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

Method 1:  Using $dx$

1.)  Find points of interesection 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 $dx$.


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

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

Method 2:  Using $dy$

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

$x = {y^2 + 2 \over 2}$ and $x = y  + 5$.

${y^2 + 2 \over 2} = y + 5$

${y^2 + 2 } = 2y + 10$

${y^2 -2y - 8} = 0$.  Hence,
$(y + 2)(y - 4) = 0$, $y = -2, 4$.

(or note from method 1, that when $x = 3, y = 3 - 5 = -2$ and when $x = 9, y = 9 - 5 = 4$

2.)  Determine which function is larger.

Between $y = -2$ and 4: 

when $y = 0$, $x = {y^2 + 2 \over 2} = {0^2 + 2 \over 2}$ and  $x = y  + 5  = 0  + 5 = 5$.  

Hence $y + 5 > {y^2 + 2 \over 2}$ when $y \in (-2, 4)$.

3.)  Write as integral(s)

Note that between -2 and 4, the height of the rectangles is $y + 5 - ({y^2 + 2 \over 2})$ and the width is $dy$.

$\int_{-2}^4 [y + 5 - ({y^2 + 2 \over 2})]dy$

4.)  Evaluate the integral:

$\int_{-2}^4 [y + 5 - ({y^2 + 2 \over 2})]dy$
= $\int_{-2}^4 [y + 5 - ({y^2\over 2} + 1)]dy$

= $\int_{-2}^4 [y + 4 - {y^2\over 2}]dy$

$ = {1 \over 2}y^2 + 4y - {1 \over 6}y^3 |_{-2}^4$

$= {1 \over 2}(4)^2 + 4(4) - {1 \over 6}(4)^3 
- [{1 \over 2}(-2)^2 + 4(-2) - {1 \over 6}(-2)^3]$

$= {1 \over 2}(16 - 4) + 4(4 + 2) - {1 \over 6}[64 + 8]$ 

$= 6 + 24 - {1 \over 6}[72] = 30 - 12 = 18$ 



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