$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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Riemann Integrals over Union of Sets

Let $S, T\subseteq\mathbb{R}^n$ be nonempty and bounded sets that satisfy $\mu(S\cap T)=0$. If $S\cup T\to\mathbb{R}$ is bounded and integrable on $S$ and on $T$, then $f$ is integrable on $S\cup T$ and $$\begin{align*} \int_{S\cup T}f(\vec{x})d\vec{x}=\int_S f(\vec{x})d\vec{x}+\int_T f(\vec{x}) d\vec{x} \end{align*}$$

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Riemann Integrals over Union of Sets

Let $S, T\subseteq\mathbb{R}^n$ be nonempty and bounded sets that satisfy $\mu(S\cap T)=0$. If $S\cup T\to\mathbb{R}$ is bounded and integrable on $S$ and on $T$, then $f$ is integrable on $S\cup T$ and $$\begin{align*} \int_{S\cup T}f(\vec{x})d\vec{x}=\int_S f(\vec{x})d\vec{x}+\int_T f(\vec{x}) d\vec{x} \end{align*}$$

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concepts
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FullPage
result
concepts
hypothesis
implications
proof

Riemann Integrals over Union of Sets

Let $S, T\subseteq\mathbb{R}^n$ be nonempty and bounded sets that satisfy $\mu(S\cap T)=0$. If $S\cup T\to\mathbb{R}$ is bounded and integrable on $S$ and on $T$, then $f$ is integrable on $S\cup T$ and $$\begin{align*} \int_{S\cup T}f(\vec{x})d\vec{x}=\int_S f(\vec{x})d\vec{x}+\int_T f(\vec{x}) d\vec{x} \end{align*}$$

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Hypothesis

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Results

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Proof

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Riemann Integrals over Union of Sets

Let $S, T\subseteq\mathbb{R}^n$ be nonempty and bounded sets that satisfy $\mu(S\cap T)=0$. If $S\cup T\to\mathbb{R}$ is bounded and integrable on $S$ and on $T$, then $f$ is integrable on $S\cup T$ and $$\begin{align*} \int_{S\cup T}f(\vec{x})d\vec{x}=\int_S f(\vec{x})d\vec{x}+\int_T f(\vec{x}) d\vec{x} \end{align*}$$

Concepts

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Hypothesis

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Proof

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result
concepts
hypothesis
implications
proof