@ Denote by $s_k$ the constant value that $s$ takes in the $k$th open subinterval,
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
$$\int_b^a s(x) \mathop{dx} = -\int_a^b s(x) \mathop{dx} \quad\text{and}\quad
\int_a^a s(x) \mathop{dx} = 0.$$
\section{\defined{Partition}}%
\label{sec:def-partition}
@ -1803,16 +1806,16 @@ If $s(x) < t(x)$ for every $x$ in $[a, b]$, then
\section{\partial{Additivity With Respect to the Interval of Integration}}%
\label{sec:step-additivity-with-respect-interval-integration}
Let $s$ be a \nameref{sec:def-step-function} on closed interval $[a, b]$.
Let $a, b, c \in \mathbb{R}$ and $s$ a \nameref{sec:def-step-function} on the
smallest closed interval containing them.
Then
$$\int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} =
\int_a^b s(x) \mathop{dx} \quad\text{if}\quad a < c < b.$$
\todo{This holds for any arrangement of values $a$, $b$, and $c$.}
$$\int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} +
\int_b^a s(x) \mathop{dx} = 0.$$
\begin{proof}
Let $s$ be a step function on closed interval $[a, b]$.
WLOG, suppose $a < c < b$ and $s$ be a step function on closed interval
$[a, b]$.
By definition of a step function, there exists a \nameref{sec:def-partition}
$P$ such that $s$ is constant on each open subinterval of $P$.
@ -1829,6 +1832,14 @@ Then
\sum_{k=i+1}^n s_k \cdot (x_k - x_{k - 1}) \\
& = \int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx}.
\end{align*}
Rearranging terms shows
\begin{align*}
0
& = \int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} -
\int_a^b s(x) \mathop{dx} \\
& = \int_a^c s(x) \mathop{dx} + \int_c^b s(x) \mathop{dx} +
\int_b^a s(x) \mathop{dx}.
\end{align*}
\end{proof}
@ -1873,9 +1884,7 @@ Then
Let $s$ be a step function on closed interval $[a, b]$.
Then
$$\int_{ka}^{kb} s \left( \frac{x}{k} \right) \mathop{dx} =
k \int_a^b s(x) \mathop{dx} \quad\text{for every } k > 0.$$
\todo{This also holds for negative values of $k$.}
k \int_a^b s(x) \mathop{dx} \quad\text{for every } k \neq 0.$$
\begin{proof}
@ -1885,22 +1894,60 @@ Then
subinterval of $P$.
Let $s_i$ denote the value of $s$ on the $i$th open subinterval of $P$.
Let $k > 0$ be a real number.
Then $t(x) = s(x / k)$ is a step function on closed interval $[ka, kb]$
with partition $Q = \{kx_0, kx_1, \ldots, kx_n\}$.
Furthermore, $t$ is constant on each open subinterval of $Q$.
Let $t_i$ denote the value of $t$ on the $i$th open subinterval of $Q$.
By construction, $t_i = s_i$.
Let $k \neq 0$ be a real number.
There are two cases to consider:
By definition of the \nameref{sec:def-integral-step-function},
\begin{align*}
\int_{ka}^{kb} s(x / k) \mathop{dx}
& = \int_{ka}^{kb} t(x) \mathop{dx} \\
& = \sum_{i=1}^n t_i \cdot (kx_i - kx_{i-1}) \\
& = k \sum_{i=1}^n t_i \cdot (x_i - x_{i-1}) \\
& = k \sum_{i=1}^n s_i \cdot(x_i - x_{i-1}) \\
& = k \int_a^b s(x) \mathop{dx}.
\end{align*}
\paragraph{Case 1}%
Suppose $k > 0$.
Then $t(x) = s(x / k)$ is a step function on closed interval $[ka, kb]$
with partition $Q = \{kx_0, kx_1, \ldots, kx_n\}$.
Furthermore $t_i = s_i$.
By definition of the \nameref{sec:def-integral-step-function},
\begin{align*}
\int_{ka}^{kb} s(x / k) \mathop{dx}
& = \int_{ka}^{kb} t(x) \mathop{dx} \\
& = \sum_{i=1}^n t_i \cdot (kx_i - kx_{i-1}) \\
& = k \sum_{i=1}^n t_i \cdot (x_i - x_{i-1}) \\
& = k \sum_{i=1}^n s_i \cdot(x_i - x_{i-1}) \\
& = k \int_a^b s(x) \mathop{dx}.
\end{align*}
\paragraph{Case 2}%
Let $k < 0$ be a real number.
Then $t(x) = s(x / k)$ is a step function on closed interval $[kb, ka]$
with partition $Q = \{kx_n, kx_{n-1}, \ldots, kx_0\}$.
Furthermore $t_i = s_i$.
By definition of the \nameref{sec:def-integral-step-function},
\begin{align*}
\int_{ka}^{kb} s(x / k) \mathop{dx}
& = -\int_{kb}^{ka} s(x / k) \mathop{dx} \\
& = -\int_{kb}^{ka} t(x) \mathop{dx} \\
& = -\sum_{i=1}^n t_i \cdot (kx_{i-1} - kx_i) \\
& = -\sum_{i=1}^n s_i \cdot (kx_{i-1} - kx_i) \\
& = -\sum_{i=1}^n s_i \cdot (-k) \cdot (x_i - x_{i-1}) \\
& = k \sum_{i=1}^n s_i \cdot (x_i - x_{i-1}) \\
& = k \int_a^b s(x) \mathop{dx}.
\end{align*}
\end{proof}
\section{\partial{Reflection Property}}%
\label{sec:step-reflection-property}
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}.$$
\begin{proof}
Let $k = -1$.
By \nameref{sec:step-expansion-contraction-interval-integration},
$$\int_{-a}^{-b} s \left( \frac{x}{-1} \right) =
-\int_a^b s(x) \mathop{dx}.$$
Simplifying the left-hand side of the above identity, and multiplying both
sides by $-1$ yields the desired result.
\end{proof}