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\title{Take Home Exam 1}
\target{Math 158: Spring 2013}

\begin{document}


\begin{exam}{Due: Tuesday, March 11}



  \bp \textbf{[2 points]} Here's a picture of the elliptic curve
  $y^2=x^3-5x+8$.  The coordinates of the black point are $(1,2)$.
  What are the coordinates of the red point?

  \begin{center}
    \includegraphics[scale=0.4]{elliptic.pdf}
  \end{center}
  \ba First, differentiate to find the slope of the tangent line to
  the curve:
$$2y \frac{dy}{dx}=3x^2-5\Rightarrow \frac{dy}{dx}=\frac{3x^2-5}{2y}.$$
So, the tangent line at $(1,2)$ has slope $-\frac{1}{2}$ and equation
$$y-2=-\frac{1}{2}(x-1).$$
Since the red point lies both on the line and on the curve, its
coordinates satisfy both $y^2=x^3-5x+8$ and $y-2=-\frac{1}{2}(x-1).$ I
solved these equations by writing $y=-\frac{1}{2}(x-1)+2$ and
substituting it into $y^2=x^3-5x+8$ to get
$$\left(-\frac{1}{2}(x-1)+2\right)^2=x^3-5x+8 \Rightarrow
4x^3-x^2-10x+7=0.$$ Realizing that $x=1$ is a solution to this
equation helps to factor this cubic
$$4x^3-x^2-10x+7=(x-1)^2(4x+7)=0$$
and we find the $x$ coordinate of the red point is $-\frac{7}{4}$.  It
follows that the $y$ coordinate is $\frac{27}{8}$.  So, the answer is
$\ds \left(\frac{7}{4},\frac{27}{8}\right)$ \ea \ep



\bp \textbf{[2 points]} Factoring the identity function.  \ben
\item Find two functions $f$ and $g$ satisfying $f(0)=0$, $g(0)=0$ and
  $x=f(x)g(x)$.  \ba There are many ways to do this.  Here's one:
  $f(x)=x^{\frac{1}{3}}$ and $g(x)=x^\frac{2}{3}$.  \ea
\item Prove that there do not exist two differentiable functions $f$
  and $g$ with $f(0)=0$, $g(0)=0$ and $x=f(x)g(x)$.  \ba If
  $x=f(x)g(x)$ and $f$ and $g$ were differentiable, we'd have
  $1=f'(x)g(x)+f(x)g'(x)$.  Setting $x=0$ yields
  $1=f'(0)g(0)+f(0)g'(0)$.  If $f(0)=g(0)=0$, we get a contradiction
  $1=0$.

  \emph{Note:} Neither $x^{\frac{1}{3}}$ nor $x^\frac{2}{3}$ are
  differentiable at $x=0$.  \ea \een

  \ep

  \bp \textbf{[2 points]} Suppose that $f:[0,1]\to [0,1]$ is
  differentiable and that $f'(x)\neq 1$ for any $x$.  Prove that there
  is one and only one number $c\in [0,1]$ with $f(c)=c.$ \ba First, we
  prove there is a number $c\in [0,1]$ with $f(c)=c$.  We only need
  continuity for this.  If $f(0)=0$ or $f(1)=1$, we're done.  So,
  assume $f(0)>0$ and $f(1)<1$.  Since $f$ is differenitable on
  $[0,1]$, it is continuous, hence $g(x)=f(x)-x$ is continuous.  Since
  $g(0)=f(0)-0>0$ and $g(1)=f(1)-1<0$, the intermediate value theorem
  implies that there exists a number $c\in [0,1]$ with $g(c)=0$.  That
  is, $f(c)=c$.

  Now suppose $f$ is differentiable and $f'(x)\neq 1$ for any $x$.
  Assume that $f(c)=c$ and $f(d)=d$.  If $c\neq d$, the mean value
  theorem says there exists a number $x$ between $c$ and $d$ with
  $f'(x)=\frac{f(d)-f(c)}{d-c}=1$, which isn't possible if there are
  no numbers $x$ with $f'(x)=1$.  Therefore $c=d$.  \ea \ep


  \bp \textbf{[2 points]} The standard normal curve is given by $$\ds
  y=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}.$$ Give a good sketch of
  this curve.  Be sure to indicate the points at which the curve
  changes concavity.  \ba Let $\ds
  f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}.$ Note that $f(x)>0$
  for all $x$.  We compute
$$f'(x)=-\frac{1}{\sqrt{2\pi}}xe^{-\frac{x^2}{2}}.$$
Note that $f(0)=0$, $f(x)>0$ if $x<0$ and $f(x)<0$ if $x>0$.
Therefore,
\begin{itemize}
\item $f$ is increasing on the interval $(-\infty, 0)$
\item $f$ attains a maximum value of $f(0)=\frac{1}{\sqrt{2\pi}}$ at
  $x=0$, and
\item $f$ is decreasing on the interval $(0,\infty)$
\end{itemize}
We compute $$f''(x)=-\frac{1}{\sqrt{2\pi}} \left(
  e^{-\frac{x^2}{2}}-x^2 e^{-\frac{x^2}{2}}\right)=
\frac{1}{\sqrt{2\pi}} \left( x^2-1 \right) e^{-\frac{x^2}{2}}.$$ Since
the exponential function is always positive, the sign of $f''(x)$
equals the sign of $x^2-1$, which is negative on $(-1,1)$ and positive
elsewhere.  Therefore
\begin{itemize}
\item $f$ is concave up on $(\infty,-1)\cup (1,\infty)$, and
\item $f$ is concave down on the interval $(-1,1).$
\item $f$ changes concavity at the points $\left(-1,\frac{1}{\sqrt{2 e
        \pi}}\right)$ and $\left(1,\frac{1}{\sqrt{2 e \pi}}\right)$.
\end{itemize}

\begin{center}
  \includegraphics[scale=0.5]{normalcurve.pdf}
\end{center}

\ea \ep



\bp \textbf{[2 points]} For any $x>0$, let $\ds A(x)=\int_2^x f(t)dt$
where $f$ is the function whose graph is sketched below:

\begin{center}
  \includegraphics[scale=0.75]{plot_of_f.pdf}
\end{center}

Give a good sketch of the graph of $A$.  Indicate the value of $A$ and
$A'$ at $x=0,1,2,3,4,5,6,7,8$ and indicate on which intervals $A$ is
concave up and concave down.

\ba From the fundamental theorem of calculus, $A'(x)=f(x)$, which can
be read from the graph.  Then, we can determine concavity by looking
at $A''(x)=f'(x)$ and see that $A$ is concave up when $f$ is
increasing (on the intervals $(0,2)$ and $(6,8)$) and see that $A$ is
concave down when $f$ is decreasing (on the intervals $(2,6)$.)

To find the values of $A$, I computed areas of triangles, keeping care
with the signs.  Here's a table, and a sketch:

\begin{center}
  \begin{tabular}{|c||c|c|c|c|c|c|c|c|c|}
    \hline
    $x$ \T \B& $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$\\
    \hline
    $A'(x)$ \T \B & $0$ & $1$ & $2$ & $0$ & DNE & $\frac{1}{2}$ & $0$  & $\frac{1}{2}$ & $1$\\
    \hline
    $A(x)$ \T \B & $-2$ & $-\frac{3}{2}$ & $0$ & $1$ & $0$ & $\frac{3}{4}$ & $1$
    & $\frac{5}{4}$ & $2$\\
    \hline
  \end{tabular}

  \bigskip

\includegraphics[scale=0.75]{plot_of_A.pdf}
\end{center}


\ea

\ep


\bp \textbf{[2 points]} Suppose $a<b<c<d<e$.  \ben
\item Find the minimum of
  $f(x)=(x-a)^2+(x-b)^2+(x-c)^2+(x-d)^2+(x-e)^2$ \ba $f$ is a degree
  two polynomial, it has one minimum where $f'(x)=0$.  A computation
  shows the minimum is at the average of the numbers $a,b,c,d,e$:
  \begin{multline*}
    f'(x)=2(x-a)+2(x-b)+2(x-c)+2(x-d)+2(x-e)=0\\
    \Rightarrow 5 x = (a+b+c+d+e) \Rightarrow x
    =\frac{1}{5}\left(a+b+c+d+e\right).
  \end{multline*}
  So, the minimum value
  is $$f\left(\frac{1}{5}\left(a+b+c+d+e\right)\right).$$ 
\ea

\addtocounter{enumi}{1}

\item \textbf{[Bonus point]} Prove that if
  $a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}+\frac{e}{5}=0$ then
  $a+bx+cx^2+dx^3+ex^4=0$ for some $x\in [0,1]$.  \ba Note if $f$ is
  continuous on $[a,b]$ and $\int_a^b f=0$, then there must exist a
  number $x\in [a,b]$ with $f(x)=0$.  If $f(x)>0$ and continuous, then
  $\int_a^b f(x) >0$.  Similarly, if $f(x)<0$ and continuous,
  $\int_a^b f(x)<0$.  Thus, if $\int_a^b f =0$ and $f$ is continuous,
  we know $f$ can't be strictly positive, nor strictly negative, and
  so by the intermediate value theorem, $f$ must be zero for at least
  one $x\in [a,b]$.

  Now, applying this to $$\int_0^1 a+bx+cx^2+dx^3+ex^4
  dx=a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}+\frac{e}{5}$$ proves the
  statement.  \ea \een \ep


  \addtocounter{problemnum}{-1}


  \bp \textbf{Continued.} 
\ben
\addtocounter{enumi}{1}

\item Find the minimum of $g(x)=|x-a|+|x-b|+|x-c|+|x-d|+|x-e|$ 
\ba The
  graph of $g$ is continuous and piecewise linear---it's a straight line with nonzero
  slope on each subinterval $(-\infty, a)$, $(a,b)$, $(b,c)$, $(c,d)$,
  $(d,e)$, $(e,\infty)$.  The minimum of $g$ must occur where the
  derivative doesn't exist; i.e. one of the points $\{a,b,c,d,e\}$.
  To figure out which, we'll evaluate $g$ at each point.   
\begin{gather*}
g(a)=(b-a)+(c-a)+(d-a)+(e-a)\\
g(b)=(b-a)+(c-b)+(d-b)+(e-b)\\
g(c)=(c-a)+(c-b)+(d-c)+(e-c)\\
g(d)=(d-a)+(d-b)+(d-c)+(e-d)\\
g(e)=(e-a)+(e-b)+(e-c)+(e-d) 
\end{gather*} 
Since $a<b$, the last three terms of $g(a)$ are greater than the
last three terms of $g(b)$, so we find that $$g(a)>g(b).$$
Since $d<e$, the first three terms of $g(e)$ are greater than the first
three of $g(d)$, so we find that 
$$g(e)>g(d).$$ 

Now, by combining the first two terms of $g(b)$, the first and third
terms of $g(c)$, and last two terms of $g(d)$ we find that 
\begin{gather*}
g(b)=(c-a)+(d-b)+(e-b)\\
g(c)=(c-a)+(d-a)+(e-c)\\
g(d)=(d-a)+(d-b)+(e-c)\\
\end{gather*} 
Comparing $g(b)$ with $g(c)$, we find 
that $$g(b)>g(c)$$ becasue the last two terms of $g(b)$ are
greater than the last two terms of $g(c)$.
Comparing $g(c)$ with $g(d)$, we find that 
$$g(c)<g(d)$$ because the first two terms of $g(d)$ are greater than
the first two terms of $g(c)$.

Conclusion:  The minimum of $g$ occurs at $c$ and the minimum value is $d+e-a-b$.

\ea
\een \ep

  \bp Here is the unit circle $x^2+y^2=1$.


  \begin{center}
    \includegraphics[scale=0.5]{circle.pdf}
  \end{center}


  \ben
\item \textbf{[1 point]} Express the area of the sector $OAC$ in terms
  of $t$.  \ba First, we compute the area of the region $ABC$ (with
  the circular part):
  \begin{align*}
    Area(ABC)&=\int_{x=\cos(t)}^{x=\cos(0)} y dx= \int_{t}^0 \sin(u)
    (-\sin(u)du)\\
    &=\int_{0}^t \sin^2(u)du  \\
    &=\frac{u}{2}-\frac{\cos(2u)}{4}\biggr]_0^t\\
    &=\frac{t}{2}-\frac{\cos(2t)}{4}.
  \end{align*}
  Now, we add the area of the triangle $OAB$ which is
  $\frac{1}{2}\sin(t)\cos(t)$ to get the area of sector $OAC$
$$Area(OBC)=Area(ABC)+Area(OAB)=\frac{t}{2}-\frac{\cos(2t)}{4}+\frac{1}{2}\sin(t)\cos(t)=\frac{t}{2}.$$
To make the last simplification, recall that
$\cos(2t)=2\sin(t)\cos(t).$ \ea
\een \ep

  \addtocounter{problemnum}{-1}


  \bp \textbf{Continued.} 
\ben
\addtocounter{enumi}{1}
\item \textbf{[1 point]} Suppose the point $A$ is moving
  counterclockwise around the circle so that the area of the sector
  $OAC$ is increasing at a constant rate.  Determine the rate is the
  area of triangle $OAC$ changing at the moment that it is
  equilateral.  
\ba It will be helpful to have some notation.  Let $a$ be the area of
triangle $OAC$ and let $b$ be the area of sector $OAC$.  Let $s$
represent time, expressed in some units.
The quantities $a$ and $b$ depend on time and so they are functions of
$s$.

The area $a$ (as a function of $t$) is given by  
$a(t)=\frac{1}{2}\left(\overline{OC}\right)\left(\overline{AB}\right)=\frac{1}{2}\sin(t).$  
The rate at which $a$ is changing is given by
$$\frac{da}{ds}=\frac{da}{dt}\frac{dt}{ds}=\frac{1}{2}\cos(t)\frac{dt}{ds}.$$

The area $b$ as a function of $t$ was computed above to be
$b(t)=\frac{t}{2}$.  The rate at which $b$ is changing (which we are
told is constant) is given by
$$\frac{db}{ds}=\frac{db}{dt}\frac{dt}{ds}=\frac{1}{2}\frac{dt}{ds}.$$

Putting these together, we see that
$\frac{da}{ds}=\cos(t)\frac{db}{ds}$.  At the moment that
the triangle is equilateral, $\cos(t)=\frac{1}{2}$ and we find that
the rate at which the area of the triangle is increasing is
$\frac{ds}{dt}=\frac{1}{4}\frac{dt}{ds}$, which is half the
rate at which the area of the sector is increasing.

  \ea \een \ep

  \addtocounter{problemnum}{-1}


  \bp \textbf{Continued.}  Here is the unit hyperbola $x^2-y^2=1$.

  \begin{center}
    \includegraphics[scale=0.5]{hyperbola.pdf}
  \end{center}

  \ben \addtocounter{enumi}{2}
\item \textbf{[1 point]} Express the area of the hyperbolic sector
  $OAC$ in terms of $t$. 
\ba
First, we compute the area of the region $ABC$ (including the curved piece):
\begin{align*}
Area(ABC)&=\int_{x=\cosh(0)}^{x=\cosh(t)} y dx= \int_{t}^0 \sinh(u)
(\sinh(u)du)\\
&=\int_{0}^t \sinh^2(u)du \\
&=-\frac{u}{2}+\frac{\sinh(2u)}{4}\biggr]_0^t\\
&=-\frac{t}{2}+\frac{\sinh(2t)}{4}.
\end{align*}
The area of the triangle $OAB$ is $\frac{1}{2}\cosh(t)\sinh(t).$
To get the answer, we subtract
$$Area(OBC)=Area(OAB)-Area(ABC)=\frac{1}{2}\sinh(t)\cosh(t)-\frac{t}{2}+\frac{\sinh(2t)}{4}=\frac{t}{2}.$$
To make the last simplification, recall that
$\sinh(2t)=2\sinh(t)\cosh(t).$
\ea
\een \ep

  \addtocounter{problemnum}{-1}


  \bp \textbf{Continued.} 
\ben
\addtocounter{enumi}{3}
\item \textbf{[Bonus point]} Suppose the point $A$ is moving away from
  the origin so that the area of the sector $OAC$ is increasing at a
  constant rate.  Determine the rate is the area of triangle $OAB$
  changing at the moment that it is isoceles.  
\ba The triangle $OAB$ will never be isoceles since
$\overline{OB}=\cosh(t)$ and $\overline{AB}=\sinh(t)$ are never equal
(and the hypotenuse $\overline{OA}$ is always longer than both
$\overline{OB}$ and $\overline{AB}$.)

So, instead, we'll compute the rate at which the area of triangle $OAC$ is
changing at the moment that it is isoceles.

It will be helpful to have some notation.  Let $a$ be the area of
triangle $OAC$ and let $b$ be the area of sector $OAC$.  Let $s$
represent time, expressed in some units.
The quantities $a$ and $b$ depend on time and so they are functions of
$s$.

The area $a$ (as a function of $t$) is given by  
$a(t)=\frac{1}{2}\left(\overline{OC}\right)\left(\overline{AB}\right)=\frac{1}{2}\sinh(t).$  
The rate at which $a$ is changing is given by
$$\frac{da}{ds}=\frac{da}{dt}\frac{dt}{ds}=\frac{1}{2}\cosh(t)\frac{dt}{ds}.$$

The area $b$ as a function of $t$ was computed above to be
$b(t)=\frac{t}{2}$.  The rate at which $b$ is changing (which we are
told is constant) is given by
$$\frac{db}{ds}=\frac{db}{dt}\frac{dt}{ds}=\frac{1}{2}\frac{dt}{ds}.$$

Putting these together, we see that
$\frac{da}{ds}=\cosh(t)\frac{db}{ds}$.  At the moment that
the triangle is isoceles, $$\overline{AC}=\overline{OC}
\Leftrightarrow \sqrt{(\cosh(t)-1)^2+\sinh(t)^2)}=1.$$   Solving for
$\cosh(t)$:
\begin{align*}
\sqrt{(\cosh(t)-1)^2+\sinh(t)^2)}=1 & \Rightarrow
(\cosh(t)-1)^2+\sinh(t)^2=1\\
& \Rightarrow
(\cosh(t)-1)^2+\cosh(t)^2-1=1\\
& \Rightarrow
2\cosh(t)^2-2\cosh(t)-1=0\\
& \Rightarrow \cosh(t)=\frac{1}{2} \left(1\pm\sqrt{3}\right)
\end{align*}
Since $\cosh(t)>0$, we find that $\cosh(t)=\frac{1}{2}
\left(1+\sqrt{3}\right)$ at the moment that the triangle $OAC$ is isoceles.

Therefore, $\frac{da}{ds}=\frac{1}{2}
\left(1+\sqrt{3}\right) \frac{db}{ds}.$

\ea
\een \ep



  \bp \textbf{[2 points]} Compute $\ds \int_0^\pi \sin^n(x)dx$ for
  $n=0,1,2,\ldots$.  
\ba
First, let us record a formula for integrating powers of the sine function:
\begin{equation*}
\int \sin^{k}(t) dt =
-\frac{1}{k}\sin^{k-1}(t)\cos(t)+\frac{k-1}{k}\int \sin^{k-2}(t)dt.
\end{equation*}
A way to deduce this formula is outlined as exercise 8 in section 5.10 in
Apostol's book, and I do it as a footnote to this problem.
Integrating over $[0,\pi]$ yields nice reduction formula:
\begin{equation}\label{reduction-eq}
\int_0^\pi  \sin^{k}(t) dt =
\frac{k-1}{k}\int_0^\pi \sin^{k-2}(t)dt.
\end{equation}

Let $c_n=\int_0^\pi \sin^n(t)dt.$  Notice that 
\begin{gather*}c_1=\int_0^\pi
\sin(t)=2\intertext{ and }c_2=\int_0^\pi \sin^2(t)dt = \int_0^\pi
\frac{1-\cos(2t)}{2} dt=\frac{\pi}{2}.
\end{gather*}  
Applying the reduction \eqref{reduction-eq}, one finds that $c_n$ satisfies the
recursion
$$c_n=\frac{n-1}{n}c_{n-1}$$ which allows us to determine $c_n$ for
every $n\in \N:$
\begin{align*}
c_1 & = 2 \\
c_2 & = \frac{\pi}{2} \\
c_3 &= \frac{2}{3} c_1  = \frac{4}{3} \\
c_4 &= \frac{3}{4} c_2  = \frac{3}{8}\pi \\
c_5 &= \frac{4}{5} c_3 = \frac{16}{15} \\
c_6 & = \frac{5}{6} c_4= \frac{5}{16}\pi \\
%c_7 & = \frac{6}{7} c_5= \frac{5}{35}\pi\\
&\vdots \\
c_k &= \begin{cases}
 2\left(\frac{2}{3}\right)\left(\frac{4}{5}\right)\left(\frac{6}{7}\right)\cdots\left(\frac{k-1}{k}\right) &\text{ if $k$ is odd,}\\
 \left(\frac{\pi}{2}\right)\left(\frac{3}{4}\right)\left(\frac{5}{6}\right)\left(\frac{7}{8}\right)\cdots
\left(\frac{k-1}{k}\right) &\text{ if $k$ is even}.
\end{cases}
\end{align*}
\ea
\ep


  \bp \textbf{[All correct: 3 points.  Eight correct: 1 points]}
  Compute.  Interpret any integrals above as limits if necessary.  For
  example,
$$ \int_0^1
\frac{dt}{\sqrt{1-t^2}}=\ds \lim_{x \to 1^{-}}\int_0^x
\frac{dt}{\sqrt{1-t^2}}.$$

\ben
\item $\ds \int_0^1 \frac{t dt}{\sqrt{1+t^2}}=\sqrt{2}-1$ \ba Let
  $u=1+t^2$.  Then $du=2tdt\Rightarrow tdt=\frac{1}{2}du$ and
  $\int_0^1$ becomes $\int_1^2$.  We have
$$ \int_0^1 \frac{tdt}{\sqrt{1+t^2}}=\int_1^2 \frac{\frac{1}{2}du}{\sqrt{u}}=\sqrt{u}\biggr]_1^2=\sqrt{2}-1.$$
\ea
\item $\ds \int_0^1 \frac{t dt}{\sqrt{1-t^2}}=1$ \ba Here, the
  integral is only defined over intervals contained in $(-1,1)$, so we
  interpret $\ds \int_0^1 \frac{tdt}{\sqrt{1-t^2}}$ as $\ds \lim_{x
    \to 1^-} \int_0^x \frac{tdt}{\sqrt{1-t^2}}$.  Now, we use the
  substitution $u=1-t^2$ which leads to $du=-2tdt\Rightarrow
  tdt=-\frac{1}{2}du$ and $\lim_{x \to 1^-}\int_0^x$ becomes $\lim_{y
    \to 0^+}\int_1^y$.
  \begin{align*}
    \int_0^1 \frac{tdt}{\sqrt{1-t^2}}&=\lim_{x\to 1^-}
    \int_0^x \frac{tdt}{\sqrt{1-t^2}}\\
    &=\lim_{y \to 0^+}
    \int_1^y -\frac{\frac{1}{2}du}{\sqrt{u}}\\
    &=-\sqrt{u}\biggr]_1^y\\
    &=\lim_{y \to 0^+}-\sqrt{y}-(-1)\\
    &=1.
  \end{align*}
  \ea
\item $\ds \int_0^1 \frac{t dt}{1+t^2}=\log(\sqrt{2})$ \ba Here, we
  use the same substitution as in part (a): $u=1+t^2$.  Then
$$ \int_0^1 \frac{tdt}{1+t^2}=\int_1^2 \frac{\frac{1}{2}du}{u}=\frac{1}{2}\log(u)\biggr]_1^2=\frac{1}{2}\log{2}.$$
\ea \een \ep

\addtocounter{problemnum}{-1}


\bp \textbf{Continued.}  \ben \addtocounter{enumi}{3}
\item $\ds \int_0^1 \frac{t dt}{1-t^2}=\infty$ \ba First, we must
  interpret this integral as $\lim_{x \to 1^-}\int_0^x
  \frac{tdt}{1-t^2}$.  Then, we make a substitution as in part (b):
  \begin{align*}
    \int_0^1 \frac{tdt}{1-t^2}&=\lim_{x\to 1^-}
    \int_0^x \frac{tdt}{1-t^2}\\
    &=\lim_{y \to 0^+}
    \int_1^y -\frac{\frac{1}{2}du}{u}\\
    &=-\frac{1}{2}\log(u)\biggr]_1^y\\
    &=\lim_{y \to 0^+}-\frac{1}{2}\log(y)-\left(-\frac{1}{2}\log(1)\right)\\
    &=\infty.
  \end{align*}
  \ea
\item $\ds \int_0^1 \frac{dt}{\sqrt{1+t^2}}=\ln(1+\sqrt{2}).$ \ba If
  you remember that $\arcsinh'(t)=\frac{1}{\sqrt{1+t^2}}$, then this
  problem is very simple.  If you don't, you can make the substitution
  $t=\sinh(u).$ Here, we change variables in the limit of integration:
  $t=0\Rightarrow u=0$, and $t=1\Rightarrow u=\arcsinh(1).$ So, we get
$$\ds \int_0^1
\frac{dt}{\sqrt{1+t^2}}=\int_0^{\arcsinh(1)}\frac{\cosh(u)du}{\cosh(u)}
= \int_0^{\arcsinh(1)}du = \arcsinh(1).$$ It's not necessary, but one
can rewrite $\arcsinh(1)$ as $\ln(1+\sqrt{2}).$

\emph{Remark:} If you want to express the $\arcsin(t)$ in terms of
logs but can't remember the formula, here's how to invert the
hyperbolic sine.  Write $u=\arcsinh(t) \Leftrightarrow
t=\sinh(u)=t=\frac{1}{2}(e^u-e^{-u})$.  Multiply by $e^u$ and simplify
to get $(e^{u})^2-2te^u-1=0$ and use the quadratic formula to get
$e^u=t+\sqrt{1+t^2}$.  \ea \een \ep

\addtocounter{problemnum}{-1}


\bp \textbf{Continued.}  \ben \addtocounter{enumi}{5}
\item $\ds \int_0^1 \frac{dt}{\sqrt{1-t^2}}=\frac{\pi}{2}$ \ba First
  note that this integral is only defined over intervals within
  $(-1,1)$, so we interpret the integral as a limit: $\ds \int_0^1
  \frac{dt}{\sqrt{1-t^2}}$ as $\ds \lim_{x \to 1^-} \int_0^x
  \frac{dt}{\sqrt{1-t^2}}.$ If you remember that $\arcsin'(t)=
  \frac{1}{\sqrt{1-t^2}}$, then this problem can be solved as follows:
$$\int_0^1 \frac{dt}{\sqrt{1-t^2}}=\lim_{x \to 1^-}\int_0^x
\frac{dt}{\sqrt{1-t^2}}=\lim_{x\ to
  1^-}\arcsin(x)-\arcsin(0)=\frac{\pi}{2}.$$

If you don't remember that $\arcsin'(t)= \frac{1}{\sqrt{1-t^2}}$, use
the substitution $t=\sin(u)$.  Then $dt=\cos(u)du$ and $\lim_{x \to
  1^-}\int_0^x$ becomes $\lim_{y \to \frac{\pi}{2}^-}\int_0^y.$
\begin{align*}
  \int_0^1 \frac{dt}{\sqrt{1-t^2}}&=\lim_{x \to 1^-}\int_0^x \frac{dt}{\sqrt{1-t^2}}\\
  &=\lim_{y \to \frac{\pi}{2}^-}\int_0^{y}\frac{\cos(u)du}{\cos(u)}\\
  &=\lim_{y \to \frac{\pi}{2}^-}\int_0^{y}du\\
  &=\lim_{y \to \frac{\pi}{2}^-}y\\
  &=\frac{\pi}{2}.
\end{align*}
\ea
\item $\ds \int_0^1 \sqrt{1-t^2}dt=\frac{\pi}{2}$ \ba This is the area
  of the upper half of the unit circle, which is $\frac{\pi}{2}.$ If
  you didn't realize this, you could make the substitution $t=\cos(u)$
  and proceed (this will lead to a computation just like the one in
  7(a)).  \ea

  \een \ep

  \addtocounter{problemnum}{-1}


  \bp \textbf{Continued.}  \ben \addtocounter{enumi}{7}
\item $\ds \int_0^1
  \sqrt{1+t^2}dt=\frac{\sqrt{2}}{2}+\frac{\log(1+\sqrt{2})}{2}.$ \ba I
  recommend using the substitution $t=\sinh(u).$ Then,
  $1+t^2=\sin^2(u)+1=\cosh^2(u)$ and $dt=\cosh(u)du.$ Let's find the
  general antiderivative first, then compute the definite integral:
  \begin{align*}
    \int \sqrt{1+t^2} \,dt & = \int \sqrt{\cosh^2(u)}\cosh(u)\,du \\
    &= \int \cosh^2(u) \, du \\
    &= \int \left(\frac{e^{2u}+2+e^{-2u}}{4}\right) \, dt \\
    &= \frac{e^{2u}+4u-e^{-2u}}{8}+C
    % \end{align*}
    \intertext{To finish, and put things in terms of $t$ again, use
      the fact that $t=\sinh(u) \Rightarrow u=\ln(t+\sqrt{1+t^2})$ and
      use a little algebra}
    &=\frac{(t+\sqrt{1+t^2})^2+4\ln(t+\sqrt{1+t^2})-(t+\sqrt{1+t^2})^{-2}}{8}+C\\
    &=\frac{t\sqrt{1+t^2}}{2}+\frac{\ln(t+\sqrt{1+t^2})}{2}+C.
  \end{align*}
  So, we have $$\int_0^1
  \sqrt{1+t^2}dt=\frac{t\sqrt{1+t^2}}{2}+\frac{\ln(t+\sqrt{1+t^2})}{2}\biggr]_0^1=\frac{\sqrt{2}}{2}+\frac{\log(1+\sqrt{2})}{2}.$$

  \ea

\item $\ds \int_0^1 \frac{dt}{1+t^2}=\frac{\pi}{4}.$ \ba Let
  $t=\tan(u)$.  Here $t=\tan(u)\Rightarrow 1+t^2=\sec^2(u)$ and $dt =
  \sec^2(u)du$.  Changing the limits of integration, $t=0 \Rightarrow
  u=0$ and $t=1 \Rightarrow u=\frac{\pi}{4}.$ We compute:
$$ \int_0^1 \frac{dt}{1+t^2} =\int_0^\frac{\pi}{4}\frac{\sec^2(u)}{\sec^2(u)}du=u\biggr]_0^\frac{\pi}{4}=\frac{\pi}{4}.$$
\ea

\een \ep

\addtocounter{problemnum}{-1}


\bp \textbf{Continued.}  \ben \addtocounter{enumi}{9}
\item $\ds \int_0^1 \frac{dt}{1-t^2}=\infty$ \ba First note that this
  integral is only defined over intervals within $(-1,1)$, so we
  interpret the integral as a limit: $\ds \int_0^1 \frac{dt}{1-t^2}$
  as $\ds \lim_{x \to 1^-} \int_0^x \frac{dt}{1-t^2}.$ Then, use the
  fact that
$$\frac{1}{1-t^2}=\frac{\frac{1}{2}}{1+t}+\frac{\frac{1}{2}}{1-t}.$$
\begin{align*}
  \int_0^1 \frac{dt}{1-t^2}&= \lim_{x \to 1^-} \int_0^x
  \frac{dt}{1-t^2}\\
  &= \lim_{x \to 1^-} \int_0^x
  \frac{\frac{1}{2}dt}{1+t}+\frac{\frac{1}{2}dt}{1-t}\\
  &= \lim_{x \to 1^-}
  \frac{1}{2}\log(1+t)-\frac{1}{2}\log(1-t) \biggr]_0^x\\
  &=\infty
\end{align*}
\ea \een

\ep




\end{exam}
\end{document}

