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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Indefinite integration}

This is also known as {}``antidifferentiation{}'', but that word is really
convoluted.\footnote{%
It sounds like {}``antidisestablishmentarianism{}'', which had its 15 minutes
of fame as the longest word in the dictionary. By the way, this is supposed
to mean the philosophy of opposing those who would change the established system.
I actually heard someone use it once, on TV. That person's show was cancelled
long ago, since no one knew what he was saying. It's not even the longest word
any more, having been beaten by some miner's lung disease.
} 

The process of finding an anti-derivative, or an \textit{indefinite integral},
of a function is usually a \textbf{lot} harder than finding a derivative. You
can tell this isn't just because we don't want to teach the methods, since you
can buy a calculator that can find derivatives of functions, but even a big
computer can't find \textit{all} antiderivatives of any function you might write
down. But, we can start off easy: 

\noindent \textit{Find the function \( F(x) \) whose derivative is \( x^{4} \).
That is, find the antiderivative of \( x^{4} \).}  

Now, you know that the derivative of \( x^{5} \) is \( 5x^{4} \), \( (x^{5})'=5x^{4} \),
so we only have to change the input a little to take care of that 5; \[
\left( \frac{1}{5}x^{5}\right) '=x^{4}.\]


The only problem is that this is not the only function with that derivative.
The others are only a trivial change from this one. For any constant \( c \),
\[
\left( \frac{1}{5}x^{5}+c\right) '=x^{4}.\]


By the way, I don't like calling these things ``antiderivatives'', so we're
going to start calling the antiderivative of a function \( f \) the \textit{indefinite
integral} of \( f \). We write the indefinite integral of \( f \), \( F \)
by: \[
F(x)=\int f(x)dx+C.\]


The \( +C \) is there to remind you that the indefinite integral is only known
up to a constant. Maybe we should call it \textit{an} indefinite integral of
\( f \), but we don't. 

\begin{thm}
The indefinite integral of \( x^{a} \) is \[
\int x^{a}dx=\frac{1}{a+1}x^{a+1}+C,\]
 if \( a\neq -1 \). 
\end{thm}
There is no need to work through a formal proof of this, since all you have
to do to check that it is correct is differentiate the right-hand side. 

\begin{xca}
Why doesn't the formula in this theorem work for \( a=-1 \)? 
\end{xca}

\subsubsection{Examples/Table}

You need to know these particular integrals. In general, I do not like to tell
students to memorize things, but for this I make an exception. Again, to verify
that these are correct, differentiate the right-hand sides of the equations. 

\vspace{0.3cm}
{\centering \begin{tabular}{|r|rclc|}
\hline 
1&
\( \int x^{n}dx \) &
=&
\( \frac{1}{n+1}x^{n+1}+C \)&
\( n\neq -1 \)\\
\hline 
2&
\( \int \frac{1}{x}dx \)&
=&
\( \ln |x|+C \)&
\( \left( =\int x^{-1}dx\right)  \)\\
\hline 
3&
\( \int e^{x}dx \)&
=&
\( e^{x}+C \)&
\\
\hline 
4&
\( \int \cos (x)dx \)&
=&
\( \sin (x)+C \)&
\\
\hline 
5&
\( \int \sin (x)dx \)&
=&
\( -\cos (x)+C \)&
\\
\hline 
6&
\( \int \sec ^{2}(x)dx \)&
=&
\( \tan (x)+C \)&
\\
\hline 
7&
\( \displaystyle \int \frac{1}{\sqrt{1-x^{2}}}dx \)&
=&
\( \arcsin (x)+C \)&
\( =\sin ^{-1}(x)+C \)\\
\hline 
8&
\( \displaystyle \int \frac{1}{1+x^{2}}dx \)&
=&
\( \arctan (x)+C \)&
\( =\tan ^{-1}(x)+C \)\\
\hline 
9&
\( \int \frac{1}{|x|\sqrt{x^{2}-1}}dx \)&
=&
\( \textrm{arcsec}(x)+C \) &
\( =\sec ^{-1}(x)+C \)\\
\hline 
\end{tabular}\par}
\vspace{0.3cm}

There are some general rules for integration as well. These are the easy ones.

\begin{prop}
~
\begin{enumerate}
\item \[
\int cf(x)dx=c\int f(x)dx\]
 
\item \[
\int f(x)\pm g(x)dx=\int f(x)dx\pm \int g(x)dx\]
 
\end{enumerate}
\end{prop}
\textbf{\emph{Proof:}} These facts all follow from the fact that these are antderivatives,
and the same rules hold for derivatives. 

\begin{example}
Find \[
\int x^{2}+3x^{4}dx.\]
 
\end{example}

\subsubsection*{Solution}

\begin{example}
\[
\int \sin x-\frac{1}{x}dx=-\cos x-\ln |x|+C\]
 

Note the \( ln|x| \) in that answer. The integral of \( 1/x \) is just \( \ln (x)+C \)
for \( x \) positive, but is \( \ln (-x)+C=\ln |x|+C \) for \( x \) negative,
as you can check by differentiating, as usual.
\end{example}

\subsubsection{Motion}

If the position of a particle at time \( t \) is given by \( s(t) \), then
we saw that its velocity was \( v=s'(t) \). The second derivative of the position,
the rate of change of the velocity, is the acceleration \[
a(t)=v'(t)=s''(t).\]


The understanding of motion problems really involves, usually, working this
relationship backwards. Many times you will know the acceleration (this will
come out of the forces involved), and you'll have to find the position as a
function of time. If the acceleration is \( a(t) \), then, using the notation
I mentioned earlier for antiderivatives, \[
v(t)=\int a(t)dt+C,\]


where I am adding the \( +C \) for emphasis, since the notation \( \int dt \)
is supposed to mean any antiderivative of whatever is inside, so the \( +C \)
is really there already. But, here you might be able to figure out, from the
problem, the velocity at some time (usually \( t=0 \)), and use that to solve
for the \( C \). Then, \[
s(t)=\int v(t)dt+C\]


and again you figure out the constant. 

\begin{example}
Assume that a thingie is moving along a line so that its acceleration is \( 4feet/sec^{2} \).
If at \( t=0 \) the thingie is moving at \( 5feet/sec \) and is \( 7feet \)
from origin, what is its position at any time \( t \)? 
\end{example}

\subsubsection*{Solution}

As long as we stay near the surface of the Earth, the acceleration of a body
downward due to gravity is \( 32\, feet/sec^{2} \), (or \( 9.8\, meters/sec^{2} \)),
which can be taken to be a constant for our purposes. If we presume that the
air resistance can be forgotten (not really a good assumption, but it makes
the problem easier), then a body in free fall will have that acceleration. Use
that in the problems below. 

\begin{xca}
I shot an arrow, into the air. I shot it at a \( 45^{\circ } \) angle to the
ground, at an initial speed of 100 feet per second. The arrow was 5 feet off
the ground when it left my bow. I was pointed due North at the time. Where did
it fall to earth? Assume that there is no air resistance, and that the ground
is perfectly flat. 
\end{xca}
%

\begin{xca}
A mischievous child (certainly \textit{not} my child) has heard about dropping
pennies off the Empire State Building. So, he goes the story one better. While
standing on the observation deck 1,250 feet above the street, he throws a penny
straight down, with an initial velocity of 75 feet per second. How fast is it
going when the penny hits the ground? Assume that there is no air in New York. \end{xca}

\end{document}
