Finish Apostol 1.22.
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@ -2670,7 +2670,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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Let $f$ be a nonnegative function, integrable on an interval $[a, b]$.
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Then the graph of $f$, that is, the set
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\begin{equation}
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\label{sec:measurability-graph-nonnegative-function-eq1}
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\label{sec:theorem-1.11-eq1}
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\{(x, y) \mid a \leq x \leq b, y = f(x)\},
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\end{equation}
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is measurable and has area equal to $0$.
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@ -2686,7 +2686,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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equal to $0$.
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\paragraph{(i)}%
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\label{par:measurability-graph-nonnegative-function-i}
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\label{par:theorem-1.11-i}
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By definition of integrability, there exists one and only one number $I$
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such that
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@ -2704,10 +2704,9 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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Let $Q$ denote the ordinate set of $f$.
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By \nameref{sec:measurability-ordinate-sets}, $Q$ is measurable with area
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equal to the integral $I = \int_a^b f(x) \mathop{dx}$.
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By \nameref{par:measurability-graph-nonnegative-function-i}, $Q'$ is
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By \nameref{par:theorem-1.11-i}, $Q'$ is
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measurable with area also equal to $I$.
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We note the graph of $f$,
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\eqref{sec:measurability-graph-nonnegative-function-eq1}, is equal to set
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We note the graph of $f$, \eqref{sec:theorem-1.11-eq1}, is equal to set
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$Q - Q'$.
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By the \nameref{sec:area-difference-property}, $Q - Q'$ is measurable and
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$$a(Q - Q') = a(Q) - a(Q') = I - I = 0.$$
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@ -2751,7 +2750,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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on every $k$th open subinterval of $P$.
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Then, by \eqref{sec:theorem-1.9-eq1}, it follows
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\begin{equation}
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\label{sec:integrability-bounded-monotonic-functions-eq1}
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\label{sec:theorem-1.12-eq1}
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\int_a^b s(x) \mathop{dx} \leq \ubar{I}(f)
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\leq \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}.
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\end{equation}
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@ -2770,7 +2769,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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& = \left[\frac{b - a}{n}\right] \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\
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& = \frac{(b - a)(f(b) - f(a))}{n}.
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\end{align*}
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By \eqref{sec:integrability-bounded-monotonic-functions-eq1},
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By \eqref{sec:theorem-1.12-eq1},
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\begin{align*}
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\ubar{I}(f)
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& \leq \bar{I}(f) \\
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@ -2790,7 +2789,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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on every $k$th open subinterval of $P$.
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Then, by \eqref{sec:theorem-1.9-eq1}, it follows
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\begin{equation}
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\label{sec:integrability-bounded-monotonic-functions-eq2}
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\label{sec:theorem-1.12-eq2}
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\int_a^b s(x) \mathop{dx} \leq \ubar{I}(f)
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\leq \bar{I}(f) \leq \int_a^b t(x) \mathop{dx}.
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\end{equation}
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@ -2809,7 +2808,7 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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& = \left[\frac{b - a}{n}\right] \sum_{k=1}^n f(x_{k-1}) - f(x_k) \\
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& = \frac{(b - a)(f(a) - f(b))}{n}.
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\end{align*}
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By \eqref{sec:integrability-bounded-monotonic-functions-eq2},
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By \eqref{sec:theorem-1.12-eq2},
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\begin{align*}
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\ubar{I}(f)
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& \leq \bar{I}(f) \\
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@ -2822,9 +2821,9 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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\end{proof}
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\section{\unverified{%
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Calculation of the Integral of a Bounded Monotonic Function}}%
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\label{sec:calculation-integral-bounded-monotonic-function}
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\section{\partial{%
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Calculation of the Integral of a Bounded Increasing Function}}%
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\label{sec:calculation-integral-bounded-increasing-function}
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\label{sec:theorem-1.13}
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\begin{theorem}[1.13]
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@ -2832,13 +2831,157 @@ If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$.
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Assume $f$ is increasing on a closed interval $[a, b]$.
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Let $x_k = a + k(b - a) / n$ for $k = 0, 1, \ldots, n$.
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If $I$ is any number which satisfies the inequalities
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$$\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k)
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\leq I \leq
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\frac{b - a}{n} \sum_{k=1}^n f(x_k)$$
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\begin{equation}
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\label{sec:theorem-1.13-eq1}
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\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k)
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\leq I \leq
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\frac{b - a}{n} \sum_{k=1}^n f(x_k)
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\end{equation}
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for every integer $n \geq 1$, then $I = \int_a^b f(x) \mathop{dx}$.
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\end{theorem}
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\begin{proof}
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Let $f$ be increasing on a closed interval $[a, b]$ and $I$ be a number
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satisfying \eqref{sec:theorem-1.13-eq1}.
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Let $s$ be the step function below $f$ with constant value $f(x_{k-1})$
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on every $k$th open subinterval of $P$.
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Let $t$ be the step function above $f$ with constant value $f(x_k)$
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on every $k$th open subinterval of $P$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_a^b s(x) \mathop{dx}
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& = \sum_{k=1}^n f(x_{k-1})\left[\frac{b - a}{n}\right] \\
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& = \sum_{k=0}^{n-1} f(x_k)\left[\frac{b - a}{n}\right] \\
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\int_a^b t(x) \mathop{dx}
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& = \sum_{k=1}^n f(x_k)\left[\frac{b - a}{n}\right].
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\end{align*}
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Therefore \eqref{sec:theorem-1.13-eq1} can alternatively be written as
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\begin{equation}
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\label{sec:theorem-1.13-eq2}
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\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}.
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\end{equation}
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By \nameref{sec:theorem-1.12}, $f$ is integrable.
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Therefore \nameref{sec:theorem-1.9} indicates $f$ satisfies
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\begin{equation}
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\label{sec:theorem-1.13-eq3}
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\int_a^b s(x) \mathop{dx}
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\leq \int_a^b f(x) \mathop{dx}
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\leq \int_a^b t(x) \mathop{dx}.
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\end{equation}
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Manipulating \eqref{sec:theorem-1.13-eq2} and \eqref{sec:theorem-1.13-eq3}
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together yields
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\begin{align*}
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I - \int_a^b f(x) \mathop{dx}
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& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}, \\
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\int_a^b f(x) \mathop{dx} - I
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& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}.
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\end{align*}
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Combining the above inequalities in turn yields
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\begin{align*}
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0
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& \leq \abs{\int_a^b f(x) \mathop{dx} - I} \\
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& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx} \\
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& = \sum_{k=1}^n f(x_k) \left[ \frac{b - a}{n} \right] -
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\sum_{k=1}^n f(x_{k-1}) \left[ \frac{b - a}{n} \right] \\
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& = \frac{b - a}{n} \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\
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& = \frac{(b - a)(f(b) - f(a))}{n}.
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\end{align*}
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The above chain of inequalities holds for all positive integers $n \geq 1$,
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meaning \nameref{sec:theorem-i.31} applies.
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Thus $$\abs{\int_a^b f(x) \mathop{dx} - I} = 0,$$ which immediately implies
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the desired result.
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\end{proof}
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\section{\partial{%
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Calculation of the Integral of a Bounded Decreasing Function}}%
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\label{sec:calculation-integral-bounded-decreasing-function}
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\label{sec:theorem-1.14}
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\begin{theorem}[1.14]
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Assume $f$ is descreasing on $[a, b]$.
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Let $x_k = a + k(b - a) / n$ for $k = 0, 1, \ldots, n$.
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If $I$ is any number which satisfies the inequalities
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\begin{equation}
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\label{sec:theorem-1.14-eq1}
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\frac{b - a}{n} \sum_{k=1}^n f(x_k)
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\leq I \leq
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\frac{b - a}{n} \sum_{k=0}^{n-1} f(x_k)
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\end{equation}
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for every integer $n \geq 1$, then $I = \int_a^b f(x) \mathop{dx}$.
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\end{theorem}
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\begin{proof}
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Let $f$ be decreasing on a closed interval $[a, b]$ and $I$ be a number
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satisfying \eqref{sec:theorem-1.14-eq1}.
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Let $s$ be the step function below $f$ with constant value $f(x_k)$
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on every $k$th open subinterval of $P$.
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Let $t$ be the step function above $f$ with constant value $f(x_{k-1})$
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on every $k$th open subinterval of $P$.
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By definition of the \nameref{sec:def-integral-step-function},
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\begin{align*}
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\int_a^b s(x) \mathop{dx}
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& = \sum_{k=1}^n f(x_k) \left[\frac{b - a}{n}\right] \\
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\int_a^b t(x) \mathop{dx}
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& = \sum_{k=1}^n f(x_{k-1}) \left[\frac{b - a}{n}\right] \\
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& = \sum_{k=0}^{n-1} f(x) \left[\frac{b - a}{n}\right].
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\end{align*}
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Therefore \eqref{sec:theorem-1.14-eq1} can alternatively be written as
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\begin{equation}
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\label{sec:theorem-1.14-eq2}
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\int_a^b s(x) \mathop{dx} \leq I \leq \int_a^b t(x) \mathop{dx}.
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\end{equation}
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By \nameref{sec:theorem-1.12}, $f$ is integrable.
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Therefore \nameref{sec:theorem-1.9} indicates $f$ satisfies
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\begin{equation}
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\label{sec:theorem-1.14-eq3}
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\int_a^b s(x) \mathop{dx}
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\leq \int_a^b f(x) \mathop{dx}
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\leq \int_a^b t(x) \mathop{dx}.
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\end{equation}
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Manipulating \eqref{sec:theorem-1.14-eq2} and \eqref{sec:theorem-1.14-eq3}
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together yields
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\begin{align*}
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I - \int_a^b f(x) \mathop{dx}
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& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}, \\
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\int_a^b f(x) \mathop{dx} - I
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& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx}.
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\end{align*}
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Combining the above inequalities in turn yields
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\begin{align*}
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0
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& \leq \abs{\int_a^b f(x) \mathop{dx} - I} \\
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& \leq \int_a^b t(x) \mathop{dx} - \int_a^b s(x) \mathop{dx} \\
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& = \sum_{k=1}^n f(x_k) \left[ \frac{b - a}{n} \right] -
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\sum_{k=1}^n f(x_{k-1}) \left[ \frac{b - a}{n} \right] \\
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& = \frac{b - a}{n} \sum_{k=1}^n f(x_k) - f(x_{k-1}) \\
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& = \frac{(b - a)(f(b) - f(a))}{n}.
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\end{align*}
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The above chain of inequalities holds for all positive integers $n \geq 1$,
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meaning \nameref{sec:theorem-i.31} applies.
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Thus $$\abs{\int_a^b f(x) \mathop{dx} - I} = 0,$$ which immediately implies
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the desired result.
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\end{proof}
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\section{\unverified{%
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Calculation of the Integral \texorpdfstring{$\int_0^b x^p \mathop{dx}$}{int-x-p}
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when \texorpdfstring{$p$}{p} is a Positive Integer}}%
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\label{sec:calculation-integral-int-x-p-p-positive-integer}
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\label{sec:theorem-1.15}
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\begin{theorem}[1.15]
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If $p$ is a positive integer and $b > 0$, we have
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$$\int_0^b x^p \mathop{dx} = \frac{b^{p+1}}{p+1}.$$
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\end{theorem}
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\begin{proof}
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TODO
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