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{\Large \bf Recitation Note} \\
{\large 17.872 TA Jiyoon Kim \\ Sep. 30, 2003}\\
\end{center}
\section{Some Integration Theorem}
\begin{enumerate}
\item A constant can be moved through an integral sign.
\[
\int cf(x) dx = c\int f(x) dx
\]
\item An integration of a sum is the sum of the integrations.
\[
\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx
\]
\item An integration of a difference is the difference of the integrations.
\[
\int [f(x) - g(x)] dx = \int f(x) dx - \int g(x) dx
\]
\end{enumerate}
\section{Integration Rules}
\begin{enumerate}
\item Power Rule
\[
\int x^n dx = \frac{1}{n+1}x^{n+1} + c
\]
\item Exponential Rule I
\[
\int e^x dx = e^x + c
\]
\item Exponential Rule II
\[
\int f'(x)e^{f(x)} dx = e^{f(x)} + c
\]
\item Logarithm Rule I
\[
\int \frac{1}{x} dx = ln x + c
\]
\item Logarithm Rule II
\[
\int \frac{f'(x)}{f(x)} dx = ln f(x) + c
\]
\end{enumerate}
Examples:\\
\\
1) $\bd \int 2e^{2x} + \frac{14x}{7x^2 + 5} dx$\\
2) $\bd \int \frac{4x}{2x^2 + 1} dx$\\
3) $\bd \int \frac{x}{3x^2 + 5} dx$\\
\\
\section{Substitution Rule}
When you have integration with high power or with complicated form, you can use substitution rule.  First, you put
the complicated part as $u$ and find $\frac{du}{dx}$.  Then, substitute the original equation with $u$ and
$\frac{du}{dx}$ to solve it in $u$ term.\\
\\
Example: \\
\\
1) $\bd \int 2x(x^2+1)^{50} dx$\\
2) $\bd \int 6x^2(x^3+2)^{99} dx$
\\
\section{Integration by Parts}
The integral of $v$ with respect to $u$ is equal to $uv$ minus the integration of $u$ with respect to $v$.
\[
\int v du = uv - \int u dv
\]
Here, you use the less complicated part (usually one term) as $v$ and the rest of the integration including
integral sign and $dx$ as $u$.  For example, if you have $\int x(x+1)^{\frac{1}{2}} dx$, $v$ should be $x$ and $u$
should be $\int (x+1)^{\frac{1}{2}} dx$.  Then, find $dv$ and $du$ to plug them in the formula.\\
\\
Example:\\
\\
1) $\bd \int xe^x dx$\\
2) $\bd \int ln \ x \ dx$\\
\\
\section{Properties of Sigma Notation}
The following properties of sigma notation will help to manipulate sums:\\
\begin{enumerate}
\item
\[\sum^{n}_{k = 1} ca_k = c\sum^{n}_{k = 1} a_k \]
\item
\[\sum^{n}_{k = 1} (a_k + b_k) = \sum^{n}_{k = 1} a_k + \sum^{n}_{k = 1} b_k \]
\item
\[ \sum^{n}_{k = 1} (a_k - b_k) = \sum^{n}_{k = 1} a_k - \sum^{n}_{k = 1} b_k \]
\end{enumerate}
{\bf Question} Are these true?\\
\\
1) $\bd \int \left[\sum^{n}_{i = 1} f_i(x)\right] dx = \sum^{n}_{i = 1}\left[\int f_i(x) dx \right]$\\
\\
2) $\bd \frac{d}{dx}\left[ \sum^{n}_{i = 1} f_i(x) \right] = \sum^{n}_{i = 1} \left[\frac{d}{dx} [f_i(x)] \right]$\\
\\
3) $\bd \sum^{n}_{i = 1} a_i b_i = \sum^{n}_{i = 1} a_i \sum^{n}_{i = 1} b_i$\\
\\
4) $\bd \sum^{n}_{i = 1} \frac{a_i}{b_i} = \frac{\sum^{n}_{i = 1}a_i}{\sum^{n}_{i = 1} b_i}$\\
\\
5) $\bd \sum^{n}_{i = 1} a_i^2 = \left(\sum^{n}_{i = 1} a_i \right)^2$\\
\\
\\
{\it Quick Question}\\
Let $\bar{x}$ denote the arithmetic average of the $n$ numbers $x_1, x_2, \cdots , x_n$.  Prove that\\
\[\sum^{n}_{i = 1}(x_i - \bar{x}) = 0\]
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