\input epsf
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\vsize 10.3truein
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\def\u{\vskip -10pt}
\def\v{\vskip 10pt}
\def\s{\vskip 3pt}
\def\r{\vskip -4pt}

Suppose $f$ integrable\hfil \break
(Note $f$ continuous implies $f$ integrable).
%%\u
%%Suppose $a = x_0 < x_1 < x_2 < .... x_{n-1} < x_n = b$, $\Delta x = x_i - x_{i-1}$ 
%%\v
%% \centerline{$\int_a^b f(x) dx = lim _{n \rightarrow \infty} \Sigma_{i = 1}^n f(x_i^*) \Delta x$ }
%%\s

If  $n$ equal subdivisions:  $\Delta x = {b - a \over n}$ and if we use right-hand endpoints: $x_i = a + i\Delta x = a + {(b - a)i \over n}$
\v\s
 \centerline{$\int_a^b f(x) dx = lim _{n \rightarrow \infty} \Sigma_{i = 1}^n f(a + {(b - a)i \over n}) ({b - a \over n})$ }
\v
\hrule


Evaluate the limit by recognizing the sum as a Riemann sum for a function defined on [0, 1]

1.)  $lim _{n \rightarrow \infty} \Sigma_{i = 1}^n sin({i \over n}){1 \over n}$ 
\vfill
2.)  $lim _{n \rightarrow \infty} \Sigma_{i = 1}^n {i^5 \over n^6}$ 
\vfill
%%3.)  $lim _{n \rightarrow \infty} \Sigma_{i = 1}^n sin({i \over n}){1 \over n}$ 


\v
\hrule

The Fundamental Theorem of Calculus:  Suppose $f$ continuous on $[a, b]$.

1.)  If $G(x) = \int_a^x f(t) dt$, then $G'(x) = f(x)$.

2.)   $\int_a^b f(t) dt = F(b) - F(a)$ where $F$ is any antiderivative of $f$, that is $F' = f$.

%%\v
%%\hrule
\eject
The Fundamental Theorem of Calculus:  Suppose $f$ continuous on $[a, b]$.

1.)  If ${d \over dx}[ \int_a^x f(t) dt] = f(x)$.

2.)   $\int_a^b F'(t) dt = F(b) - F(a)$.

\v
%%\hrule
%%\eject
Examples:
\vskip -3pt

1.)  If $G_1(x) = \int_0^x t^2 dt$, then $G_1'(x) = \underline{\hskip 1in}$.

2.)  If $G_2(x) = \int_5^x t^2 dt$, then $G_2'(x) = \underline{\hskip 1in}$.

3.)  If $G_3(x) = \int_{-2}^x sin(t^2) dt$, then $G_3'(x) = \underline{\hskip 1in}$.


4.)  If $G_4(x) = \int_4^x tan({t^3 \over t+ 1}) dt$, then $G_4'(x) = \underline{\hskip 1in}$.


5.)  If $G_5(x) = \int_1^x \sqrt{3t - 5} dt$, then $G_5'(x) = \underline{\hskip 1in}$.


\eject

The Fundamental Theorem of Calculus:  Suppose $f$ continuous on $[a, b]$.

1.)  If $G(x) = \int_a^x f(t) dt$, then $G'(x) = f(x)$.



Proof

$G'(x) = lim_{h \rightarrow 0} {G(x + h) - G(x) \over h}$,

$= lim_{h \rightarrow 0} {\int_a^{x+h} f(t) dt - \int_a^x f(t) dt \over h}$

$= lim_{h \rightarrow 0} {\int_x^{x+h} f(t) dt \over h}$


$\leq lim_{h \rightarrow 0} {\int_x^{x+h} M_h dt \over h}$ 

~~~~~~~~~where $M_h = max\{ f(t) ~|~ x \leq t \leq t+h \}$ 

~~~~~~~~~(Note $M_h$ exists by extreme value thm)


$\leq lim_{h \rightarrow 0} {(M_h)(h)  \over h}$ 

$\leq lim_{h \rightarrow 0} M_h = f(x)$ 

Similarly $G'(x) \geq f(x)$ 

(using $m_h = min\{ f(t) ~|~ x \leq t \leq t+h \}$ 
\eject
The Fundamental Theorem of Calculus:  Suppose $f$ continuous on $[a, b]$.

\r

2.)   $\int_a^b f(t) dt = F(b) - F(a)$ where $F$ is any antiderivative of $f$, that is $F' = f$.

Proof

Let $G(x) = \int_a^x f(t) dt$. Then $G'(x) = f(x)$  (ie, $G$ is an antiderivative of $f$). 

Let $F$ be any antiderivative of $f$.

Then $F(x) = G(x) + C = \int_a^x f(t) dt + C$ for some constant $C$.

Thus $F(b) - F(a) = G(b) + C - [G(a) + C] $

$= G(b) - G(a) = \int_a^b f(t) dt - \int_a^a f(t) dt = \int_a^b f(t) dt $.  

\v\hrule

Find the average of 3, 2, 5, 6:
\vfill
\vfill
\vfill
\vfill

The average value of $n$ values, $f(t_1), ..., f(t_n)$ is $${f(t_1) + 
f(t_2) + ... + 
f(t_n) \over n} = {\Sigma_{i = 1}^n f(t_i) \over n}$$  



\end

