[Analysis] Riemann Integral

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/00_line_integrals.tex

@@ -34,4 +34,26 @@ we have $\displaystyle\int_{\gamma} f(s) \cdot \dx \vec{s} = \int_{\sigma} f(s)
 \shorttheorem Let $X$ be open set, $f$ conservative vector field. Then $\exists C^1$ function $g$ s.t. $f = \nabla g$.
 If any two points of $X$ can be joined by a parametrized curve, then $g$ is unique up to a constant: if $\nabla g_1 = f$, then $g - g_1$ is constant on $X$
 
-\shortremark If $f$ vec. field on $X$, then $g$ is called a \bi{potential} for $f$
+\shortremark Two points $x, y \in X$ can be joined by parametrized curve $\gamma$ if $\gamma(a) = x$ and $\gamma(b) = y$. In that case, $X$ is called \bi{path-connected}.
+It is true when $X$ is \textit{convex} (e.g. when $X$ is a disc or a product of intervals).
+If $f$ is a vector field on $X$, then $g$ is called a \bi{potential} for $f$ and it is not unique, since we can add a constant to $g$ without changing the gradient.\\
+%
+\stepLabelNumber{all}
+\shortproposition For a vectorfield to be conservative, a \textit{necessary condition} is that $\displaystyle\frac{\partial f_i}{\partial x_j} = \frac{\partial f_j}{x_i}$
+for any $1 \leq i \neq j \leq n \in \N$\\
+%
+\stepLabelNumber{all}
+\compactdef{Start Shaped Set} $X \subseteq \R^n$ is star shaped if $\exists x_0 \in X$ s.t. $\forall x \in X$, the line segment from $x$ to $x_0$ is contained in $X$,
+and we also say that $X$ is \textit{star shaped around} $x_0$\\
+%
+\stepLabelNumber{all}
+\shorttheorem Let $X$ start shaped and open, $f$ a $C^1$ vector field fulfilling Proposition \ref{all:4-1-13}. Then $f$ is conservative.
+
+\drmvspace
+\setLabelNumber{all}{20}
+\shortdef Let $X \subseteq \R^3$ open and $f$ a $C^1$ vector field. Then the \bi{curl} of $f$ is the conservative vector field
+$\text{curl}(f) = \begin{bmatrix}
+        \partial_y f_3 - \partial_z f_2 \\
+        \partial_z f_1 - \partial_x f_3 \\
+        \partial_x f_2 - \partial_y f_1
+    \end{bmatrix}$

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/01_int_in_rn.tex

@@ -1,2 +1,38 @@
 \newsectionNoPB
 \subsection{Riemann integral in Vector Space}
+The integral of a continuous function $f: X \rightarrow \R$ with $X \subseteq \R^n$ bounded and closed, is denoted $\int_X f(x) \dx x$ with properties:
+\rmvspace
+\begin{enumerate}[label=(\arabic*), noitemsep]
+    \item \bi{(Compatibility)} If $n = 1$ and $X = [a, b]$, integral is the indefinite integral as per Analysis I
+    \item \bi{(Linearity)} If $f$, $g$ are continuous on $X$ and $a, b \in \R$, then $\displaystyle \int_X (a f(x) + b g(x)) \dx x = a \int_X f(x) \dx x + b \int_X g(x) \dx x$
+    \item \bi{(Positivity)} If $f \leq g$, then so is the integral and if $f \geq 0$, so is the integral and if $Y \subseteq X$, then int. over $Y$ is $\leq$ over $X$
+    \item \bi{(Upper bound \& Triangle Inequality)} $\displaystyle \left| \int_{X} f(x) \dx x \right| \leq \int_{X} |f(x)|\dx x$ and 
+        $\displaystyle \left| \int_{X} (f(x) + g(x)) \dx x \right| \leq \int_{X} |f(x)| \dx x \int_X |g(x)|$
+    \item \bi{(Volume)} The integral of $f$ is the volume of $\{ (x, y) \in X \times \R : 0 \leq y \leq f(x) \} \subseteq \R^{n + 1}$.
+        If $X$ is a bounded rectangle, e.g. $X = [a_1, b_1] \times \ldots \times [a_n, b_n] \subseteq \R^n$ and $f = 1$, then $\int_{X} \dx x = (b_n - a_n) \dots (b_1 - a_1)$.
+        We write $\text{Vol}(X)$ or $\text{Vol}_n(X)$
+    \item \bi{(Multiple integral)} \textit{(Fubini)} If $n_1, n_2 \in \Z$ s.t. $n = n_1 + n_2$,
+        then for $x_1 \in \R^{n_1}$, let $Y_{x_1} = \{ x_2 \in \R^{n_2} : (x_1, x_2) \in X \} \subseteq \R^{n_2}$.
+        Let $X_1$ be the set of $x_1 \in \R^n$ such that $Y_{x_1}$ is not empty. Then $X_1$ and $Y_{x_1}$ are compact.\\
+        If $\displaystyle g(x_1) = \int_{Y_{x_1}} f(x_1, x_2) \dx x_2$ is continuous on $X_1$, then
+        \dnrmvspace
+        \begin{align*}
+            \int_{X} f(x_1, x_2) \dx x = \int_{X_1} g(x_1) \dx x = \int_{X_1} g(x_1) \dx x_1 = \int_{X_1} \left( \int_{Y_{x_1}} f(x_1, x_2) \dx x_2 \right) \dx x_1
+        \end{align*}
+        
+        \rmvspace
+        Exchanging the role of $x_1$ and $x_2$ we have (with $Z_{x_2} = \{ x_1 : (x_1, x_2) \in X \}$) if integral over $x_1$ is continuous.
+        \rmvspace
+        \begin{align*}
+            \int_{X} f(x_1, x_2) \dx x = \int_{X_2} \left( \int_{Z_{x_2}} f(x_1, x_2) \dx x_1 \right) \dx x_2
+        \end{align*}
+        \drmvspace
+    \item \bi{(Domain additivity)} If $X_1$ and $X_2$ are compact and $f$ continuous on $X = X_1 \cup X_2$, then (for $Y = X_1 \cap X_2$)
+        \rmvspace
+        \begin{align*}
+            \int_X f(x) \dx x + \int_Y f(x) \dx x = \int_{X_1} f(x) \dx x + \int_{X_2} f(x) \dx x
+        \end{align*}
+
+        \rmvspace
+        In particular, if $Y$ empty (or size is ``negligible''), then $\int_{X} f(x) \dx x = \int_{X_1} f(x) \dx x + \int_{X_2} f(x) \dx x$
+\end{enumerate}