
{\Large The definite integral \pz and the fundamental theorem of calculus}\pz

Areas naturally correspond to aggregate amounts.  \pz As we see on
the overhead, if the bars on a bar graph of profit are of unit width, \pz
then the area of the bar graph represents total profit.  \pz
We will see that for a continuous(ly modelled) revenue stream, the
area under the marginal revenue curve represents total revenue.  \pz

The Fundamental Theorem of Calculus shows 
that anti-derivative computes area under a curve. \pz

{\large Approximating areas by rectangles} \pz

While we do not know how to find areas of regions with curved boundaries,
we do know how to find areas of regions with straight boundaries, 
especially rectangles.  \pz  We approximate the area of a curved region
by trying to fill it (or cover it, with some ``spillage'') with rectangles. \pz
This method for finding curved areas, such as that inside a circle,
was known to some Greek mathematicians \pz but was lost until
the Renaissance and the (re?)invention of integral calculus. \pb

\begin{example}
Approximate the area under the curve $f(x) = x^2$ using ten rectangles.\pz

We see on the overhead that ten rectangles gives a pretty good fit.  \pz
The area of each rectangle is computed as base times height.  Each rectangle
has base of $\frac{1}{10} = 0.1$ \pz and the $i$th rectangle has height equal to 
$f(0.1 \cdot i)$. \pz Thus, the total area is 

\begin{multline*}
0.1 (0.1)^2 + 0.1 (0.2)^2 + 0.1 (0.3)^2 + \cdots + 0.1 (1.0)^2  \pz \\
= 0.1 \left[(0.1)^2 + (0.2)^2 + \cdots + (0.9)^2 + (1.0)^2 \right]  \pz \\
= 0.1 \left[ 0.01 + 0.04 + \cdots + 0.21 + 1.00 \right] \pz \\
= 0.1 (3.85) \pz = 0.385 \pz
\end{multline*}
\end{example}


We will see soon that the exact answer is $\frac{1}{3}$, so this approximation 
is not bad \pz (but not spectacularly good either). \pb

The general technique is as follows: \pz we find the area under $f(x)$ between
$x=a$ and $x=b$ by
approximating it by $N$ rectangles, each of width $\frac{b-a}{N}$.  \pz  
We let $\Delta x = \frac{b-a}{N}$ to save writing, \pz and also let
$x_i = a + \Delta x \cdot i$, the place of the dividing line between the $i$th
rectangle and the $i+1$st.  \pz Then the area of these $N$ rectangles is \pz
\begin{multline*}
 f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_N) \Delta x \\
 = \Delta x\left[ f(x_1) + f(x_2) + \cdots + f(x_N) \right] 
 \end{multline*} 
 
 \pb

If we let $N$ get larger and larger, we get a better approximation to the
area, leading to the following. \pz

\begin{theorem}
If $f(x)$ is continuous, the area under the graph of $f$ between $x=a$ 
and $x = b$ is $\lim_{N \to \infty} \left( \frac{b-a}{N}\right) \left[ f(x_1) + \cdots + f(x_N)\right]$. \pz
\end{theorem}

As we know well, computing limits can be a difficult process, \pz and in this case
the function whose limit we take is complicated. \pz  One of the best theorems in
all of mathematics gives us an easy way to find this area using the anti-derivative. \pz

\begin{theorem}[First Fundamental Theorem]
Let $F(x)$ be an anti-derivative of $f(x)$.  \pz Then the area under $f(x)$ between
$x= a$ and $x=b$ is equal to $F(b) - F(a)$. \pz
\end{theorem}

\begin{example}
\begin{itemize}
\item Find the area under $f(x) = x^2$ between $x= 0$ and $x=1$ \pz and between
$x = 2$ and $x = 5$. \pz
\item Find the area under $f(x) = e^x$ between $x = 0$ and $x = 3$. \pz
\item Find the area under $f(x) = \frac{1}{x}$ between $x=1$ and $x=2$ \pz
and between $1$ and $4$ \pz and between $1$ and $10$ \pz and in general
between $1$ and $a$ where $a > 1$ is any number. \pz
\end{itemize}

\end{example}

Question: why would using any anti-derivative give the same answer when
using the Fundamental Theorem?



 