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Section 5.2

Find the area under the curve $f(x) = -{1 \over 2}x + 4$, above the $x-$axis and between $x 
= 2$ and $x = 6$.

\includegraphics[width=48ex]{graphint1}


Method 1:  In this case our function is very simple, so we can determine the area without calculus:

\vskip .7in
\eject
Method 2:  Estimate using rectangles.
\u
Inscribed rectangles with $\Delta x = 1$:

\includegraphics[width=48ex]{graphint}

\vskip .9in


$f(3)(1) + f(4)(1) + f(5)(1) + f(6)(1)$ =


$= [-{1 \over 2}(3) + 4](1) + [-{1 \over 2}(4) + 4](1)$

\rightline{$ + [-{1 \over 2}(5) + 4](1) + [-{1 \over 2}(6) + 4](1)$}

= ${5 \over 2} (1) + 2(1) + {3 \over 2} (1) + 1(1)$ = 7


\eject

Inscribed rectangles with $\Delta x = {1\over 2}$:

\includegraphics[width=48ex]{graphint}

\vskip .7in


$f({5 \over 2})({1\over 2}) + f(3)({1\over 2}) + f({7 \over 2})({1\over 2}) + f(4)({1\over 2})$

\vskip 5pt

\rightline{$ + f({9 \over 2})({1\over 2}) + f(5)({1\over 2}) + f({11 \over 2})({1\over 2}) + f(6)({1\over 2}) $ }


$=  [-{1 \over 2}({5 \over 2}) + 4]({1\over 2}) + [-{1 \over 2}(3) + 4]({1\over 2}) +  [-{1 \over 2}({7 \over 2}) + 4]({1\over 2}) $

$+ ~[-{1 \over 2}(4) + 4]({1\over 2}) + [-{1 \over 2}({9 \over 2}) + 4]({1\over 2})$ $  + ~ 
[-{1 \over 2}(5) + 4]({1\over 2})$ 

+ $ [-{1 \over 2}({11 \over 2}) + 4]({1\over 2}) + [-{1 \over 2}(6) + 4]({1\over 2})$


= $ {11 \over 4} ({1\over 2}) +  {5 \over 2} ({1\over 2}) +
{9 \over 4} ({1\over 2})+ {2}({1\over 2}) + {7 \over 4} ({1\over 2})  + {3 \over 2}({1\over 2})  + {5 \over 4} ({1\over 2}) + 1({1\over 2})$ 

$= {15 \over 2}$

Inscribed rectangles with $\Delta x = {6 - 2\over n} ={4 \over n}$:

\includegraphics[width=48ex]{graphint}


\eject
Circumscribed rectangles with $\Delta x = 1$:

\includegraphics[width=48ex]{graphint}

\vskip .9in


$f(2)(1)  + f(3)(1) + f(4)(1) + f(5)(1) $ =


$= [-{1 \over 2}(2) + 4](1) + [-{1 \over 2}(3) + 4](1)$

\rightline{$ + [-{1 \over 2}(4) + 4](1) + [-{1 \over 2}(5) + 4](1)$}

= $3 + {5 \over 2} (1) + 2(1) + {3 \over 2} (1)$ = 9

\eject


Estimate the distance traveled between $t = 2$ and $t = 6$ if  the velocity is given by the function $f(t) = -{1 \over 2}t + 4$.


Estimate using inscribed rectangles with $\Delta t = 1$:

\includegraphics[width=48ex]{graphint}

\vskip .9in


$f(3)(1) + f(4)(1) + f(5)(1) + f(6)(1)$ =


$= [-{1 \over 2}(3) + 4](1) + [-{1 \over 2}(4) + 4](1)$

\rightline{$ + [-{1 \over 2}(5) + 4](1) + [-{1 \over 2}(6) + 4](1)$}

= ${5 \over 2} (1) + 2(1) + {3 \over 2} (1) + 1(1)$ = 7


\eject

%%Circumscribed rectangles with $\Delta x = {6-2 \over n} = {4 \over n}$:

\vskip .9in


Defn:  $\int_a^b f(x) dx 
= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(x_i) \Delta x$

If $f$ is continuous, 
can use  inscribed rectangles, circumscribed rectangles,  all left-hand endpoints, all right-hand 
endpoints, or all midpoints, etc.  



If $\Delta x = {b-a \over n} $ and if right-hand endpoints are used, then 
$x_i = a + i \Delta x = a + {(b - a)i \over n}$ 

\vskip 5pt

\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}) $}

%%$\int_2^6  [-{1 \over 2}x + 4] dx =  $


\vfill
\vfill
\vfill
%%\eject



\eject 

Find the distance traveled between $t = 2$ and $t = 6$ if  the velocity is given by the function $f(t) = -{1 \over 2}t + 4$.

\includegraphics[width=24ex]{graphint1}


Method 1:  In this case our function is very simple, so we can determine the area without calculus:

\vskip .7in
\vfill
Method 2:  Use calculus by estimating with rectangles and taking limit.

\centerline{Area =
$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(a + {(b - a)i \over n} ) ({b-a \over n}) $}
\centerline{$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(2 + {4i \over n} ) ({4 \over n}) $}
\centerline{$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ -{1 \over 2}(2 + {4i \over n}) + 4 ] ({4 \over n})  = 8$}

\vfill

Method 3 (section 5.3):  Use calculus by integrating.

\centerline{$\int_2^6 (-{1 \over 2}t + 4) dt =  (-{1 \over 4}t^2 + 4t)|_2^6$}

\centerline{$ =  (-{1 \over 4}(6)^2 + 4(6)) -  (-{1 \over 4}(2)^2 + 4(2)) $}

\centerline{$ =  -9 + 24 - (-1 + 8) =  15 - 7 = 8$}

\end
