\documentclass[letter,12pt]{article}
\usepackage[margin=0.5in]{geometry}
\pagestyle{headings}
\usepackage{amssymb,amsmath,hyperref,enumerate}
\usepackage{graphicx}
\begin{document}
\providecommand{\R}{\mathbb{R}}
\providecommand{\at}{\restriction}
\providecommand{\evalat}[3]{\left[ #1 \right]_{#2}^{#3}}
\providecommand{\ddx}{\frac{d}{dx}}
\providecommand{\dudx}{\frac{du}{dx}}
\providecommand{\dthdx}{\frac{d\theta}{dx}}
\renewcommand{\o}{\circ}
\providecommand{\rcp}[1]{\frac{1}{#1}}
\providecommand{\dom}{\operatorname{dom}}
\def\dividelen#1#2{\@tempdima=#1 \divide\@tempdima by #2 \@tempdima}

\section*{Fundamental Theorem of Calculus (FTC) Reiterated}

\paragraph{Theorem [FTC]:}
\begin{enumerate}[(I)]\item Let $f$ be continuous on $[a,b]$. Then
	\[ F\left(x\right) = \int_a^x f\left(t\right) dt \]
	is differentiable on $\left(a,b\right)$ and continuous on $[a,b]$, and for $x$ in
	$\left(a,b\right)$ we have
	\[ F'\left(x\right) = f\left(x\right) .\]
	
	So for continuous $f$,
	    \[ \ddx \int_a^x f\left(t\right) dt = f\left(x\right) .\]

\item  If $f$ is continuous on $[a,b]$ and $F'=f$ on $[a,b]$, then
	\[ \int_a^b f\left(t\right) dt = F\left(b\right)-F\left(a\right) .\]
    
	In other words:
	if $g$ is differentiable on $[a,b]$ with continuous derivative, then
	\[ \int_a^b g'\left(t\right) dt = \evalat{g\left(t\right)}{a}{b} = g\left(b\right)-g\left(a\right) .\]
\end{enumerate}
\paragraph{Idea of proof:}
\begin{enumerate}[(I)]\item \begin{equation*}\begin{aligned} F'\left(x\right) &= \lim_{h\rightarrow 0} &&\frac{F\left(x+h\right)-F\left(x\right)}{h}\\
	%&= \lim_{h\rightarrow 0} &\rcp{h} &\left(F\left(x+h\right)-F\left(x\right)\right)\\
	&= \lim_{h\rightarrow 0} &\rcp{h} &\left(\int_a^{x+h} f\left(t\right)dt - \int_a^x f\left(t\right)dt\right)\\
	&= \lim_{h\rightarrow 0} &\rcp{h} &\int_x^{x+h} f\left(t\right)dt\\
	&= \lim_{h\rightarrow 0} &\rcp{h} &\textrm{[signed area xetween t=x and t=x+h]}\\
	&&&\;\;\approx hf\left(x\right)\\
	&&&\textrm{\;\;\;\;\;\;\;\;\;(since $f$ is
	  continuous at $x$ and integrable around $x$)} \\
	  &= f\left(x\right)  \end{aligned}\end{equation*}

\item  By (I), $\int_a^x g'\left(t\right)dt$ is an antiderivative of $g'\left(t\right)$ on $\left(a,b\right)$.
      So by the MVT, $g\left(x\right) = \int_a^x g'\left(t\right)dt + C$ on $\left(a,b\right)$, and hence by
      continuity on $[a,b]$.
      
      Since $\int_a^a g'\left(t\right)dt=0$, we must have $C=g\left(a\right)$. So
	\[ \int_a^b g'\left(t\right) dt = g\left(b\right)-g\left(a\right) .\]
	
\end{enumerate}

Note that FTC-I makes sense of the indefinite integral notation for
antiderivatives, since
      \[ \ddx \int f\left(x\right) dx = f\left(x\right) ,\]
	

\section*{Substitution}

\subsection*{Examples and intuition}
  We know $\int_2^3 \cos\left(x\right) dx = \sin\left(3\right)-\sin\left(2\right) \left( =-0.768\right)$.

  Consider $\int_2^3 \cos\left(2x\right)$.

  \[\includegraphics[width=0.5\textwidth]{graphs/cos2.pdf} \]
  \[\includegraphics[width=0.5\textwidth]{graphs/coswide.pdf} \]

  Let $u=2x$. The area between $x=2$ and $x=3$ for $\cos\left(2x\right)$ corresponds to the
  area between $u=4$ and $u=6$ for $\cos\left(u\right)$ - but the former is squashed by a
  constant factor of 2 relative to the latter.

  We can compensate for the squashing by multiplying $\cos\left(2x\right)$ by a constant
  factor of 2, so we expect:
      \[ \int_2^3 2\cos\left(2x\right)dx = \int_4^6\cos\left(u\right)du = \sin\left(6\right)-\sin\left(4\right) .\]

  Indeed, by the chain rule, $\sin\left(2x\right)$ is an antiderivative of $2\cos\left(2x\right)$, so
  this is right.

  We deduce \[ \int_2^3 \cos\left(2x\right)dx = \rcp{2} \int_2^3 2\cos\left(2x\right)dx =
  \frac{\sin\left(6\right)-\sin\left(4\right)}{2} = 0.239 \]

  Now consider $\int_2^3 \cos\left(x^2\right)$.
  \[\includegraphics[width=0.5\textwidth]{graphs/cososq.pdf} \]
  \[\includegraphics[width=0.5\textwidth]{graphs/coswide.pdf} \]

  Let $u=x^2$. The area between $x=2$ and $x=3$ for $\cos\left(x^2\right)$ corresponds to
  the area between $u=4$ and $u=9$ for $\cos\left(u\right)$ - but the former is squashed by
  a factor of $\ddx x^2 = 2x$ relative to the latter.

  We can compensate for the squashing by multiplying $\cos\left(x^2\right)$ by a
  factor of $2x$, so we expect:
      \[ \int_2^3 2x\cos\left(x^2\right)dx = \int_4^9\cos\left(u\right)du = \sin\left(9\right)-\sin\left(4\right) .\]

  Again, we can confirm this using the chain rule: $\sin\left(x^2\right)$ is an
  antiderivative of $2x\cos\left(x^2\right)$.

  Note that we have actually discovered {\bf nothing} about $\int_2^3 \cos\left(x^2\right)$!
  Instead, we have found an entirely {\bf different} integral, namely
      \[ \int_2^3 2x\cos\left(x^2\right)dx .\]
  There is no way to get from that to any information about $\int_2^3\cos\left(x^2\right)$!

\newpage
\subsection*{Formal formulation}

\paragraph{Theorem [substitution rule]:}
\begin{enumerate}[(a)]\item For indefinite integrals:
	Suppose $f$ is continuous and $g$ is differentiable. Then
      \[ \int f\left(g\left(x\right)\right)g'\left(x\right) dx = \int f\left(u\right) du \]
	where $u=g\left(x\right)$.
\item For definite integrals:
	Suppose further that $g'$ is continuous. Then
      \[ \int_a^b f\left(g\left(x\right)\right)g'\left(x\right) dx = \int_{g\left(a\right)}^{g\left(b\right)} f\left(u\right) du \]
\end{enumerate}
\paragraph{Proof:}
\begin{enumerate}[(a)]\item If $F$ is an antiderivative of $f$, then by the chain rule
	\[\ddx F\left(g\left(x\right)\right) = f\left(g\left(x\right)\right)g'\left(x\right)\]
    so $F\left(u\right) = F\left(g\left(x\right)\right)$ is an antiderivative of $f\left(g\left(x\right)\right)g'\left(x\right)$.

\item Now by FTC-II 
	\[ \int_a^b f\left(g\left(x\right)\right)g'\left(x\right) dx = F\left(g\left(b\right)\right) - F\left(g\left(a\right)\right) = \int_{g\left(a\right)}^{g\left(b\right)}
	f\left(u\right) du \]
\end{enumerate}

\subsection*{Further examples}

\begin{align*} \int_{-1}^2 x^3e^{x^4}dx &= \rcp{3} \int_{-1}^2 e^{x^4} 3x^3 dx \\
&= \rcp{3} \int_{\left(-1\right)^4}^{2^4} e^{u} du  & \left(u=x^4, \dudx=4x^3\right) \\
&= \evalat{\rcp{3} e^u}{\left(-1\right)^4}{2^4} \\
&= \evalat{\rcp{3} e^u}{1}{16} \\
&= \frac{e^{16}-e}{3} \\
&= 2.96 * 10^6 \end{align*}

\begin{align*} \int x^3e^{x^4}dx &= \rcp{3} \int e^{x^4} 3x^3 dx \\
&= \rcp{3} \int e^{u} du  & \left(u=x^4, \dudx=4x^3\right) \\
&= \rcp{3} e^u + C\\
&= \frac{e^{x^4}}{3} + C \end{align*}

\begin{align*} \int \frac{1}{\sqrt{2+3x}} dx
&= \rcp{3} \int \frac{1}{\sqrt{2+3x}}3dx \\
&= \rcp{3} \int \rcp{\sqrt{u}}du &\left(u=2+3x, \dudx=3\right) \\
&= \rcp{3} 2\sqrt{u} + C \\
&= \frac{2}{3} \sqrt{2+3x} + C \end{align*}

\begin{equation*}\begin{aligned} \int_1^e \frac{2x+\ln x}{x} dx
&= \int_1^e \frac{2x}{x} dx &&+ \int_1^e \frac{\ln x}{x} dx \\
&= \int_1^e 2dx &&+ \int_0^1 u du &\left(u=\ln x, \dudx=\rcp{x}\right) \\
&= \evalat{x}{x=1}{x=e} &&+ \evalat{\frac{u^2}{2}}{u=0}{u=1} \\
&= e-1 &&+ \rcp{2}-0 \\
&= e-\rcp{2} \end{aligned}\end{equation*}

\subsection*{Exploiting symmetry}

Suppose $f\left(x\right)$ is an odd function, i.e.~ $f\left(-x\right)=-f\left(x\right)$

\[ \includegraphics[width=0.33\textwidth]{graphs/oddpoly.pdf}
\includegraphics[width=0.33\textwidth]{graphs/sin.pdf}
\includegraphics[width=0.33\textwidth]{graphs/xcos.pdf}
\]

Then 
    \begin{align*} \int_{-a}^a f\left(x\right) dx &=
    \int_{-a}^0 f\left(x\right) dx + \int_0^a f\left(x\right) dx \\
    &= \int_{-a}^0 -f\left(-x\right) dx + \int_0^a f\left(x\right) dx \\
    &= \int_a^0 f\left(u\right) du + \int_0^a f\left(x\right) dx & \left(u=-x, \dudx=-1\right) \\
    &= -\int_0^a f\left(u\right) du + \int_0^a f\left(x\right) dx
    &= 0
    \end{align*}

Similarly, if $f\left(x\right)$ is an even function, i.e. $f\left(-x\right)=f\left(x\right)$

\[ \includegraphics[width=0.33\textwidth]{graphs/evenpoly.pdf}
\includegraphics[width=0.33\textwidth]{graphs/cos.pdf}
\includegraphics[width=0.33\textwidth]{graphs/xsin.pdf}
\]

Then 
    \begin{align*} \int_{-a}^a f\left(x\right) dx &=
    \int_{-a}^0 f\left(x\right) dx + \int_0^a f\left(x\right) dx \\
    &= \int_{-a}^0 f\left(-x\right) dx + \int_0^a f\left(x\right) dx \\
    &= -\int_a^0 f\left(u\right) du + \int_0^a f\left(x\right) dx & \left(u=-x, \dudx=-1\right) \\
    &= \int_0^a f\left(u\right) du + \int_0^a f\left(x\right) dx \\
    &= 2\int_0^a f\left(x\right) dx
    \end{align*}

\subsection*{$ln|x|$ as an antiderivative of $\rcp{x}$}

Recall that $\ddx ln x = \rcp{x}$. So e.g. it does follow that
    $\int_1^2 \rcp{x} dx = ln 2 - ln 1$.

But $ln x$ is only defined for $x>0$, while $\rcp{x}$ is also defined for
$x<0$.

\paragraph{Cunning trick:}
    When $x > 0$:
    \[ \ddx ln|x| = \ddx ln x = \rcp{x} \]
    When $x < 0$:
    \[ \ddx ln|x| = \ddx ln \left(-x\right) = -\rcp{-x} = \rcp{x} \]

So we can write
    \[ \int \rcp{x} dx = \ln |x| + C ,\]
(meaning that this is family of all antiderivatives when we restrict to an
interval not containing 0)

\paragraph{Warning:}
    $\rcp{x}$ is {\bf not} integrable on any interval containing 0. So e.g.
     
    \[ \int_{-1}^1 \rcp{x} dx \]

    does not exist (and in particular is {\bf not} equal to $\ln|1| - \ln| -1| = 0$,
    even though the function is odd!).

\paragraph{Example:}
  \begin{align*} \int \tan\left(x\right) dx &= \int\frac{\sin\left(x\right)}{\cos\left(x\right)} dx \\
  &= -\int\rcp{u}\dudx dx  & \left(u=\cos\left(x\right), \dudx = -\sin\left(x\right)\right) \\
  &= -\int\rcp{u}du \\
  &= -\ln|u| + C \\
  &= -\ln|cos\left(x\right)| + C \end{align*}

  (but again, you can only integrate $\tan\left(x\right)$ on intervals on which it is
  defined!)

\newpage
\section*{Area between curves}

\paragraph{Example:}
    Find the area of the region enclosed by the graphs of $x^2$ and
    $6-3\left(x-2\right)^2$.
\[ \includegraphics[width=\textwidth]{graphs/intersect1.pdf} \]

\paragraph{Solution:}
    We first find the $x$-values of the intersection points :
    \begin{align*} &\phantom{\Leftrightarrow }x^2 = 6-3\left(x-2\right)^2 \\
    &\Leftrightarrow  4x^2-12x+6=0 \\
    &\Leftrightarrow  x=
    \frac{6 \pm \sqrt{12}}{4} \\
    &\Leftrightarrow  x=0.634 \textrm{\;or\;} x=2.37 \end{align*}

    Then the area between the graphs is the difference between the area
    between top one and the x-axis and the area between the bottom one and the
    x-axis. So the area is
\[ \includegraphics[width=0.5\textwidth]{graphs/intersect1.pdf} \]
    {
    \renewcommand{\a}{\frac{6 - \sqrt{12}}{4}}
    \renewcommand{\b}{\frac{6 + \sqrt{12}}{4}}
    
    \begin{align*} \int_\a^\b \left(6-3\left(x-2\right)^2\right) dx - \int_\a^\b x^2 dx
	&= \int_\a^\b \left(6-3\left(x-2\right)^2 - x^2\right) dx \\
	&= \int_\a^\b \left(-4x^2+12x-6\right)dx \\
	&= \evalat{-\frac{4x^3}{3}+6x^2-6x}{\a}{\b} \\
	&= 3.46 \end{align*}
    }

  It wasn't important that the area was above the x-axis, and so we get in
  general:

\paragraph{Formula:}
    If $f$ and $g$ are continuous functions on $[a,b]$, and if $f\left(x\right)\geq g\left(x\right)$ on
    $[a,b]$, then the area of the region enclosed by the graphs of $f\left(x\right)$,
    $g\left(x\right)$ and the lines $x=a$ and $x=b$ is
	\[ \int_a^b \left(f\left(x\right)-g\left(x\right)\right) dx \]


\newpage

\paragraph{Midterm remarks:}
    Review sessions {\bf tonight} - see yellow website.

    No table of integrals will be given: learn the table on p398 of Stewart.

    ko na tcica bilma

\paragraph{Examples:}
    \begin{minipage}{0.5\textwidth}
    Area of the unit circle:
	We can think of this as the area enclosed by the graphs of
	$\sqrt{1-x^2}$ and $-\sqrt{1-x^2}$. So:
    \end{minipage}
    \begin{minipage}{0.4\textwidth}
	$\includegraphics[width=\textwidth]{graphs/circle.pdf}$
    \end{minipage}

	\begin{align*} C &= \int_{-1}^1 \left(\sqrt{1-x^2} - \left(-\sqrt{1-x^2}\right)\right) dx \\
	&= 2\int_{-1}^1 \sqrt{1-x^2} dx \\
	&= 2\int_{-1}^1 \sqrt{1-\cos\left(\arccos\left(x\right)\right)^2} dx \\
	&= 2\int_{-1}^1 \sqrt{1-\cos\left(\theta\right)^2} dx & \left(\theta=\arccos\left(x\right);
	\dthdx = -\rcp{\sqrt{1-x^2}} = \rcp{-\sin\left(\theta\right)}\right) \\
	&= 2\int_{-1}^1 \sin\left(\theta\right) dx \\
	&= 2\int_{-1}^1 -\sin^2\left(\theta\right) \rcp{-\sin\left(\theta\right)} dx \\
	&= 2\int_{\pi}^0 -\sin^2\left(\theta\right) d\theta \\
	&= 2\int_0^{\pi} \sin^2\left(\theta\right) d\theta \\
	&= \int_0^{\pi} \left(1-\cos\left(2\theta\right)\right) d\theta \\
	&= \left(\evalat{\theta}{0}{\pi} - \rcp{2}\evalat{\sin\left(2\theta\right)}{0}{\pi}\right) \\
	&= \pi - 0 \\
	&= \pi \end{align*}

	\newpage
    Area of an ellipse:
	After rotating and translating to the origin, any ellipse can be
	represented as the solutions to $ax^2 + by^2=1$, i.e.
	  \[ y=\pm\rcp{\sqrt{b}}\sqrt{1-ax^2} .\]
	So the area is
	\begin{align*} E_{b,a} &= \int_{-\rcp{\sqrt{a}}}^{\rcp{\sqrt{a}}} 
	  \left(\rcp{\sqrt{b}}\sqrt{1-ax^2} - \left(-\rcp{\sqrt{b}}\sqrt{1-ax^2}\right)\right) dx \\
	  &= 2 \int_{-\rcp{\sqrt{a}}}^{\rcp{\sqrt{a}}} \rcp{\sqrt{b}}\sqrt{1-ax^2} dx \\
	  &= \frac{2}{\sqrt{b}} \int_{-\rcp{\sqrt{a}}}^{\rcp{\sqrt{a}}} \sqrt{1-ax^2} dx \\
	  &= \frac{2}{\sqrt{b}} \int_{-\rcp{\sqrt{a}}}^{\rcp{\sqrt{a}}} \sqrt{1-ax^2} dx & \left(u=\sqrt{a}x; \dudx=\sqrt{a}\right) \\
	  &= \frac{2}{\sqrt{b}\sqrt{a}} \int_{-1}^1 \sqrt{1-u^2} dx \\
	  &= \frac{\pi}{\sqrt{ab}} ,\end{align*}

	which note makes a lot of sense: $ax^2+by^2=1 \leftrightarrow  \left(\sqrt{a}x\right)^2 +
	\left(\sqrt{b}y\right)^2 = 1$, so our ellipse is a circle scaled horizontally by
	$\rcp{\sqrt{a}}$ and vertically by $\rcp{\sqrt{b}}$.

	\includegraphics[width=0.8\textwidth]{graphs/ellipse.pdf}

    \begin{minipage}{0.5\textwidth}
    Area of a crescent:

	As seen from the earth, the disc of the sun has approximately the same
	radius as the disc of the moon. During a solar eclipse, the latter
	slides over the former. When the disc of the moon is centred at the
	edge of the disc of the sun, what proportion of the sun's disc is
	covered?
      \end{minipage}
    \begin{minipage}{0.4\textwidth}
	\includegraphics[width=\textwidth]{graphs/crescent.pdf}
      \end{minipage}

	Intersection points:
	\[\sqrt{1-x^2} = 1-\sqrt{1-x^2} \leftrightarrow  2\sqrt{1-x^2} = 1 \leftrightarrow 
	x=\pm\frac{\sqrt{3}}{2} \]

	Area of covered area:
	\begin{align*} C &= \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}
	\left(\sqrt{1-x^2}-\left(1-\sqrt{1-x^2}\right)\right) dx \\
	&= \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}
	\left(2\sqrt{1-x^2}-1\right) dx \\
	&= 2\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}
	\sqrt{1-x^2} dx - \sqrt{3} \\
	&= \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}
	\left(1-\cos\left(2\theta\right)\right) d\theta - \sqrt{3} \\
	&= \evalat{\theta}{\frac{\pi}{6}}{\frac{5\pi}{6}}
	-\rcp{2}\evalat{\sin\left(2\theta\right)}{\frac{\pi}{6}}{\frac{5\pi}{6}}
	- \sqrt{3} \\
	&= \frac{2\pi}{3}
	-\rcp{2} \left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right)
	- \sqrt{3} \\
	&= \frac{2\pi}{3}
	+ \frac{\sqrt{3}}{2}
	- \sqrt{3} \\
	&= \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \\
	&= 0.391 \pi
	\end{align*}

	So 0.391 of the sun's disc is blocked.

\newpage
\section*{Volumes}

\paragraph{Example - Volume of a sphere:}
    Consider a sphere of radius $r$, centred at the origin (0,0,0).

    Chop it perpendicular to the x-axis into $n$ slivers of equal width.

    The volume of the sphere is the sum of the volumes of the slivers.

    For large $n$, i.e. for thin slivers, each sliver is roughly a cylinder of
    width $\Delta_n=\frac{2r}{n}$. The radius depends on $x$: the $i^{th}$ sliver has
    radius $\sqrt{r^2-\left(x_i^*\right)^2}$ on its right face, where $x_i^* = -r + i\Delta_n$.

    So we can estimate the volume of the $i^{th}$ sliver as
    \[ \Delta_n \pi \left(\sqrt{r^2-\left(x_i^*\right)^2}\right)^2 = \Delta_n \pi \left(r^2-\left(x_i^*\right)^2\right)  \]

    So our estimate for the volume with $n$ slivers is
    \[ V_n = \sum_{i=1}^{n} \Delta_n \pi \left(r^2-\left(x_i^*\right)^2\right) .\]

    As $n\rightarrow \infty$, our estimates converge to the actual volume. So the volume
    of the sphere is
    \begin{align*} V &= \lim_{n\rightarrow \infty} V_n \\
    &= \lim_{n\rightarrow \infty} \sum_{i=1}^{n} \Delta_n \pi \left(r^2-\left(x_i^*\right)^2\right) \\
    &= \int_{-r}^r \pi \left(r^2-x^2\right) dx \\
    &= \pi \evalat{r^2x-\frac{x^3}{3}}{-r}{r} \\
    &= \pi \left(\left(r^3-\frac{r^3}{3}\right) - \left(-r^3 - \frac{-r^3}{3}\right)\right) \\
    &= \frac{4\pi r^3}{3} .\end{align*}

\paragraph{General formula:}
    If a shape lies between $x=a$ and $x=b$, and the area of a cross-section
    perpendicular to the x-axis is a continuous function $A\left(x\right)$, then the
    volume is
	\[ \int_a^b A\left(x\right) dx .\]

    Indeed, the argument above indicates that the volume is
	\[ \lim_{n\rightarrow \infty} \sum_1^n A\left(x_i^*\right)\Delta_n \]
    where we divide $[a,b]$ into $n$ intervals of equal width, and $x_i^*$ is
    a point in the $i^{th}$ interval and $\Delta_n$ is the width of an
    interval. But this is precisely the definition of the integral
    $\int_a^b A\left(x\right) dx$, which exists since $A$ is continuous.

\paragraph{Volumes of revolution:}
    We can think of the sphere as the result of taking the semicircle which is
    the region under the graph of $y=\sqrt{r^2-x^2}$, and rotating it around the
    x axis.

    The same trick for finding the volume works with any shape of that form:

    Suppose $f\left(x\right)$ is continuous and non-negative on $[a,b]$. Consider the
    ``solid of revolution'' of the region under the graph of $f$ between
    $x=a$ and $x=b$, being the space that shape passes through as you rotate
    it through $2\pi$ around the x-axis. Then the cross-sectional area at $x$
    is $\pi f\left(x\right)^2$, so the volume is given by
	\[ \pi \int_a^b f\left(x\right)^2 dx .\]

    Example:
	Find the volume of the surface of revolution of $e^x$ between 0 and 1.

	Solution:
	    \begin{align*} V &= \pi \int_0^1 \left(e^x\right)^2 dx \\
	    &= \pi \int_0^1 e^{2x} dx \\
	    &= \frac{\pi}{2} \evalat{e^{2x}}{0}{1} dx \\
	    &= \frac{\pi}{2} \left(e^2-1\right) dx \end{align*}

\end{document}
