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\fancyfoot[CO,CE]{\it Calculus Review by A. A. Tovar, Ph. D., Created Jan. 2009, not Amended.}
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\begin{document}
\title {Differential and Integral Calculus Review and Tutorial}
\author{Anthony A. Tovar, Ph. D.
\\Eastern Oregon University
\\1 University Blvd.
\\La Grande, Oregon, 97850 }
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\maketitle
\dominitoc
\faketableofcontents
\chapter{Calculus Review}
\minitoc
\vspace{12 pt}
\noindent
This tutorial is a review of the basic results of differentiation and integration.
Of course some of the results may be new to some of the readers. Hopefully, those
readers will find the new results interesting as well as informative.
\vspace{12 pt}
\noindent
There are many things one could say about the history of calculus, but one of the most
interesting is that integral calculus was first developed by Archimedes of Syracuse
OVER 2250 YEARS AGO! He was a very interesting guy. You can google him to learn more,
but I highly recommend the (historical fiction) book "The Sand Reckoner" by Gillian Bradshaw
which is a story of his life.
\section{What are Elementary Functions?}
The Elementary functions are:
\begin{enumerate}
\item Polynomials (of integer and complex order)
\item Exponential and Logarmithmic Functions
\item Sinusoidal and Inverse Sinusoidal Functions
\item Hyberbolic Sinusoidal and Inverse Hyperbolic Sinusoidal Functions
\item Any finite number of Sums, Products, or Compositions of Elementary Functions
\end{enumerate}
\noindent
Here are some examples of elementary functions:\\
\begin{tabular}{|l|c|}
\hline
Elementary Function & Examples\\
\hline
Polynomials & $a_3 x^3 + a_2 x^2 + a_1 x + a_0$ \\
Exponential and Logarmithmic Functions& $e^{ax}$, $ln(ax)$\\
Sinusoidal and Inverse Sinusoidal Functions& $cos(ax)$, $tan^{-1}(ax)$\\
Hyberbolic Sinusoidal and Inverse Hyperbolic Sinusoidal Functions&$cosh(ax)$, $tanh^{-1}(ax)$ \\
Composite Elementary Function& $\frac{e^{sin(x) +x^2}}{cosh(x)} + ln(7x)$\\
\hline
\end{tabular}
\vspace{12 pt}
\noindent
So, what isn't an elementary function? There are certain integrals and differential
equations that ``can't be solved" so instead of solving them, we name them. For
example, Bessel functions
are solutions to ``Bessel's Equation." Nonelementary functions that have a special
name are known as {\bf Special Functions}. Another common example of a special function
is the Error Function which is the solution to the integral
\begin{equation}
erf(x) = \frac{2}{\sqrt{\pi}} \int_0^{x} e^{-\hat{x}^2} d\hat{x}
\end{equation}
\section{Differential Calculus}
Fast Facts:
\begin{enumerate}
\item Definition: $f^\prime (x) = lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$.
\item Derivative is an operator (it operates on functions).
\item In particular, the derviative is the slope operator. Thus, it represents
a ``rate of change." When the independent variable is time, the derivative
becomes a time rate of change. The time rate of change of position is
velocity, the time rate of change of velocity is acceleration, and
the time rate of change of acceleration is jerk. It can be readily
seen that the units of the time derivative of $f$ are $f/time$.
\item The inverse operator is the antiderivative or integral (This is the Fundamental Theorem
of Calculus).
\item The Integral is the Area Operator.
\item The Derivative of any Elementary Function is an Elementary Function.
\end{enumerate}
\subsection{Some Derivatives}
Here is a very short table of derivatives:
\begin{center}
\begin{tabular}{|c|c|}
\hline
Function & Derivative\\
\hline
$x^n$ & $nx^{n-1}$\\
$e^x$ & $e^x$\\
$ln (x)$ & $\frac{1}{x}$\\
\hline
$sin (x)$ & $cos (x) $\\
$cos (x)$ & $-sin (x) $\\
$-sin (x)$ & $-cos (x) $\\
$-cos (x)$ & $sin (x) $\\
\hline
\end{tabular}
\end{center}
\noindent
One may notice that the derivatives for $sin$ and $cos$ follow a simple
pattern (and that this pattern reminds one of the unit circle):
\begin{center}
\includegraphics[scale=0.5]{graphics/differentiation.eps}
\end{center}
\subsection{Basic Theorems}
\begin{center}
\begin{tabular}{|c|c|}
\hline
Theorem Name & Theorem\\
\hline
Chain Rule & $\frac{d}{dx} \left( {A(B(x)} \right) =\frac{dA(B)}{dB} \frac{dB(x)}{dx}
$\\
Linearity & $
\frac{d}{dx} \left( aA(x) + bB(x) \right) = a\frac{dA}{dx} + b\frac{dB}{dx}
$\\
Product Rule & $\frac{d}{dx} \left( A(x) B(x) \right) =
A(x) \frac{dB}{dx} + \frac{dA}{dx} B(x)$\\
Quotient Rule & $\frac{d}{dx} \left( \frac{A(x)}{B(x)} \right) =
\frac{B(x) \frac{dA}{dx} - A(x)\frac{dB}{dx}}{B^2(x)}$\\
\hline
\end{tabular}
\end{center}
\subsection{Implicit Differentiation}
Implicit differentiation is used when you do not have an explicit solution for
the dependent variable of interest.
Here is an example:\\
Find $dy/dx$ of the following function (it's the equation of a circle):
\begin{equation}
x^2 + y^2 = 1
\end{equation}
Differentiating each of the terms yields
\begin{equation}
2x dx + 2y dy = 0
\end{equation}
or
\begin{equation}
\frac{dy}{dx} = -x/y
\end{equation}
A typical application would be the max/min problem, which could be accomplished
by setting $dy/dx = 0$ which yields $x=0$. This occurs when $y^2=1$ or $y=\pm 1$,
as expected.
\subsection{Logarithmic Differentiation}
One would like to extend the product rule for more than two functions. This can
be achieved with Logarithmic Differentiation. Suppose you want to find the
derivative of
\begin{equation}
y(x) = A(x)B(x)C(x)D(x)
\end{equation}
You could apply the product rule many times, or take the logarithm of both sides first:
\begin{equation}
ln(y(x)) = ln(A(x)) + ln(B(x)) + ln(C(x)) + ln(D(x))
\end{equation}
where a simple property of the logarithm has been used. Now, taking the derivative
of both sides yields
\begin{equation}
\frac{1}{y(x)} \frac{dy}{dx} =
\frac{1}{A(x)} \frac{dA}{dx} +
\frac{1}{B(x)} \frac{dB}{dx} +
\frac{1}{C(x)} \frac{dC}{dx} +
\frac{1}{D(x)} \frac{dD}{dx}
\end{equation}
Finally, multiplying both sides by $y(x)$ yields
\begin{equation}
\frac{dy}{dx} =
B(x)C(x)D(x)\frac{dA}{dx} +
A(x)C(x)D(x) \frac{dB}{dx} +
A(x)B(x)D(x) \frac{dC}{dx} +
A(x)B(x)C(x) \frac{dD}{dx}
\end{equation}
\noindent
Of course, if one is so inclined, one could generalize the results. If $y(x)$ is written
\begin{equation}
y(x) = \prod_{i=1}^{N} A_i(x)
\end{equation}
then
\begin{equation}
\frac{dy}{dx} =
\sum_{j=1}^{N} \frac{1}{A_j (x)} \frac{dA_j (x)}{dx}
\prod_{i=1}^{N} A_i(x)
\end{equation}
\subsection{Pascal's Triangle}
Consider Pascal's Triangle:
\begin{center}
\includegraphics[scale=0.5]{graphics/pascals_triangle.eps}
\end{center}
Each of the numbers are obtained by adding the two adjacent numbers in the
row above it. For example, 10 is below it's adjacent 4 and 6.
The rows of Pascal's Triangle represent the binomial coefficients, so that
\begin{equation}
(a+b)^3 =
\mathbf{1} \cdot a^3 b^0 +
\mathbf{3} \cdot a^2 b^1 +
\mathbf{3} \cdot a^1 b^2 +
\mathbf{1} \cdot a^0 b^3
\end{equation}
or more simply
\begin{equation}
(a+b)^3 =
a^3 + 3a^2 b + 3a b^2 + b^3
\end{equation}
There is a formula for the binomial coefficients, given the ``Choose Function":
\begin{equation}
{n \choose k} = \frac{n!}{k!(n-k)!}
\end{equation}
where $n$ is the row and $k$ goes from 1 to $n$. For example, the third
number on the 5th row of Pascal's Triangle is a 10, but
\begin{equation}
{5 \choose 3} = \frac{5!}{3!(5-3)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10
\end{equation}
\noindent
Note that the top solitary ``1" is considered to be in the zeroth row and
that $0!=1$ by convention.
This relates to differential calculus, because there is a similar relationship
that applies to the product rule to multiple derivatives. Suppose we are
interested in the $4^{th}$ derivative of a product:
\begin{equation}
y^{(4)} (x) = \left ( A(x) B(x) \right)^{\prime \prime \prime \prime}
\end{equation}
While the derivative terms that must exist are straightforward, it also happens
that the coefficients of the derivatives are given by Pascal's Triangle, so that
in this case
\begin{equation}
y^{(4)} (x) =
A^{\prime \prime \prime \prime}(x) B(x) +
4A^{\prime \prime \prime }(x) B^{\prime}(x) +
6A^{\prime \prime }(x) B^{\prime \prime}(x) +
4A^{\prime }(x) B^{\prime \prime \prime}(x) +
A(x) B^{\prime \prime \prime \prime}(x)
\end{equation}
Thus, by inspection it can be seen that the $4^{th}$ derivative of $xe^x$ is
\begin{equation}
\frac{d^4}{dx^4} (xe^x) = (4+x) e^x
\end{equation}
as the first 3 terms are zero. By inspection, can you determine the $4^{th}$ derivative
of $x^2e^x$? (See Bottom of Page for answer)\footnote{$(12+8x+x^2 )e^x$}
The general case for the $n^{th}$ derivative of a product of two functions $A(x)$
and $B(x)$ may be written
\EQ
y^{(n)} (x) =
\sum_{k=0}^{n} {n \choose k} A^{(k)}(x) B^{(n)}(x)
\EN
It is left for the enthusiastic reader to obtain a general expression for the
$n^{th}$ derivative of the product of $m$ functions.
\section{Integral Calculus}
Fast Facts:
\begin{enumerate}
\item Definition of a Definite Integral: $\int_a^b f(x) dx =lim_{N \rightarrow \infty}
\left( \frac{b-a}{N} \right)
\sum_0^{N-1} f\left( a+ i \left( \frac{b-a}{N} \right) \right) $.
\item Definition of an Indefinite Integral: $\int f(x) dx = \int_a^x f(\hat{x}) d\hat{x} + C$
\item The Indefinite Integral is an operator (it operates on functions).
\item In particular, the Indefinite Integral is the Accumulated Area Operator. The area is
achieved by summing many rectangles of length $\Delta x = (b-a)/N$ and height
$f(a+i\Delta x )$. Thus, the units of $\int_a^b f(x) dx$ is the units of $f(x)$ multiplied
by the units of $x$. The integral introduces the peculiar-to-some idea of {\bf Negative Area}. For example in integral calculus the area of a circle centered at the origin is NOT $\pi r^2$, it's ZERO as the bottom half of the circle is said to have negative area!
\item The Indefinite Integral of any Elementary Function {\bf may or may not be} an Elementary Function.
\end{enumerate}
\subsection{Some Integrals}
\begin{center}
\begin{tabular}{|c|c|c|}
\hline
Function & Simple Form & Advanced Form\\
\hline
$x^n \;\; (n\neq -1)$ & $\frac{x^{n+1}}{n+1}$ &
$\int f^n(x) \frac{df}{dx} dx = \frac{f^{n+1}(x)}{n+1}$\\
$x^{-1}$ & $ln |x| $ &
$\int \frac{df/dx}{f} dx = ln |f(x)| $\\
$e^x$ & $e^x$ & $\int e^{f(x)} \frac{df}{dx} dx = e^{f(x)} $\\
$ln (x)$ & $x ln (x) - x $ &
$\int ln(f(x))\frac{df}{dx} dx = f(x) ln (f(x)) - f(x) $\\
\hline
$sin(x)$ & $-cos(x)$ & $\int sin(f(x)) \frac{df}{dx} dx = cos(f(x))$\\
$-cos(x)$ & $-sin(x)$ & \\
$-sin(x)$ & $cos(x)$ & \\
$cos(x)$ & $sin(x)$ & \\
\hline
\end{tabular}
\end{center}
\subsection{Techniques of Integration}
\begin{tabular}{|l|l|}
\hline
Technique & When to Use\\
\hline
u-Substitution & When it's obvious or when you're stuck.\\
Integration by Parts & When you have a product of two functions, and
you know \\
& the derivative of one and the integral of the other.\\
Trigonometric Substitution & When you have $(a + x^2)$ or $(a - x^2)$ terms (especially in the denominator).\\
Synthetic Division/Partial Fraction& When you have a ratio of polynomials.\\
Series Solution & When you stuck but realize a Taylor Series is easy to calculate.\\
\hline
\end{tabular}
\vspace{12 pt}
\noindent
As you may recall, the formula for integration by parts is
\begin{equation}
\int u dv = uv-\int v du
\end{equation}
A common mistake when using integration by parts on a definite integral is to
forget to evaluate the $uv$ term with the integration limits.
\subsection{Examples of Integration Techniques}
{\bf Example (Advanced Forms)} \\
\begin{equation}
\int \frac{x^2}{1 + x^3} dx = \frac{1}{3} \int \frac{3x^2}{1 + x^3} dx = \frac{1}{3}
ln |1 + x^3|
+C
\end{equation}
\noindent
Comment: The ``Advanced Forms" involve what I call the ``Hope Method." You see a
complicated function and hope that the derivative of the inside of the equation is
sitting on top. In this case it is. One is sometimes taught to use u-Substitution
here, but this integral should be done in your head. Notice how the integral would
be much harder if it is $\int \frac{1}{1 + x^3} dx$. That's why it's the Hope method,
you hope that derivative is there!
\vspace {24 pt}
\noindent
{\bf Example (Advanced Forms)} \\
\begin{equation}
\int x^5 ln(x) dx = \frac{1}{6} \int x^5 ln(x^6) dx = \frac{1}{36} \int 6 x^5 ln(x^6) dx =
\frac{1}{36} \left[ x^6 ln |x^6| -x^6 \right] =
\frac{1}{36} \left[ 6 x^6 ln |x| -x^6 \right]
+C
\end{equation}
\noindent
Comment: One of my favorite integrals. By using the properties of the logarithm, we can
make the derivative show up by just multiplying by a constant. The integral could be done
by integrating by parts, but it would be longer and much less cool!!
\vspace {24 pt}
\noindent
{\bf Example (Integration by Parts)} \\
\begin{equation}
\int \underbrace{x}_{u} \underbrace{e^x dx}_{dv} = \underbrace{x}_u \underbrace{e^x}_v -
\int \underbrace{e^x}_v \underbrace{dx}_{du} = e^x (x-1)
+C
\end{equation}
\noindent
Comment: You can see that if we had $\int x^n e^x dx$ we would perform integration
by parts $n$ times, as the power of the monomial decreases by 1 every time, but the
exponential stays firm. It is a good idea to take the derivative of your result to
insure a correct answer. In this case the derivative is obtained by the product rule:
$e^x (1) + (x-1)e^x = xe^x$ as it must.
\vspace {24 pt}
\noindent
{\bf Example (Synthetic Division)} \\
\begin{equation}
\int \frac{x^3 + 2x^2 + 2x +1}{x+2} dx = \int x^2 + 2 -\frac{3}{x+2} dx =
\frac{x^3}{3} + 2x -3ln|x+2|
+C
\end{equation}
\noindent
Comment: Use synthetic division on a rational function when the degree of the polynomial
on top is greater than or equal to that on the bottom.
Some students haven't done synthetic division in a while. It's just the
same as long division you first learned in grade school. Unfortunately, you may
have not done that in a while either!
\vspace {24 pt}
\noindent
{\bf Example (Partial Fraction Expansion)} \\
The integrand of
\begin{equation}
\int \frac{1}{x^2 + 6x+8} dx
\end{equation}
can be reduced by using a technique in ALGEBRA known as Partial Fraction Expansion. In
particular, the integrand can be rewritten
\begin{equation}
\frac{1}{x^2 + 6x+8} = \frac{1}{(x + 2)(x+4)} = \frac{A}{x+2} + \frac{B}{x+4}
\end{equation}
Solving for $A$ and $B$ yields $A=1/2$, $B=-1/2$. The solution is
\begin{equation}
\int \frac{1}{x^2 + 6x+8} dx = \frac{1}{2} [ln |x+2| - ln |x+4|] = \frac{1}{2}
ln \left| \frac{x+2}{x+4} \right|
+C
\end{equation}
\noindent
Comment: For the integral of a fifth degree polynomial divided by a second degree
polynomial, one would use synthetic division first, and use Partial Fractions on
the remaining integral. There is a technique where the partial fractions coefficients
can be determined by inspection.
\vspace {24 pt}
\noindent
{\bf Example (Series Solution)} \\
The following integral does not exist as a finite elementary function:
\begin{equation}
\int e^{-x^2} dx
\end{equation}
However, we may proceed by noting that the Taylor Series expansion for an exponential function is
\begin{equation}
e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}
\end{equation}
Thus, the integral becomes
\begin{equation}
\int e^{-x^2} dx = \int \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}
\int x^{2n} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \frac{x^{2n+1}}{2n+1}
\end{equation}
\noindent
Comment: If you multiply our result by $2/\sqrt{\pi}$, we just found the Taylor Series
expansion for erf(x).
\vspace {24 pt}
\noindent
{\bf Example (Trigonometric Substitution)} \\
Find the ``area" of a cirle. The equation of a circle is $x^2 +y^2 = r^2$. To find the area
we will double the area of the top half of the circle, which can be found be integration:
\begin{equation}
A = 2\int_{-r}^r \sqrt{r^2 - x^2} dx
\end{equation}
Making the substitution $x=rcos( \phi )$, it follows that $dx = -r sin( \phi ) d\phi$, and
that
\begin{equation}
A = 2\int_{\pi}^0 \sqrt{r^2 - r^2 cos^2 (\phi)} (-r sin ( \phi )) d\phi =
-2 r^2 \int_{\pi}^0 sin^2 ( \phi )) d\phi
= -2 r^2 \int_{\pi}^0 \left( \frac{1}{2} - \frac{1}{2} cos (2 \phi ) \right) d\phi = \pi r^2
\end{equation}
\noindent
Comment: Trig sub gives expected answer.
\vspace {24 pt}
\noindent
{\bf Example (Bonus Fun - Coordinate System Conversion)} \\
\begin{equation}
y_0 = \int_{0}^\infty e^{-x^2} dx = \int_{0}^\infty e^{-y^2} dy
\end{equation}
Thus
\begin{equation}
y_0^2 = \int_{0}^\infty e^{-x^2} dx \int_{0}^\infty e^{-y^2} dy =
\int_{0}^\infty \int_{0}^\infty e^{-x^2} e^{-y^2} dxdy
\end{equation}
Using a rectangular to polar conversion yields
\begin{equation}
y_0^2 = \int_{0}^\infty \int_{0}^{\pi /2} e^{-r^2} rd\phi dr=
\frac{\pi}{2} \left( -\frac{1}{2} \right)
\int_{0}^\infty e^{-r^2} (-2r)dr =
\frac{\pi}{4}
\end{equation}
or
\begin{equation}
y_0 = \frac{\sqrt{\pi}}{2}
\end{equation}
\noindent
Comment: You can see why erf(x) has the $\frac{2}{\sqrt{\pi}}$ normalized factor in front -
it's so that the area under the erf(x) is 1. This solution is a fun little trick that only
works when the integral is extended to infinity. In general, though, if you see an
integral whose limits are $-\infty$ to $\infty$, look to see if the integrand is an odd
function (Hope Method). If it is, the integral is zero. Finally, notice that the area under
the curve is finite, but that the length of the curve is infinite. This deserves some thought.
If you had such a box, it would mean that you could fill the box with paint, but you would
never have enough paint to paint the box!! Interestingly, It is also possible to have figures
which have infinite perimeter with a finite area. Fractals have these properties.
\section{Types and Methods of Solution}
Since not all integrals have solutions in terms of Elementary functions, there are
different types of solutions that may be obtained. Not all of these have equal merit.
\begin{enumerate}
\item Exact Analytical Explicit Solutions. (Best)
\item Exact Analytical Implicit Solutions. (Good)
\item Series Solutions. (Not as good)
\item Numerical Solutions. (OK, if nothing else works)
\end{enumerate}
\vspace{12 pt}
\noindent
Students should familiarize themselves with the following solution methods.
\begin{enumerate}
\item Analytical Methods (such as the ``Techniques of Integration", above).
\item Analytical and Numerical Solutions using Computer Algebra Systems (such as Maple).
\item Numerical Solutions using Spreadsheets.
\item Numerical Solutions using a programming language (such a C++).
\end{enumerate}
\vspace{12 pt}
\noindent
As added motivation, students should be aware that
\begin{itemize}
\item These types and methods of solution apply to differential equations as well
as integrals.
\item Integrals are a special case of a differential equation.
\item The laws of physics which apply to all of nature and devices created by
man are governed by differential equations, {\bf which, under many circumstances reduce
to integrals}.
\item Many of the algorithms for numerical integration and solution to differential
equations are available (without charge) on the Internet (as is a host of related
material). Try a web search for ``Numerical Recipes in C++."
\end{itemize}
\section{Applications of Calculus}
There are very many applications of calculus. Below is a short table of functions and paramters of
general interest to the mathematical and scientific community.
\begin{center}
\begin{tabular}{|c|c|}
\hline
Application & Formula\\
\hline
Area under curve & $\int_a^b f(x) dx$\\
Slope of curve & $\frac{df}{dx}$\\
Extremum of a curve (max, min, inflection pt.) & $\frac{df}{dx} = 0$\\
Inflection point of a curve & $\frac{d^2f}{dx^2} = 0$\\
Arc Length & $\int_a^b \sqrt{1 + \left( \frac{df}{dx} \right) ^2} dx$\\
Curvature (Radius) & $\frac{\left[ 1+ \left( \frac{df}{dx} \right)^2 \right]^{3/2}}{\left|
\frac{d^2f}{dx^2} \right|}$\\
Average of a function & $\frac{1}{b-a} \int_a^b f(x) dx$\\
RMS of a function & $\sqrt{\frac{1}{b-a} \int_a^b f^2(x) dx}$\\
\hline
Center of Mass& $\int_a^b xf(x) dx$\\
Variance (Second Central Moment) & $\int_a^b (x-\bar{x})^2f(x) dx$\\
Skewness (Third Central Moment) & $\int_a^b (x-\bar{x})^3f(x) dx$\\
Kurtosis (Fourth Central Moment) & $\int_a^b (x-\bar{x})^4f(x) dx$\\
Function Squared Norm &$\int_a^b |f(x)|^2 dx$\\
\hline
Fourier Transform & $\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(t)e^{-i \omega t} dt$\\
Convolution Integral &$\int_{-\infty}^\infty f(\tau ) g(t- \tau ) d\tau$\\
Taylor Coefficients & $\frac{1}{n!} \frac{d^n f}{dx^n} |_{x=a}$\\
Fourier Sine Series & $\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) sin(nt) dt$\\
Fourier Cosine Series & $\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) cos(nt) dt$\\
\hline
\end{tabular}
\end{center}
\end{document}