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\institution{SIAST Palliser Campus}
\department{Mathematics}
\course{MAT 235}
\unittitle{Introduction to Integration}
\unitnumber{1}
\date{September 2011}
\title{STUDY GUIDE}
\author{Alan Dill and Robert G. Petry}
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\publisher{the Department of Mathematics, SIAST Palliser Campus}
\begin{document}
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\section{A Simple Painting Problem}

In this section I will try to motivate what the main component of this
course is all about, integration.  This is intended as a sketchy
overview - the details will be fleshed out in the subsequent material.
You may want to read this again when you have finished the course to
see that it makes sense!

Suppose I was deciding to make a ``feature wall'' in a room in my
house by painting the wall a unique colour.  Since I would only be
buying paint for the one wall it would make sense to buy only exactly
the amount of paint I needed.  This amount would be determined by the
area of my wall.  For a rectangular wall this area is
\[A=HB\] where $H$ is the height of the wall and $B$ is its base
length. See Figure~\vref{fig:feature_rectangle}. So if the wall has
height $H=4$ $\mathrm{m}$ and base $B=3$ $\mathrm{m}$ we'd have an area
$A=12$ $\mathrm{m}^2$ to cover.  If a litre of paint covers $15$
$\mathrm{m}^2$, I would need
$\frac{12~\mathrm{m}^2}{15~\mathrm{m}^2/\mathrm{litre}}=.8~\mathrm{litres}$
of paint, so a single $1$-litre can would be sufficient.
\begin{figure}[h]
\begin{center}
\includegraphics[width=.2\textwidth,keepaspectratio=,clip=]{feature_rectangle.pdf}
\caption{\label{fig:feature_rectangle}A rectangular wall}
\end{center}
\end{figure}

Now suppose my feature wall was not a rectangle but looked like
Figure~\vref{fig:feature_irreg}.  Here the height $H$ is not fixed,
but varies across the wall.  How would we calculate the
area\footnote{One might question that such an irregular shape even has
  a clearly defined area.  An intuitive way to define the area would
  be to paint the wall and see how much paint we used.  If it took
  $1.2~\mathrm{litres}$ to paint it, then its area would be
  $(1.2~\mathrm{litres})\cdot(15~\mathrm{m}^2/\mathrm{litre})=18~\mathrm{m}^2$.}
now?  One way would be to try to estimate the area by covering it with
shapes for which we know how to calculate the area such as rectangles
and triangles.  In Figure~\vref{fig:feature_irreg_poly} we see one way
to estimate the area by adding a rectangle as before and two triangles
for which we could also calculate the areas
($A_{\Delta}=\frac{(\mathrm{base})\cdot(\mathrm{altitude})}{2}$).  A
second way to estimate the area is shown where we add several thin
rectangles whose height varies with the wall height.

\begin{figure}[h]
\begin{center}
\includegraphics[width=.25\textwidth,keepaspectratio=,clip=]{feature_irreg.pdf}
\caption{\label{fig:feature_irreg}An irregular-shaped wall}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
\includegraphics[width=.4\textwidth,keepaspectratio=,clip=]{feature_irreg_poly.pdf}
\caption{\label{fig:feature_irreg_poly}Two ways to estimate the irregular area with a) one rectangle and two triangles and b) seven thin rectangles.}
\end{center}
\end{figure}

These approaches are only approximate however and if we were using
expensive ``paint'' such as gilding a wall of a temple with gold, or
if we were constructing a residential subdivision for which we had
$20$ identical feature walls to paint, the exact area of this
wall would be desirable.  If we order too much paint we are wasting
money but if we don't order enough we would lose money on our painters
as they wait for more paint to be delivered! This course will allow us
to solve such problems like calculating such an irregular area (and
many more things) exactly or at least to excellent approximation.

The first thing we need to do is represent the problem mathematically.
For instance, in the case of our irregular wall let's draw an $x$-axis
along the base of the wall and represent the varying height $H$ with a
\emph{function} $f(x)$ as shown in
Figure~\vref{fig:feature_irreg_func} so that we know the height at
every point $x$ along the wall.  If we let the wall start at $x=a=1$
$\mathrm{m}$ and end at $x=b=4$ $\mathrm{m}$ (so the base $B=b-a=3$
$m$), then the function needs to be properly defined over the domain
of values $[1,4]$ (outside of this it can be whatever we want).

\begin{figure}[h]
\begin{center}
\parbox[c]{.87\textwidth}{\vspace{0em}\resizebox{.87\textwidth}{!}{\input{feature_irreg_func.pdf_t}}}
\caption{\label{fig:feature_irreg_func}We can represent the varying wall height $H$ with a function $f(x)$.  The exact wall area will be the sum of rectangles of height $f(x)$ and infinitessimal width $dx$ between $x=a$ and $x=b$.}
\end{center}
\end{figure}

Now what's the area?  We take our direction from case b) in
Figure~\vref{fig:feature_irreg_poly} and add up thin rectangles of
\emph{varying} height $H=f(x)$ and \emph{equal} base width $B=dx$ (see
Figure~\vref{fig:feature_irreg_func}), and we write the area
as\footnote{Here $dx$ is taken to be a single entity.  The $d$ is like
  the $\Delta$ in an expression like $\Delta x$.  We write a $d$ to
  distinguish the infinitessimal nature of the width.}
\[\int_a^b f(x)\;dx=\underbrace{\int_a^b}_{\stackrel{\scriptstyle \mathrm{Sum~over~rectangles}}{\mathrm{from~}a~\mathrm{to}~b}}\underbrace{\underbrace{f(x)}_H~~\cdot~~\underbrace{dx\vspace{4em}}_B}_{\stackrel{\scriptstyle \mathrm{Area~of~rectangle}}{\mathrm{at~} x}}\]
Here the $\int$ sign is an $S$ which reminds us we want to \emph{sum}
(or add) all the rectangles.  We need to put the limits $a$ and $b$ on
the sum so that we know the domain over which we want to add the
rectangles.  This expression is called a \emph{definite integral} and
it becomes mathematically precise if we make the base width $dx$
``infinitely thin''.  We'll see how to do that in this course.

Now taking limits as things get infinitely small is what calculus is
about.  You've seen how to take a \emph{derivative} of a function like
our $f(x)$. Geometrically the derivative represents the slope of a
tangent to the curve $y=f(x)$ at a point $x$.  We wrote that
derivative as $f'(x)$ or $\frac{dy}{dx}$.  Now it turns out that
evaluating the definite integral above (finding our area) can be done
exactly if we calculate something called the \emph{antiderivative} of
$f(x)$ which, like the derivative $f'$ of $f$, is itself a function,
now called $F(x)$.  Given the name ``antiderivative'' you could guess
correctly that finding this $F$ is related to how you take
derivatives.  We'll see how to do this.  Once you know how to find
$F(x)$ given a function $f$, you can get the definite integral (our
area) easily by evaluating the expression
\[\int_a^b f(x)\;dx=F(b)-F(a)\]
We can find the area just by subtracting two numbers, $F$ evaluated at
\mbox{$x=b$} \mbox{minus} $F$ evaluated at $x=a$!  Obviously figuring
out how to get these antiderivatives will be pretty useful.  The last
equation is known as the \emph{fundamental theorem of calculus}.

Now we need a symbol for this antiderivative $F$ to make it clear
which function $f$ it is associated with.  Just like we write the
derivative of $f$ by $f'$, we similarly write our antiderivative with
a new symbol involving $f$,
\[F =\int f(x)\;dx\] The symbol on the right hand side is chosen
because of the relation of $F$ to the definite integral above.  In
fact we call $F$ not only an antiderivative but also an
\emph{indefinite integral} for this reason.  (Indefinite since there
aren't any $a$'s or $b$'s.)

Sadly we can't always find an antiderivative function $F(x)$ for every
function $f(x)$ and then we have to use some \emph{numerical methods}
(i.e. adding up small areas by hand) to estimate our definite
integrals.  We'll see how to do that too.

So in summary in our course we will learn to do the following:
\begin{itemize}
\item Figure out how to calculate the antiderivative/indefinite
  integral $\int f(x)\;dx$ for a given function $f(x)$.
\item Calculate the definite integral $\int_a^b f(x)\;dx$ using
  antiderivatives when they exist and numerically otherwise.
\item Apply these integrals to other problems besides area.
\end{itemize}

Finally we'll introduce a different mathematical concept,
\emph{linear algebra}, in a final unit.

If you feel overwhelmed, this is only a road map.  Hopefully at least
we're now motivated to take the journey.
\clearpage

\section{The Indefinite Integral}

If we are given a function, we can differentiate the function,
i.e. find its derivative.  The derivative of a function is itself a
function.

For example, if the given function is $x^4$, its derivative is $4x^3$.

Just as we can reverse a process such as squaring (by taking the
square root), we can also reverse the process of differentiation.

This means that if we are given a function, we can find another
function that has the given function as its derivative.

We see from the above example that the derivative of $x^4$ is $4x^3$.
Thus, if we are asked to find a function that has $4x^3$ as its
derivative, we could give $x^4$ as an answer.  In this case we say
that $x^4$ is an \emph{antiderivative} of $4x^3$.

It should be noted that $x^4$ is not the only possible antiderivative
of $4x^3$.  Since the derivative of a constant is zero, adding
\emph{any constant} to $x^4$ will give us another function whose
derivative is $4x^3$.  For example, $x^4+5$, $x^4-7$ and
\mbox{$x^4+\pi$} are all antiderivatives of $4x^3$.

To show that $x^4 + $\emph{any constant} can be an antiderivative of
$4x^3$, we write the general form of the antiderivative (also called
the \emph{indefinite integral}) of $4x^3$ as $x^4+ C$.  Here $C$ is
called the \emph{constant of integration}, and can have any value.  We
will see later on that, if additional information about the given
function is available (such as its value for a given value of $x$), we
can assign a specific value to the constant $C$.

We use a special symbol called an \emph{integral sign} ($\int$) to
denote the indefinite integral of a function.  If $f(x)$ is a
function whose indefinite integral is to be found, we write
\[\int f(x)\;dx = F(x)+ C\]
where $f(x) = F'(x)$ and $C$ is the constant of
integration.\footnote{The student will note that we neglected this
  constant in our casual discussion in the introductory section.}

What this means is, the function whose derivative is $f(x)$ (or
equivalently, the function whose differential $dF$ is $f(x)dx$) is
$F(x)+C$.

The process of finding the indefinite integral of a function is called
\emph{integration}.

We can derive some basic rules for integration simply by reversing
some of the basic rules for differentiation.  These rules are given in
Figure~\vref{fig:intrules1}.
\clearpage

\begin{figure}[h]
\begin{boxedeqnarray}
\int f(x)\;dx & = & F(x)+ C\ \ \ \ \left(\ \mathit{where}\;f(x)=F'(x)\ \right) \\
\int du & = & u + C \\
\int a\;f(x)\;dx& = &a\int f(x)\;dx=a\;F(x)+C \label{eq:multiplicativeconstant}\\
\int\left[f(x)+g(x)+\ldots\right]\;dx & = &\int f(x)\;dx+\int g(x)\;dx+\ldots \label{eq:summation}\\
\int x^n\;dx& = &\frac{x^{n+1}}{n+1}+C\ \ \ \ (n\neq-1)\\
\nonumber
\end{boxedeqnarray}
\caption{\label{fig:intrules1}Rules for Integration}
\end{figure}

\textbf{Examples}

Perform the following integrations:
\begin{enumerate}
\item $\displaystyle \int 3x\;dx$
\item $\displaystyle\int x^4\;dx$
\item $\displaystyle\int 2x^{\frac{3}{2}}\;dx$
\item $\displaystyle\int 6\sqrt[3]{x}\;dx$
\item $\displaystyle\int \frac{4}{\sqrt{x}}\;dx$
\item $\displaystyle\int\left(\frac{1}{x^{3}}+\frac{1}{2}\right)\;dx$
\item $\displaystyle\int (x\sqrt{x}-5x^{2})\;dx$
\item $\displaystyle\int (1+2x)^{2}\;dx$
\end{enumerate}

If we are given the derivative of a function in the form of an equation
(this is a simple form of \emph{differential equation}), we can find
the original function (i.e. we can solve the differential equation) by
integration, as in the example on the following page.\clearpage

\textbf{Additional Example} (Text, p. 864)
\begin{itemize}
\item Problem 34
\end{itemize}

\textassignment{Sec. 30-1, pp. 857-863}{Ex. 1, p. 863 \# 1-38 (all)}

\section{More Rules of Integration}
\subsection{Integral of a Power of a Function}

Suppose we are given a function  $y=\frac{u^{n+1}}{n+1}$, where
$u$ is a function of $x$.

We can take the derivative of both sides using the rule for the
derivative of a constant times a power of a function.

\begin{eqnarray}
\frac{dy}{dx} & = & \frac{1}{n+1}\,(n+1)\,u^{n}\,\frac{du}{dx} \nonumber \\
\frac{dy}{dx} & = & u^{n}\,\frac{du}{dx} \nonumber
\end{eqnarray}

We can put this into differential form by multiplying both sides by
$dx$, then integrate both sides.
\begin{eqnarray}
dy & = & u^{n}\;du \nonumber \\
\int dy & = & \int u^{n}\;du \nonumber \\
y+C & = & \int u^{n}\;du \nonumber
\end{eqnarray}

But  $y=\frac{u^{n+1}}{n+1}$.  Substituting this expression for $y$
gives us the following formula for the integral of a power of a
function:
\begin{boxedequation}
\int u^{n}\;du = \frac{u^{n+1}}{n+1}+C\ \ \ \ (n\neq-1)
\label{eq:powerrule}
\end{boxedequation}
Note that the left hand side of this formula contains $du$, which
is the differential of the function $u$.  This differential is
given by
\[du=\frac{du}{dx}dx\]

The quantity under the integral sign in Equation~(\ref{eq:powerrule})
\textbf{must} contain the differential $du$ in addition to $u^n$. If
it does not contain $du$, but contains a constant multiple of $du$, we
can multiply by a constant to obtain the required expression for $du$,
\emph{provided we compensate by dividing by the same constant outside
  the integral sign.}  The objective is to have both the function $u$
and the differential $du$ present following the integral sign.  This
must be done before the above integration formula can be used.

\clearpage
\textbf{Examples}

Integrate the following:
\begin{enumerate}
\item $\displaystyle\int (5x+1)^{2}\;dx$
\item $\displaystyle\int (x^{2}+1)^{3}x\;dx$
\item $\displaystyle\int (x^{3}-2)(5x^{2})\;dx$
\item $\displaystyle\int \frac{dx}{\sqrt{3x+1}}$
\item $\displaystyle\int \sqrt[3]{4-3x}\;dx$
\item $\displaystyle\int \frac{x+1}{\sqrt{x^{2}+2x}}\;dx$
\newline
\newline
Find the function whose derivative is given (i.e. solve each
differential equation):
\item $\displaystyle\frac{dy}{dx}=\sqrt[4]{1-2x}$
\item $\displaystyle\frac{dy}{dx}=x(x^{2}-5)^2$
\end{enumerate}

It should be noted that the solution of a differential equation, such as
the last two examples, consists not of a number, but a function that
makes the differential equation true.

\subsection{Integral of a Reciprocal Function}

You may have wondered why the rule (\ref{eq:powerrule}) cannot be
applied when $n=-1$.  In that case the right hand side of the equation
would then have zero in the denominator and we know dividing by zero
is meaningless.  However we can still integrate $u^{-1}$.  If $u$ is a
function of $x$, the indefinite integral of $\frac{du}{u}$ is given by
\begin{boxedequation}
\int \frac{du}{u}=\ln|u|+C
\end{boxedequation}
where $\ln|u|$ is the natural logarithm of the absolute value of $u$.

Note that, just as in the case of the power formula, the differential
{du} \emph{must be present} before the formula can be
applied.

\clearpage
\textbf{Examples}

\begin{enumerate}
\item $\displaystyle\int \frac{dx}{1-x}$
\item $\displaystyle\int \frac{x\;dx}{3x^{2}+1}$
\item $\displaystyle\int \frac{(x-1)\;dx}{x^{2}-2x}$
\item $\displaystyle\int \frac{x^2\;dx}{1-x^{3}}$
\item $\displaystyle\int \frac{x^2+1}{x^3+3x}\;dx$
\end{enumerate}

\subsection{Using a Table of Integrals or a Computer}

The Table of Integrals (Textbook Appendix C, p. A-44) can be used to
find many more integrals.\footnote{Note that there is nothing magic
  about a table of integrals, they arise simply by considering the
  forms that derivatives of known functions can take.}  The trick is
to choose a given function $u$ of $x$ which makes your indefinite
integral match an expression in the table.  (The differential $du$
must appear correctly as well up to a multiplicative constant which
can be factored out of the integral as usual using
Formula~(\ref{eq:multiplicativeconstant}).)  In this course students
are not expected to solve these other possible indefinite integrals,
but if they would like may try some by looking at Examples 29-43, on
page 869 of the text.\footnote{Being able to only integrate $x^n$ is
  still quite powerful.  Formulas~(\ref{eq:multiplicativeconstant})
  and (\ref{eq:summation}) then allow us to calculate the indefinite
  integral of any \emph{polynomial} $f(x)=a_n x^n+a_{n-1}
  x^{n-1}+\ldots+a_1x+a_0$.  The \emph{Weierstrass approximation theorem}
  states that every continuous function defined on an interval $[a,b]$
  can be approximated as closely as desired by such a polynomial
  function.  As such our minimal integration tools can, in theory,
  find an approximate indefinite integral of arbitrary accuracy to any
  function by first approximating it with a polynomial.}

In practice one will often use computer programs to
\emph{symbolically} evaluate an integral in which you are interested.
(i.e. to find an indefinite integral by searching for patterns as
being done by hand here.)  Figure~\vref{fig:wxmaxima} shows the
computer program \emph{wxMaxima} computing an indefinite integral.
\mbox{wxMaxima} is a free (see \texttt{wxmaxima.sourceforge.net})
cross-platform (i.e. runs on Linux, Windows, or Mac) mathematics
software package.  Proprietary packages that perform similar
calculations also exist (\emph{Maple}, \emph{Mathematica}, etc.).
Note, however, that computers can only calculate what is possible.
Unlike taking derivatives of common functions, not all such functions
have an antiderivative which can be written in a closed form in terms
of familiar functions.


\begin{figure}
\includegraphics[width=\textwidth,keepaspectratio=,clip=]{wxmaxima.jpg}
\caption{\label{fig:wxmaxima}Screenshot of the computer program
  \emph{wxMaxima} computing the \mbox{indefinite} integral \mbox{$\int
  (5x^2+2x+3+\frac{1}{x})\;dx$}.}
\end{figure}

\textassignment{Sec. 30-2, pp. 864-868}{Ex. 2, p. 869 \# 1-28 (all)}

\section{The Constant of Integration}

When we integrate, we are finding a function whose derivative (or
differential) is given.  But since there is always a constant of
integration, $C$, the resulting function is not unique, but
actually represents a whole family of functions, each of which has a
different value of $C$, but is otherwise the same (see Fig. 30-2
on page 871 of the text).

In order to ``pin down'' the value of $C$ and get a unique function,
we must be given some additional information, such as the coordinates
of a point through which the function passes.  We can use this
information to evaluate the constant of integration.

If we are given the second derivative of a function, we can get the
original function by integrating twice in succession.  For each step,
we get a constant of integration.  To evaluate both constants, we need
two additional pieces of information (for example, the coordinates of a
point and the slope of the tangent line at that point).

\textbf{Examples}
\begin{enumerate}
\item Find the equation of the curve whose tangent slope at $x$ is
  $\sqrt{6x-3}$ and which passes through $(2, -1)$.
\item The second derivative of a function is $12x^2$.  The curve
  passes through $(1,-1)$ and the tangent slope at that point is 9.
  Find the function.
\end{enumerate}
\textbf{Additional Examples} (Text, p. 873)
\begin{itemize}
\item Problem 8
\item Problem 10
\item Problem 12
\end{itemize}

\textassignment{Sec. 30-3, pp. 870-872}{Ex. 3, p. 873 \# 5-11 (odd)}

%The following section was omitted for MAT 201.
%\section{Equations of Motion}
%
%\subsection{Straight-Line Motion}
%
%Recall that velocity is the derivative of displacement, i.e.
%\[v=\frac{ds}{dt}\]
%
%In differential form, this can be written as follows:
%\[ds=v\;dt\]
%
%Integrating both sides, we get
%\[s=\int v\;dt\] 
%
%Acceleration is the derivative of velocity, i.e.
%
%\[a=\frac{dv}{dt}\]
%
%In differential form,
%
%\[dv=a\;dt\]
%
%Integrating gives
%
%\[v=\int a\;dt\]
%
%\subsection{Curvilinear Motion}
%
%Applying the above process to the $x-$ and $y-$components of
%displacement, velocity and acceleration gives the following formulas
%for $x$, $y$, $v_x$ and $v_y$:
%
%\begin{center}
%\begin{tabular}{cc}
%$\displaystyle x=\int v_x\;dt$ & $\displaystyle y=\int v_y\;dt$ \\
%$\displaystyle v_x=\int a_x\;dt$ & $\displaystyle v_y=\int a_y\;dt$
%\end{tabular}
%\end{center}
%
%\clearpage
%\textbf{Example}
%\begin{itemize}
%\item A cannonball is fired vertically with an initial upward velocity
%  of $125$ m/s.  find its velocity and displacement after $15$
%  seconds. (Take $g = -9.8$ $\textrm{m}/\textrm{s}^2$.)
%\end{itemize}
%\textbf{Additional Example} (Text, p. 889)
%\begin{itemize}
%\item Problem 8
%\end{itemize}
%
%\subsection{Rotational Motion}
%Reversing the formulas for angular velocity
%($\omega$) and angular acceleration ($\alpha$),
%\begin{eqnarray}
%\omega & = & \frac{\theta}{dt} \nonumber \\
%\alpha & = & \frac{d\omega}{dt} \nonumber
%\end{eqnarray}
%gives the following formulas for angular displacement and angular
%velocity:
%\begin{eqnarray}
%\theta =\int \omega\;dt \nonumber \\
%\omega =\int \alpha\;dt \nonumber
%\end{eqnarray}
%
%\textbf{Example} (Text, p. 889)
%\begin{itemize}
%\item Problem 12
%\end{itemize}
%
%\textassignment{Sec. 31-1, pp. 886-889}{Ex. 1, p. 889 \# 1-13 (odd)}
\section{Review Problems: Introduction to \mbox{Integration}}
\textbf{Questions:}
\begin{enumerate}
\item Integrate the following:
  \begin{enumerate}
  \item $\displaystyle\int (2-4x^2)\;dx$
  \item $\displaystyle\int x^2(x^3-3)^4\;dx$
  \item $\displaystyle\int \frac{t+2}{t}\;dt$
  \end{enumerate}
% \item A particle starts from the origin.  Its $x$-component of
%  velocity is \mbox{$t^2-3$ mm/s}, and its $y$-component of velocity
%  is $3t$ mm/s.
%  \begin{enumerate}
%  \item Write equations for $x$ and $y$ as functions of time.
%  \item Find the distance of the particle from the origin after $2.00 s$.
  \item Chapter 30 Review Problems, p. 883 \# 1-9 (all), 24
%\item Chapter 31 Review Problems, p. 909 \# 9, 11, 18
\end{enumerate}
\clearpage
\textbf{Answers:}
\begin{enumerate}
\item \begin{enumerate}
  \item $2x-\frac{4}{3}x^3+C$
  \item $\frac{1}{15}(x^3-3)^5+C$
  \item $t+2\ln|t|+C$
  \end{enumerate}
% \item \begin{enumerate}
%   \item $x=\frac{t^3}{3}-3t$, $y=\frac{3}{2}t^2$
%   \item 6.86 mm
%   \end{enumerate}
\item Chapter 30  (Answers to Even-Numbered Review Problems):
  \begin{description}
  \item[2.] $\displaystyle \frac{x^2}{2}+x-\frac{1}{2x^2}+C$
  \item[4.] $\displaystyle -\frac{2}{t}+C$
  \item[6.] $\displaystyle \frac{1}{9}(x^3-4)^3+C$
  \item[8.] $\displaystyle \frac{1}{2}\ln|x^2+3|+C$
  \item[24.] $y=\frac{x^3}{6}+\frac{9}{2}x-18$
  \end{description}
%\item Chapter 31:
%  \begin{description}
%  \item[18.] $v=\frac{3t^2}{2}$;\ \ $s=\frac{t^3}{2}+s_0$
%  \end{description}
\end{enumerate}
\clearpage
\section{Formulas}
\begin{center}
\textbf{\large Indefinite Integrals}
\end{center}
\setcounter{equation}{0}
\begin{boxedeqnarray}
\int f(x)\;dx & = & F(x)+ C\ \ \ \ \left(\ \mathit{where}\;f(x)=F'(x)\ \right) \\
\int du & = & u + C \\
\int a\;f(x)\;dx& = &a\int f(x)\;dx=a\;F(x)+C \label{eq:ap1}\\
\int\left[f(x)+g(x)+\ldots\right]\;dx & = &\int f(x)\;dx+\int g(x)\;dx+\ldots \\
\int x^n\;dx& = &\frac{x^{n+1}}{n+1}+C\ \ \ \ (n\neq-1)\\
\int u^{n}\;du& = &\frac{u^{n+1}}{n+1}+C\ \ \ \ (n\neq-1) \label{eq:ap2}\\
\int \frac{du}{u}& = &\ln|u|+C \label{eq:ap3}\\
\nonumber
\end{boxedeqnarray}
Note that $a$ is a \emph{constant} in (\ref{eq:ap1}). 
In formulas (\ref{eq:ap2}) and (\ref{eq:ap3}) $u$ is a function of $x$ (for example $u=2x^2+9$) and $du=\frac{du}{dx}dx$, where $\frac{du}{dx}$ is the derivative of $u$ with respect to $x$.
\begin{center}
\textbf{\large Derivatives}
\end{center}
\begin{boxedeqnarray}
\frac{d}{dx}\left(a\right) & = & 0 \nonumber \\
\frac{d}{dx}\left(x^n\right) & = & nx^{n-1} \nonumber\\
\frac{d}{dx}\left(au\right) & = & a\frac{du}{dx} \nonumber  \\
\frac{d}{dx}\left(u+v\right) & = & \frac{du}{dx}+\frac{dv}{dx} \nonumber \\
\frac{d}{dx}\left(uv\right) & = & u\frac{dv}{dx}+v\frac{du}{dx} \nonumber \\
\frac{d}{dx}\left(\frac{u}{v}\right) & = & \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2} \nonumber \\
\frac{dy}{dx} & = & \frac{dy}{du}\cdot\frac{du}{dx} \nonumber \\
\frac{d}{dx}\left(u^n\right)& = & n u^{n-1}\frac{du}{dx} \nonumber \\
\nonumber
\end{boxedeqnarray}
In these formulas $a$ is a \emph{constant}, $u$ and $v$ are
\emph{functions} of $x$.
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