\input epsf
\input graphicx
\font\bfsl=cmr8

%%\input ../../PAPERS/GOLD/psfig
\magnification 2000
\vsize 9.5 truein
\voffset -0.1truein
\hoffset -0.3truein

\hsize 6.9 truein
\nopagenumbers
\parskip=8pt
\parindent= 0pt
\def\u{\vskip -5pt}
\def\v{\vskip 2.2in}
\def\w{\vskip 7pt}
\def\z{\vskip 1.2in}
\def\s{\vskip -11pt}

Properties of the definite integral


${\int_a^a}f(x)dx = 0$ 
\vfill

${\int_a^b}f(x)dx = - {\int_b^a}f(x)dx$
\vfill


{$ {\int_a^b}kf(x)dx =k {\int_a^b}f(x)dx$}
\vfill


{$ {\int_a^b}(f_1 + f_2)(x)dx = {\int_a^b}f_1(x)dx +
{\int_a^b}f_2(x)dx$}
\vfill


$ {\int_a^c}f(x)dx = {\int_a^b}f(x)dx + {\int_b^c}f(x)dx$

\vfill


 If $f_1(x) \leq f_2(x)$, then
$ {\int_a^b}f_1(x)dx \leq {\int_a^b}f_2(x)dx$
\vfill



If $m \leq f(x) \leq M$ then $m(b-a) \leq {\int_a^b}f(x)dx \leq M(b-a)$
\vfill


\eject

Find the area between the curve $y^2 = 2x - 2$ and $y = x - 5$.

Use vertical rectangles:

1.)  Find points of intersection between the two curves.

$y^2 = 2x - 2$ and $y = x - 5$.

$(x-5)^2 = 2x - 2$

$x^2 - 10x + 25 = 2x - 2$

$x^2 - 12x + 27 = 0$


$(x - 3)(x - 9) = 0$.  Hence $x = 3, 9$.

2.)  Determine which is larger.

Between 1 and 3:  $\sqrt{2x - 2} > -\sqrt{2x - 2}$

Between 3 and 9:  $\sqrt{2x - 2} > x - 5$


3.)  Write as integral(s)

Note that between 1 and 3, the height of the rectangles is $\sqrt{2x - 2} 
- (-\sqrt{2x - 2})$ and the width is $dx$.

Note that between 3 and 9, the height of the rectangles is $\sqrt{2x - 2} 
- (x-5)$ and the width is $dx$.


$\int_1^3 [\sqrt{2x - 2} - (-\sqrt{2x - 2})]dx + \int_3^9 [\sqrt{2x - 2} - 
(x-5)]dx$  

\eject

4.)  Evaluate the integral

$\int_1^3 [2\sqrt{2x - 2}]dx + \int_3^9 [\sqrt{2x - 2} - (x-5)]dx$  

$= \int_1^3 [2\sqrt{2x - 2}]dx + \int_3^9 (\sqrt{2x - 2})dx - 
\int_3^9(x-5)dx$  


Let $u = 2x - 2$, $du = 2dx$, \hfil \break
$x = 1:  u =  2(1) - 2 = 0$; \hfil \break
$x = 3:  u = 2(3) - 2 = 4$; \hfil \break
$x = 9:  u = 2(9) - 2 = 16$ 

$= \int_0^4 u^{1 \over 2}du + \int_4^{16} {1 \over 2}u^{1 \over 2}du + 
\int_3^9(-x+5)dx$ \hfil \break

$= {2 \over 3} u^{3 \over 2}|_0^4 + {1 \over 3} u^{3 \over 2}|_4^{16}
+ (-{1 \over 2}x^2 + 5x)|_3^{9}$


$ = {2 \over 3} (4^{3 \over 2} - 0^{3 \over 2}) + {1 \over 3}(16^{3 \over 
2} 
- 4^{3 \over 2}) + (-{1 \over 2}(9)^2 + 5(9)) $

\rightline{$- (-{1 \over 2}(3)^2 + 5(3)) $}


$ = {1 \over 3}[2(8) + 64 - 8] - {81 \over 2} + 45 + {9 \over 2} - 15 = 
16$
 
$ = {72\over 3} - {72 \over 2} + 30$
$ = 24 - {36} + 30 = 18$

\eject


Find the area bounded by the functions $y = 2x^3$ and $y = 2x^{1 \over 
3}$.

1.)  Find points of intersection:

$2x^3 = 2x^{1 \over 3}$ implies $x^3 = x^{1 \over 3}$ implies  $x^9 = x$. 
 Thus   $x^9 - x =  x(x^8 - 1)= 0$.  
\hfil \break
Hence $x = 0$ and $x^8 - 1 = 0$.  $x^8 = 1$ implies $x = 1, -1$


Hence the functions $y = 2x^3$ and $y = 2x^{1 \over 3}$ intersect when $x 
= -1, 0, 1$

2.)  Draw a rough graph:

\vskip -20pt
\centerline{\includegraphics[width=24ex]{6_2}}

\vskip -20pt

3.) Find area:

Use vertical rectangles:

$\int_{-1}^0 [2x^3 - 2x^{1 \over 3}]dx + \int_0^1 [2x^{1 \over 3} - 
2x^{3}]dx$

\vfill

\end
