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\usepackage{calculus}
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\title{Math 165: Honors Calculus I\\
Exam I Solutions}
\date{October 1, 1998}

\begin{document}

\maketitle

\bigskip

\begin{enumerate}

\item Use the axioms for the real numbers to prove the following statements.
  \begin{enumerate}[a)]
  \item $0\cdot a = 0$ for all $a\in\R$

\emph{Proof.}
\[
\begin{array}{rcll}
            0 + 0 &=& 0       &[\mbox{Axiom 4}]\\
a(0 + 0)          &=& a\cdot0 &[\mbox{multiply both sides by $a$]}\\
a\cdot0 + a\cdot0 &=& a\cdot0 &[\mbox{Axiom 3}]\\
-a\cdot0 + a\cdot0 + a\cdot0 &=& -a\cdot0 + a\cdot0
                             &[\mbox{add $-a\cdot0$ to both sides}]\\
      0 + a\cdot0 &=& 0       &[\mbox{Axiom 5}]\\
          a\cdot0 &=& 0       &[\mbox{Axiom 4}]\\
\end{array}
\]

  \item $(-1)(-1) = 1$

\medskip
\emph{Proof.}
\[
\begin{array}{rcll}
            1 + (-1) &=& 0      &[\mbox{Axiom 5}]\\
      (-1)(1 + (-1)) &=& (-1)0  &[\mbox{multiply both sides by $(-1)$}]\\
      (-1)(1 + (-1)) &=& 0      &[\mbox{by a)}]\\
    (-1)1 + (-1)(-1) &=& 0      &[\mbox{Axiom 3}]\\
    (-1) + (-1)(-1)  &=& 0      &[\mbox{Axiom 4}]\\
 1 + (-1) + (-1)(-1) &=& 1 + 0  &[\mbox{add $1$ to both sides}]\\
        0 + (-1)(-1) &=& 1      &[\mbox{Axioms 4, 5}]\\
            (-1)(-1) &=& 1      &[\mbox{Axiom 5}]\\
\end{array}
\]

  \item $1>0$

\medskip
\emph{Proof.}
If $1 \notin \R^+$, then $-1\in\R^+$ by Axiom~8 ($1\ne0$ by Axiom~4). Then
Axiom~7 implies that $(-1)(-1)\in\R^+$. But $(-1)(-1) = 1$ by b), a
contradiction. Therefore, $1\in\R^+$.

  \end{enumerate}

\problemspace
\item \begin{enumerate}[a)]
  \item Define an inductive set,~$S$.

\medskip
\emph{Definition.}
A subset $S\subset\R$ is inductive if
\begin{enumerate}[1)] \item $1\in S$, and \item if $x\in S$ then $x+1\in S$.
\end{enumerate}
\medskip

  \item Define the positive integers,~$\N$.

\medskip
\emph{Definition.}
$n \in \N$ if and only if $n$ is in every inductive set.
\medskip

  \item Prove by induction that for $n\in \N$,
    \[\sum_{k=1}^n k(k-1) = \frac{n^3-n}{3} \]

\medskip
\emph{Proof.}
\begin{enumerate}[1)]
\item The formula holds for $n = 1$:
\[\sum_{k=1}^1 k(k-1) = 1(1-1) = 0 = \frac{1^3-1}{3}\]

\item Assume the formula holds for $n$. Then
  \begin{eqnarray*}
  \sum_{k=1}^{n+1} k(k-1) &=& \sum_{k=1}^n k(k-1)  + (n+1)n\\
       &=& \frac{n^3 - n}{3} + n^2+n\quad[\mbox{induction hypothesis}]\\
       &=& \frac{n^3 - n + 3n^2 + 3n}{3}\\
       &=& \frac{(n^3 + 3n^2 + 3n+1) - (n+1)}{3}\\
       &=& \frac{(n+1)^3 - (n+1)}{3}\\
  \end{eqnarray*}
So the formula holds for $n+1$. Therefore, by the \textsc{Principle of
Mathematical Induction}, the formula holds for all $n\in\N$.

\end{enumerate}

  \end{enumerate}

\problemspace
\item 
  \begin{enumerate}[a)]
  \item Define completely ${n\choose k}$.

\medskip
\emph{Definition.}
\[ {n\choose k} = \frac{n!}{(n-k)!k!} \]
where $n!$ is defined inductively by $0! = 1$ and for $n\ge 1$,
$n! = n\cdot(n-1)!$
\medskip

  \item State the {\sc Binomial Theorem}.

\medskip
\emph{Theorem.}
For any $x$, $y\in\R$ and $n\in\N$,
\[
(x + y)^n = \sum_{k=0}^n{n\choose k}x^{n-k}y^k
\]
  \item Prove that for any positive integer~$n$,
  \[ \sum_{k=0}^n (-1)^k{n\choose k}2^{n-k} = 1 \]

\medskip
\emph{Proof.} By the \textsc{Binomial Theorem}, 
\[ 1 = (2 - 1)^n = \sum_{k=0}^n {n\choose k}2^{n-k}(-1)^k \]
  \end{enumerate}

\problemspace
\item 
  \begin{enumerate}[a)]
  \item For a subset of $S\subset\R$, define $\sup S$ and $\inf S$.

\medskip
\emph{Definition.}
\begin{enumerate}[1)]
\item $\sup S$ is the least upper bound for $S$, i.e., $x \le \sup S$
for all $x\in S$ and if $x \le B$ for all $x \in S$, then $\sup S \le B$.

\item $\inf S$ is the greatest lower  for $S$, i.e., $\inf S \le x$ for all
$x\in S$ and if $B \le x$ for all $x \in S$, then $B \le \inf S$.
\medskip

\end{enumerate}

  \item Let $S = \bigg\{ \frac{n}{n+1}\, \big|\, n\in \N\bigg\}$. 
  Prove that $\sup S = 1$.

\medskip
\emph{Proof.}
Since $0 < 1$, $n < n+1$ for all $n\in \N$. Thus, $\frac{n}{n+1} < 1$ for
all $n\in\N$ showing that 1 is an upper bound for~$S$. In particular,
$\sup S \le 1$.  Suppose $k = \sup S < 1$. Then $\frac{n}{n+1} \le
k < 1$ for all $n\in\N$.  But then, solving for $n$ and using the
fact that $1-k>0$, we get $n \le \frac{k}{1-k}$ for all $n\in\N$, a
contradiction. In fact, $\left[\frac{k}{1-k}\right]+1$ is a positive
integer that is greater than $\frac{k}{1-k}$.

  \end{enumerate}

\problemspace
\item Give precise definitions of the following.
  \begin{enumerate}[a)]
  \item $\int_a^b s(x)\,dx$ where $s$ is a step function on~$[a,b]$.

\medskip
\emph{Definition.}
Let $P = \{x_0,x_1,\ldots,x_n\}$ be a partition of $[a,b]$ on whose
subintervals the step function $s$ is constant:
\[
s(x) = s_k\quad\textrm{for}\ x_{k-1}<x<x_k,\quad k = 1,\ldots,n.
\]
Then the integral of $s$ from $a$ to $b$ is defined to be
\[
\int_a^b s(x)\,dx = \sum_{k=1}^n s_k(x_k - x_{k-1})
\]

  \item The lower integral, $\underline{I}(f)$, of a bounded
  function $f$ on $[a,b]$.

\medskip
\emph{Definition.}
Let $S = \bigg\{ \int_a^b s(x)\,dx\;\big|\; s \le f \bigg\}$. Here $s \le
f$ means that $s$ is a step function on $[a,b]$ such that $s(x) \le f(x)$
for all $x\in[a,b]$. Since $f$ is bounded on $[a,b]$, there is a constant
$M$ such that $f(x)\le M$ for $x\in[a,b]$. This implies that $\int_a^b
s(x)\,dx \le M(b-a)$ for any step function $s \le f$, so the set $S$
is bounded from above. By Axiom~10, the supremum of $S$ exists and the
lower integral is defined to be
  \[ \underline{I}(f) = \sup S \]
\medskip

  \item An integrable function $f$ on $[a,b]$.

\medskip
\emph{Definition.}
A function $f$ is integrable on $[a,b]$ if $f$ is bounded on $[a,b]$ and
there is \emph{exactly one} number $I$ satisfying
\[
\int_a^b s(x)\,dx \le I \le \int_a^b t(x)\,dx
\]
for all step functions $s$, $t$ such that $s(x) \le f(x) \le t(x)$ for
$x\in[a,b]$. (This number $I$ is also denoted by $\int_a^b f(x)\,dx$.)

  \end{enumerate}

\problemspace
\item Give precise statements of the following theorems for integrals.
  \begin{enumerate}[a)]
  \item \textsc{Linearity with Respect to the Integrand}.

\medskip
\emph{Theorem.}
If $f$ and $g$ are integrable on $[a,b]$, then for any constants
$c_1$, $c_2\in\R$, the function $c_1 f + c_2 g$ is also integrable on
$[a,b]$ and
  \[
  \int_a^b c_1f(x)+c_2g(x)\,dx = c_1\int_a^b f(x)\,dx + c_2\int_a^b g(x)\,dx
  \]

  \item \textsc{Additivity with Respect to the Interval of
  Integration}.

\medskip
\emph{Theorem.}
Let $a \le c \le b$. Then $f$ is integrable on $[a,b]$ if and only if $f$
is integrable on $[a,c]$ and on $[c,b]$. Moreover,
  \[ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx \]

  \item \textsc{Comparison Theorem}

\medskip
\emph{Theorem.}
If $f$ and $g$ are integrable on $[a,b]$ and $f(x) \le g(x)$ for all
$x\in[a,b]$, then \[ \int_a^b f(x)\,dx \le \int_a^b g(x)\,dx \]

  \end{enumerate}

\problemspace
\item
  \begin{enumerate}[a)]
  \item By \textsc{Invariance Under Translation},
\[
\int_{2}^3 (x-3)^{16}\,dx = \int_{-1}^0 x^{16}\,dx
 = \frac{0^{17}}{17} - \frac{(-1)^{17}}{17} = \frac{1}{17}
\]

  \item $\int_{-1}^1 \frac{x}{\sqrt{4-x^4}}\,dx = 0$.
The integral is 0 because $f(x) = \frac{x}{\sqrt{4-x^4}}$ is an odd
function, $f(-x) = - f(x)$, and the integral is over a balanced
interval, $[-1,1]$.
\medskip

  \item Find the area between the graphs of $f(x) = x^3$ and $g(x)
  = 3x^2-2x$ on the interval $[0,2]$.

\medskip
\emph{Solution.}
The area is defined to be \[ A = \int_0^2 | f(x) - g(x) | \,dx \]
Since $f(x) - g(x) = x^3 - 3x^2 + 2x = x(x-1)(x-2)$, we see that
$f(x)-g(x)$ is positive on $[0,1]$ and negative on $[1,2]$.
Therefore,
  \begin{eqnarray*}
  A &=& \int_0^1 x^3 - 3x^2 + 2x \,dx - \int_1^2 x^3 - 3x^2 + 2x \,dx \\
    &=&  \frac{1}{4}x^4 - x^3 + x^2\bigg|_0^1
         -  \big(\frac{1}{4}x^4 - x^3 + x^2\big)\bigg|_1^2\\
    &=& \frac{1}{4} - 1 + 1 - \big(4 - \frac{1}{4} - 8 + 1 + 4 - 1\big)\\
    &=& \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\\
  \end{eqnarray*}

  \end{enumerate}

\problemspace
\item Use step functions that are constant on subintervals of
equal length to compute a numerical approximation for
$\int_0^1 \frac{1}{1+x^2}\,dx$ that is accurate to within $\pm0.25$.

\medskip
\emph{Solution.}
Since $f(x) = \frac{1}{1+x^2}$ is a decreasing function on $[0,1]$,
we know from Theorem~1.14 that
  \[
  B_n \le \int_0^1\frac{1}{1+x^2}\,dx \le B_n + E_n
  \]
where 
  \[
  B_n = \frac{1 - 0}{n}\sum_{k=1}^n\frac{1}{1+x_k^2}
  \]
and
  \[
  E_n = \frac{(f(0) - f(1))(1 - 0)}{n} = \frac{1}{2n}
  \]
The partition points are given by $x_k = \frac{k}{n}$. If we choose
the midpoint, $B_n+\frac{1}{2}E_n$, of the interval $[B_n,B_n+E_n]$
for the approximation, then the approximation will be within $\pm 0.25$
of the true value of the integral when $\frac{1}{2}E_n  = \frac{1}{4n} <
0.25$. This  holds for $n > 1$ (!), so we may take $n = 2$.  The midpoint
is then
  \begin{eqnarray*}
  B_2 + \frac{1}{2}E_2 &=& \frac{1}{2}\bigg(\frac{1}{1+(\frac{1}{2})^2} + 
  \frac{1}{1+1^2}\bigg) + \frac{1}{8}\\
  &=& \frac{13}{20} + \frac{1}{8} = 0.65 + 0.125 = 0.775
  \end{eqnarray*}
(The true value of the integral is $\frac{\pi}{4} \approx 0.785$).
\[
\resizebox{2in}{!}{\includegraphics{exam1solns.eps}}
\]


\end{enumerate}

\end{document}
