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Find the area between $y = 0$, ~~$y = f(x)$, ~~$x = a$, ~~$x = b$.
\vskip -10pt
Special case:  Suppose $f$ is continuous and $f > 0$.

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\includegraphics[width=65ex]{5_2}

\centerline{$\matrix{ \hbox{area of } n \cr \hbox{ inscribed} \cr  \hbox{ 
rectangles } }$ 
~~~$\leq$~~~~~
 {actual area} 
~~~~$\leq$~~~
$\matrix{ \hbox{area of } n \cr \hbox{  circumscribed}  \cr \hbox{ 
rectangles } }$}

\v
\vskip -7pt

\centerline{$\displaystyle \lim_{n \to \infty} $ $\left(\matrix{ 
\hbox{area of } n \cr \hbox{inscribed} \cr  \hbox{rectangles}}\right)$ 
~~$\leq$~~ actual area ~$\leq$ ~
$\displaystyle \lim_{n \to \infty} $ $\left(\matrix{ \hbox{area of } n \cr 
\hbox{circumscribed} \cr  \hbox{rectangles}}\right)$}

\vskip 10pt
\hrule

Theorem: 

\centerline{$L =   \displaystyle \lim_{n \to \infty} $ $\left(\matrix{ 
\hbox{area of } n \cr \hbox{inscribed} \cr  \hbox{rectangles}}\right) =$ 
$\displaystyle \lim_{n \to \infty} $ $\left(\matrix{ \hbox{area of } n \cr 
\hbox{circumscribed} \cr  \hbox{rectangles}}\right) = U$.}

\vskip 10pt
\hrule
\vskip 5pt

\centerline{$\matrix{ \hbox{area of } n \cr \hbox{ inscribed} \cr  \hbox{ 
rectangles } }$ 
~~~$\leq$~~~~~
 $\Sigma_{i = 1}^n f(x_i) \Delta x$
~~~~$\leq$~~~
$\matrix{ \hbox{area of } n \cr \hbox{  circumscribed}  \cr \hbox{ 
rectangles } }$}
\vskip -5pt
where $x_i$ could be right end-point, left end-point, mid-point, or etc.

$\displaystyle \lim_{n \to \infty}\left(\matrix{ \hbox{area of } n \cr 
\hbox{inscribed} \cr  \hbox{rectangles}}\right)$ $\leq$ $\displaystyle 
\lim_{n \to \infty} 
\displaystyle\sum_{i = 1}^n f(x_i)\Delta x$ $\leq$ 
$\displaystyle \lim_{n \to \infty}\left(\matrix{ \hbox{area of } n \cr 
\hbox{circumscribed} \cr  \hbox{rectangles}}\right)$

 
 Theorem: If $f$ continuous, $f> 0,$ actual area = $\displaystyle \lim_{n 
\to \infty} 
\displaystyle\sum_{i = 1}^n f(x_i) \Delta x$
 
 
 \vskip 5pt
\hrule
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Cor:  If $f$ is continuous,  $\int_a^b f(x)dx =$ NET area = $\displaystyle 
\lim_{n \to \infty} \displaystyle\sum_{i = 1}^n f(x_i)\Delta x$


\eject

Example:

$\int_2^6 (-{1 \over 2}t + 4)dt$ =
$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(t_i ) \Delta t$

~~~~~~~~~$\Delta t = {6 - 2 \over n} = {4 \over n}$  (using $n$ equal 
subintervals)

~~~~~~~~~$t_i = 2 + i \Delta t = 2 + {4i \over n}$ (using right-hand 
endpoints)

{$\int_2^6 (-{1 \over 2}t + 4)dt$ $= lim_{n \rightarrow \infty} \Sigma_{i 
= 1}^n f(2 + {4i \over n} ) ({4 
\over n})$

$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ -{1 \over 2}(2 + {4i 
\over n}) + 4 ] ({4 \over n})$

$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ -1  - {2i \over n} + 4 ] 
({4 \over n})$

$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ 3  - {2i \over n} ] ({4 
\over n})$


$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ {12 \over n}  - {8i \over 
n^2} ]$

$= lim_{n \rightarrow \infty} (\Sigma_{i = 1}^n {12 \over n}  - \Sigma_{i 
= 1}^n {8i \over n^2})$

$= lim_{n \rightarrow \infty} (12  -{8\over n^2}\Sigma_{i = 1}^n i)$

$= lim_{n \rightarrow \infty} (12  - {8\over n^2}  {n(n+1) \over 2})$

$= lim_{n \rightarrow \infty} (12  -  {4n^2+4n \over n^2})$

$= lim_{n \rightarrow \infty} (12  -  4 - {4 \over n})$
  = 8

\end 

