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{\it PROBLEM 1:}  \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=x $, $ y=2x $, and $ x=4
$.

\vskp \vskp

\noindent {\it SOLUTION 1:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y=x $, $ y=2x $
and $ x=4 $ . Begin by finding the points of intersection of the
two graphs. From $ y=x $ and $ y=2x $  we get that

\vskp

$ x = 2x \ \ \ \ \longrightarrow $

\vskp

$ - x = 0 \ \ \ \ \longrightarrow $

\vskp

$ \displaystyle{-x} = 0 \ \ \ \ \longrightarrow $


\vskp

$ x = 0 \ \ \ \ \longrightarrow $

\vskp


\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{0}^{4} (top \ - \ bottom) \ dx }$

\vskp

$ = \displaystyle { \int_{0}^{4} (2x - x) \ dx } $

\vskp

$ = \displaystyle { \int_{0}^{4} x \ dx } $

\vskp

$ = \displaystyle { \frac{x^{2}}{2} \Big\vert_{0}^{4} } $

\vskp

$ = \displaystyle { \frac{4^{2}}{2} - \frac{0^{2}}{2} } $

\vskp

$ = \displaystyle { 8 - 0 } $

\vskp

$ = \displaystyle { 8 } $

\vskp \vskp \vskp


{\it PROBLEM 2:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=x^2 $ and $ y=x+2 $ .

\vskp \vskp

\noindent {\it SOLUTION 2:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y=x^2 $ and $
y=x+2 $ . Begin by finding the points of intersection of the two
graphs.  From $ y=x^2 $ and $ y=x+2 $  we get that

\vskp

$ x^2=x+2 \ \ \ \ \longrightarrow $

\vskp

$ x^2-x-2=0 \ \ \ \ \longrightarrow $

\vskp

$ (x-2)(x+1)=0 \ \ \ \ \longrightarrow $

\vskp

$ x=2 $ \ or \ $ x=-1 $

\vskp

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{-1}^{2} (top \ - \ bottom) \ dx }$

\vskp

$ = \displaystyle{ \int_{-1}^{2} ((x+2)-x^2) \ dx }  $

\vskp

$  = \displaystyle{ \Big( {x^2 \over 2}+2x-{x^3 \over 3} \Big)
\Big\vert_{-1}^{2} } $


\vskp

$  = \displaystyle{ \Big( {2^2 \over 2}+2(2)-{2^3 \over 3} \Big)
        - \Big( {(-1)^2 \over 2}+2(-1)-{(-1)^3 \over 3} \Big) } $

\vskp

$  = \displaystyle{ \Big(2+4-{8 \over 3}\Big) - \Big({1 \over
2}-2+{1 \over 3}\Big) } $

\vskp

$  = \displaystyle{ \Big(6-{8 \over 3}\Big) - \Big({3 \over 6}-{12
\over 6}+{2 \over 6}\Big) } $

\vskp

$  = \displaystyle{ \Big({36 \over 6}-{16 \over 6}\Big) - \Big({-7
\over 6}\Big) } $

\vskp

$  = \displaystyle{ {20 \over 6}+{7 \over 6} } $

\vskp

$  = \displaystyle{ {27 \over 6} } $ .

\vskp \vskp \vskp


{\it PROBLEM 3:}  \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=e^x, y=e^{-x},  $ and
$ x=\ln 3 $  .

\vskp \vskp

\noindent {\it SOLUTION 3:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y=e^x $, $
y=e^{-x} $, and $ x=\ln 3 $ . Begin by finding the points of
intersection of the two graphs. From $ y=e^{x} $ and $ y=e^{-x} $
we get that

\vskp

$ e^{x} = e^{-x} \ \ \ \ \longrightarrow $

\vskp

$ e^{2x} = 1 \ \ \ \ \longrightarrow $

\vskp

$ x = 0 \ \ \ \ \longrightarrow $

\vskp

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{0}^{\ln 3} (top \ - \ bottom) \ dx
}$

\vskp

$ = \displaystyle { \int_{0}^{\ln 3} (e^{x} - e^{-x}) \ dx } $

\vskp

$ = \displaystyle { \Big(e^{x} - (-e^{-x})\Big) \Big\vert_{0}^{\ln
3} } $

\vskp

$ = \displaystyle { \Big(e^{x} + e^{-x}\Big) \Big\vert_{0}^{\ln 3}
} $

\vskp

$ = \displaystyle { \Big(e^{\ln 3} + e^{-\ln 3}\Big) - \Big(e^{0}
+ e^{-0}\Big) } $

\vskp

$ = \displaystyle { \Big(3 + \frac{1}{3}\Big) - \Big(1 + 1\Big) }
$

\vskp

$ = \displaystyle { \Big(\frac{10}{3}\Big) - \Big(2\Big) } $

\vskp

$ = \displaystyle { \frac{10}{3} - \frac{6}{3} } $

\vskp

$ = \displaystyle { \frac{4}{3} } $

\vskp \vskp \vskp

{\it PROBLEM 4:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=x^2$ and $ y=-x^2+4x+6
$ .

\vskp \vskp

\noindent {\it SOLUTION 4:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y=x^2 $ and $
y=-x^{2}+4x+6 $ . Begin by finding the points of intersection of
the two graphs.  From $ y=x^2 $ and $ y=-x^{2}+4x+6 $  we get that

\vskp

$ x^{2} = -x^{2} + 4x + 6 \ \ \ \ \longrightarrow $

\vskp

$ 2x^{2} - 4x - 6 = 0 \ \ \ \ \longrightarrow $

\vskp

$ x^{2} - 2x - 3 = 0 \ \ \ \ \longrightarrow $

\vskp

$ (x-3)(x+1) = 0 \ \ \ \ \longrightarrow $

\vskp

$ x = 3 $ or $ x = -1 $

\vskp

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{-1}^{3} (top \ - \ bottom) \ dx }$

\vskp

$ = \displaystyle { \int_{-1}^{3} ((-x^{2}+4x+6)-(x^{2})) \ dx } $

\vskp

$ = \displaystyle { \int_{-1}^{3} (-x^{2}+4x+6) \ dx } $

\vskp

$ = \displaystyle { \Big(\frac{-2x^{3}}{3} + \frac{4x^{2}}{2} + 6x
\Big) \Big\vert_{-1}^{3} } $

\vskp

$ = \displaystyle { \Big(-\frac{2}{3}x^{3} + 2x^{2} + 6x \Big)
\Big\vert_{-1}^{3} } $

\vskp

$ = \displaystyle { \Big( -\frac{2}{3}(3)^{3} + 2(3)^{2} + 6(3)
\Big) - \Big( -\frac{2}{3}(-1)^{3} + 2(-1)^{2} + 6(-1) \Big) } $

\vskp

$ = \displaystyle { \Big( -18 + 18 + 18 \Big) - \Big( \frac{2}{3}
+ 2 - 6 \Big) } $

\vskp

$ = \displaystyle { \Big( 18 \Big) - \Big( \frac{2}{3} -
\frac{6}{3} - \frac{18}{3} \Big) } $

\vskp

$ = \displaystyle { \Big( 18 \Big) - \Big( - \frac{22}{3} \Big) }
$

\vskp

$ = \displaystyle { 18 + \frac{22}{3} } $

\vskp

$ = \displaystyle { \frac{54}{3} + \frac{22}{3} } $

\vskp

$ = \displaystyle {\frac{76}{3}} $

\vskp \vskp \vskp


{\it PROBLEM 5:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=x^3+x^2$ and $
y=3x^2+3x $ .

\vskp \vskp

\noindent {\it SOLUTION 2:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y=x^3+x^2 $ and $
y=3x^2+3x $ . Begin by finding the points of intersection of the
two graphs. From $ y=x^3+x^2 $ and $ y=3x^2+3x $  we get that

\vskp

$ x^{3} + x^{2} = 3x^{2} + 3x \ \ \ \ \longrightarrow $

\vskp

$ x^{3} - 2x^{2} - 3x = 0 \ \ \ \ \longrightarrow $

\vskp

$ x(x^{2} - 2x - 3) = 0 \ \ \ \ \longrightarrow $

\vskp

$ x(x-3)(x+1) = 0 \ \ \ \ \longrightarrow $

\vskp

$ x=0 $, $ x=3 $, or $ x=-1 $

\vskp

\noindent Using vertical cross-sections we get that

\vskp




AREA $ = \displaystyle{ \int_{-1}^{0} (top \ - \ bottom) \ dx +
\int_{0}^{3} (top \ - \ bottom) \ dx }$

\vskp

$ = \displaystyle { \int_{-1}^{0} ((x^{3}+x^{2})-(3x^{2}+3x)) \ dx
+ \int_{0}^{3} ((3x^{2}+3x) -(x^{3}+x^{2})) \ dx } $

\vskp

$ = \displaystyle { \int_{-1}^{0} (x^{3}-2x^{2}-3x) \ dx +
\int_{0}^{3} (-x^{3}+2x^{2}+3x) \ dx } $

\vskp

$ = \displaystyle { \Big( \frac{x^{4}}{4} - \frac{2x^{3}}{3} -
\frac{3x^{2}}{2} \Big) \Big\vert_{-1}^{0} + \Big( -\frac{x^{4}}{4}
+ \frac{2x^{3}}{3} + \frac{3x^{2}}{2} \Big) \Big\vert_{0}^{3}} $

\vskp

$ = \displaystyle { \Big( \frac{0^{4}}{4} - \frac{2(0)^{3}}{3} -
\frac{3(0)^{2}}{2} \Big) - \Big( \frac{(-1)^{4}}{4} -
\frac{2(-1)^{3}}{3} - \frac{3(-1)^{2}}{2} \Big) + \Big(
-\frac{3^{4}}{4} + \frac{2(3)^{3}}{3} + \frac{3(3)^{2}}{2} \Big)  } $

\vskp

$ \displaystyle { \ \ \ \ - \Big( -\frac{0^{4}}{4} +
\frac{2(0)^{3}}{3} + \frac{3(0)^{2}}{2} \Big) } $

\vskp

$ = \displaystyle { \Big( 0 \Big) - \Big( \frac{1}{4} +
\frac{2}{3} - \frac{3}{2} \Big) + \Big( -\frac{81}{4} + 18 +
\frac{27}{2} \Big) - \Big( 0 \Big) } $

\vskp

$ = \displaystyle { - \Big( \frac{3}{12} + \frac{8}{12} -
\frac{18}{12} \Big) + \Big( -\frac{81}{4} + \frac{72}{4} +
\frac{54}{4} \Big) } $

\vskp

$ = \displaystyle { - \Big( -\frac{7}{12} \Big) + \Big(
\frac{45}{4} \Big) } $

\vskp

$ = \displaystyle { \frac{7}{12} + \frac{45}{4} } $

\vskp

$ = \displaystyle { \frac{7}{12} + \frac{135}{12} } $

\vskp

$ = \displaystyle { \frac{142}{12} } $

\vskp

$ = \displaystyle { \frac{71}{6} } $

\vskp \vskp \vskp


{\it PROBLEM 6:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y= \ln x, y=1,  $ and $
x=e^2 $ .

\vskp \vskp

\noindent {\it SOLUTION 6:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y=\ln x $, $ y=1 $
and $ y=e^{2} $ . Begin by finding the points of intersection of
the two graphs. From $ y=\ln x $ and $ y=1 $  we get that

\vskp

$ \ln x = 1 \ \ \ \ \longrightarrow $

\vskp

$ e^{\ln x} = e^{1} $

\vskp

$ x = e $

\vskp

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{e}^{e^2} (top \ - \ bottom) \ dx
}$

\vskp

$ = \displaystyle { \int_{e}^{e^{2}} (\ln x - 1) \ dx } $

\vskp

$ = \displaystyle { \int_{e}^{e^{2}} \ln x \ dx - \int_{e}^{e^{2}}
1 \ dx } $

\vskp

\noindent Let $ A = \displaystyle { \int \ln x \ dx } $.

\vskp

\noindent Use integration by parts. Let $ u = \ln x $ and $ dv =
dx $ such that $ du = \frac{1}{x} \ dx $ and $ v = x $.

\vskp

$ A = \displaystyle { x \ln x - \int (\frac{1}{x})(x) \ dx } $

\vskp

$ = \displaystyle { x \ln x - \int \ dx } $

\vskp

$ = \displaystyle { x \ln x - x + C } $

\vskp

$ \displaystyle { A - \int_{e}^{e^{2}} 1 \ dx } $

\vskp

$ = \displaystyle { \Big(  x \ln x - x - x \Big)
\Big\vert_{e}^{e^{2}}
 } $

\vskp

$ = \displaystyle { \Big(  x \ln x - 2x \Big)
\Big\vert_{e}^{e^{2}} } $

\vskp

$ = \displaystyle { \Big( e^{2} \ln e^{2} - 2e^{2} \Big) - \Big( e
\ln e - 2e \Big) } $

\vskp

$ = \displaystyle { \Big( e^{2}(2) - 2e^{2} \Big) - \Big( e - 2e
\Big) } $

\vskp

$ = \displaystyle { \Big( 0 \Big) - \Big( -e \Big) } $

\vskp

$ = \displaystyle { e } $


\vskp \vskp \vskp


{\it PROBLEM 7:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y= \cos x, y= \sin x,  $
and $ x=0 $ .

\vskp \vskp

\noindent {\it SOLUTION 7:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y = \cos x $, $ y
= \sin x $ and $ x = 0 $ . Begin by finding the points of
intersection of the two graphs. From $ y=x^2 $ and $ y=x+2 $  we
get that

\vskp

$ \cos x = \sin x \ \ \ \ \longrightarrow $

\vskp

$ \frac{\cos x}{\sin x} = 1 \ \ \ \ \longrightarrow $

\vskp

$ \cot x = 1 \ \ \ \ \longrightarrow $

\vskp

$ x = \frac{\pi}{4} $

\vskp

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{0}^{\pi / 4} (top \ - \ bottom) \ dx
}$

\vskp

$ = \displaystyle { \int_{0}^{\pi / 4} ( \cos x - \sin x ) \ dx }
$

\vskp

$ = \displaystyle { \Big( \sin x - (-\cos x) \Big)
\Big\vert_{0}^{\pi / 4} } $

\vskp

$ = \displaystyle { \Big( \sin x + \cos x \Big) \Big\vert_{0}^{\pi
/ 4} } $

\vskp

$ = \displaystyle { \Big( \sin \frac{\pi}{4} + \cos \frac{\pi}{4}
\Big) - \Big( \sin 0 + \cos 0 \Big) } $

\vskp

$ = \displaystyle { \Big( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}
\Big) - \Big( 0 + 1 \Big) } $

\vskp

$ = \displaystyle { \sqrt{2} - 1 } $

\vskp \vskp \vskp


{\it PROBLEM 8:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=8/x, y=2x,  $ and $ y=2
$ .

\vskp \vskp

\noindent {\it SOLUTION 8:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y = \frac{8}{x} $,
$ y = 2x $ and $ y = 2 $ . Begin by finding the points of
intersection of the two graphs.  From $ y = \frac{8}{x} $ and $ y
= 2x $  we get that

\vskp

$ \frac{8}{x} = 2x \ \ \ \ \longrightarrow $

\vskp

$ 8 = 2x^{2} \ \ \ \ \longrightarrow $

\vskp

$ 4 = x^{2} \ \ \ \ \longrightarrow $

\vskp

$ x = 2 $ or $ x = -2 \ \ \ \ \longrightarrow $

\vskp

$ y = 4 $ or $ y = -4 $

\vskp

\noindent Using horizontal cross-sections such that $ x =
\frac{8}{y} $ and $ x = \frac{y}{2} $ we get that

\vskp

AREA $ = \displaystyle{ \int_{2}^{4} (right \ - \ left ) \ dy }$

\vskp

$ \displaystyle { \int_{2}^{4} ( \frac{8}{y} - \frac{y}{2} ) \ dy
} $

\vskp

$ = \displaystyle { \Big( 8 \ln y - \frac{y^{2}}{4} \Big)
\Big\vert_{2}^{4} } $

\vskp

$ = \displaystyle { \Big( 8 \ln 4 - \frac{4^{2}}{4} \Big) - \Big(
8 \ln 2 - \frac{2^{2}}{4} \Big) } $

\vskp

$ = \displaystyle { \Big( 8 \ln 4 - 4 \Big) - \Big( 8 \ln 2 - 1
\Big) } $

\vskp

$ = \displaystyle { \Big( 8 \ln 2^{2} - 4 \Big) - \Big( 8 \ln 2 -
1 \Big) } $

\vskp

$ = \displaystyle { \Big( 16 \ln 2 - 4 \Big) - \Big( 8 \ln 2 - 1
\Big) } $

\vskp

$ = \displaystyle { 8 \ln 2 - 3 } $

\vskp \vskp \vskp


{\it PROBLEM 9:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=\ln x $ and $ y= (\ln
x)^2 $ .

\vskp \vskp

\noindent {\it SOLUTION 9:} \ \ \ Compute the area of the enclosed
region bounded by the graphs of the equations $ y=\ln x $ and $ y=
(\ln x)^2 $ . Begin by finding the points of intersection of the
two graphs. From $ y=\ln x $ and $ y= (\ln x)^2 $  we get that

\vskp

$ \ln x = (\ln x)^{2} \ \ \ \ \longrightarrow $

\vskp

$ \ln x - (\ln x)^{2} = 0 \ \ \ \ \longrightarrow $

\vskp

$ \ln x (1 - \ln x) \ \ \ \ \longrightarrow $

\vskp

$ \ln x = 0 $ or $ \ln x = 1 $ $ \ \ \ \ \longrightarrow $

\vskp

$ x = 1 $ or $ x = e $

\vskp

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{1}^{e} (top \ - \ bottom) \ dx }$

\vskp

$ = \displaystyle { \int_{1}^{e} (\ln x - (\ln x)^{2}) \ dx } $

\vskp

$ = \displaystyle { \int_{1}^{e} \ln x \ dx - \int_{1}^{e} (\ln
x)^{2} \ dx } $

\vskp

\noindent Let $ A = \int_{1}^{e} \ln x \ dx $ and $ B =
\displaystyle { \int (\ln x)^{2} \ dx } $.

\vskp

\noindent (Recall from {\it PROBLEM 6} that $ \displaystyle { \int
\ln x \ dx = x\ln - x + C } $ )

\vskp

$ A = \displaystyle { \Big( x \ln x - x \Big) \Big\vert_{1}^{e} }
$

\vskp

$ = \displaystyle { \Big( e \ln e  - 1 \ln 1 \Big) - \Big( e - 1
\Big) } $

\vskp

$ = \displaystyle { \Big( e - 0 \Big) - \Big( e - 1 \Big) } $

\vskp

$ = \displaystyle { 1 } $

\vskp

\noindent Use integration by parts. Let $ u = (\ln x)^{2} $ and $
dv = dx $ such that $ du = 2 \ln x (\frac{1}{x}) \ dx $ and $ v =
x $.

\vskp

$ B = \displaystyle { \Big( x(\ln x)^{2} \Big) \Big\vert_{1}^{e} -
\int_{1}^{e} 2 \ln x \ dx } $

\vskp

$ = \displaystyle { \Big( x (\ln x)^{2} \Big) \Big\vert_{1}^{e} -
2 \Big( x \ln x - x \Big) \Big\vert_{1}^{e} } $

\vskp

$ = \displaystyle { \Big( e(\ln e)^{2} - 1(\ln 1)^{2} \Big) - 2
\Big( (e \ln e - e) - (1 \ln 1 - 1) \Big) } $

\vskp

$ = \displaystyle { \Big( (e - 0) \Big) - 2 \Big( (e - e) - (0 -
1) \Big) } $

\vskp

$ = \displaystyle { e - 2 } $

\vskp

$ A - B = 1 - (e - 2) $

\vskp

$ = 3 - e $

\vskp \vskp \vskp


{\it PROBLEM 10:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=\tan^2 x $, $ y=0 $ and
$ x=1 $ .

\vskp \vskp

\noindent {\it SOLUTION 10:} \ \ \ Compute the area of the
enclosed region bounded by the graphs of the equations $ y =
\tan^2 x $, $ y = 0 $, and $  x =1 $ . Begin by finding the points
of intersection of the two graphs.  From $ y = \tan^{2} x$ and $ y
= 0 $ we get that

\vskp

$ \tan^{2} x = 0  \ \ \ \ \longrightarrow $

\vskp

$ \tan x = 0  \ \ \ \ \longrightarrow $

\vskp

$ x = 0 $

\vskp

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{0}^{1} (top \ - \ bottom) \ dx }$

\vskp

$ = \displaystyle { \int_{0}^{1} \tan^{2} x \ dx } $

\vskp

$ = \displaystyle { \int_{0}^{1} ( \sec^{2} x - 1 ) \ dx } $

\vskp

$ = \displaystyle { \Big( \tan x - x \Big) \Big\vert_{0}^{1} } $

\vskp

$ = \displaystyle { \Big( \tan 1 - 1 \Big) - \Big( \tan 0 - 0
\Big) } $

\vskp

$ = \displaystyle { \tan 1 - 1 } $

\vskp \vskp \vskp


{\it PROBLEM 11:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y=x, y=2x,  $ and $ y=6-x
$ .

\vskp \vskp

\noindent {\it SOLUTION 11:} \ \ \ Compute the area of the
enclosed region bounded by the graphs of the equations $ y=x $, $
y=2x $, and $ y=6-x $ . Begin by finding the points of
intersection of the two graphs.

\vskp

\noindent From $ y=x $ and $ y=2x $  we get that

\vskp

$ x = 2x \ \ \ \ \longrightarrow $

\vskp

$ -x = 0 \ \ \ \ \longrightarrow $

\vskp

$ x = 0 $

\vskp

\noindent From $ y=x $ and $ y=6-x $  we get that

\vskp

$ x = 6-x \ \ \ \ \longrightarrow $

\vskp

$ 2x = 6 \ \ \ \ \longrightarrow $

\vskp

$ x = 3 $

\vskp

\noindent From $ y=2x $ and $ y=6-x $  we get that

\vskp

$ 2x = 6-x \ \ \ \ \longrightarrow $

\vskp

$ 3x = 6 \ \ \ \ \longrightarrow $

\vskp

$ x = 2 $

\vskp

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{0}^{2} (top \ - \ bottom) \ dx +
\int_{2}^{3} (top \ - \ bottom) \ dx }$

\vskp

$ = \displaystyle { \int_{0}^{2} (2x - x) \ dx + \int_{2}^{3}
((6-x) - x) \ dx } $

\vskp

$ = \displaystyle { \int_{0}^{2} x \ dx + \int_{2}^{3} (6-2x) \ dx
} $

\vskp

$ = \displaystyle { \frac{x^{2}}{2} \Big\vert_{0}^{2} + \Big( 6x -
\frac{2x^{2}}{2} \Big) \Big\vert_{2}^{3} } $

\vskp

$ = \displaystyle { \frac{x^{2}}{2} \Big\vert_{0}^{2} + \Big( 6x -
x^{2} \Big) \Big\vert_{2}^{3} } $

\vskp

$ = \displaystyle { \Big( \frac{2^{2}}{2} - \frac{0^{2}}{2} \Big)
+ \Big( (6 \cdot 3 - 3^{2}) - (6 \cdot 2 - 2^{2}) \Big) } $

\vskp

$ = \displaystyle { \Big( 2 - 0 \Big) + \Big( 9 - 8 \Big) } $

\vskp

$ = \displaystyle { 2 + 1 } $

\vskp

$ = \displaystyle { 3 } $

\vskp \vskp \vskp


{\it PROBLEM 12:}  \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ y = \sin \sqrt{x} $, $
y=0 $, and $ x = \pi / 2 $ .

\vskp \vskp

\noindent {\it SOLUTION 12:} \ \ \ Compute the area of the
enclosed region bounded by the graphs of the equations $ y = \sin
\sqrt{x} $, $ y=0 $, and $ x = \pi / 2 $. Begin by finding the
points of intersection of the two graphs. From $ y = \sin \sqrt{x}
$ and $ y=0 $ we get that

\vskp

$ \sin \sqrt{x} = 0 \ \ \ \ \longrightarrow $

\vskp

$ x =0 $

\noindent Using vertical cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{0}^{\pi / 2} (top \ - \ bottom) \ dx
}$

\vskp

$ = \displaystyle { \int_{0}^{\pi / 2} ( \sin \sqrt{x} ) \ dx } $

\vskp

\noindent Use integration by parts. Let $ u = \sin \sqrt{x} $ and
$ dv = dx $ such that $ du = \cos \sqrt{x} \cdot \frac{1}{2
\sqrt{x}} \ dx $ and $ v = x $.

\vskp

$ = \displaystyle { x \sin \sqrt{x} \Big\vert_{0}^{\pi / 2} -
\int_{0}^{\pi / 2} \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \cdot
x \ dx } $

\vskp

$ = \displaystyle { \Big( (\frac{\pi}{2}) \sin
\sqrt{\frac{\pi}{2}} - (0) \sin \sqrt{0} \Big) - \int_{0}^{\pi /
2} \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \cdot x \ dx } $

\vskp

$ = \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} -
\int_{0}^{\pi / 2} \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \cdot
x \ dx } $

\vskp

\noindent Use substitution. Let

\vskp

$ w = \sqrt{x} $

\vskp

\noindent so that

\vskp

$ dw = \frac{1}{2 \sqrt{x}} \ dx = \frac{\sqrt{x}}{2x} \ dx $

\vskp

$ x \ dw = \frac{\sqrt{x}}{2} \ dx \ \ \ \ \longrightarrow $

\vskp

$ w^{2} \ dw = \frac{\sqrt{x}}{2} \ dx $

\vskp

$ = \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} -
\int_{0}^{\sqrt{\pi / 2}} \cos w \cdot w^{2} \ dw } $

\vskp

\noindent Use integration by parts. Let $ u' = w^{2} $ and $ dv' =
\cos w \ dw $ such that $ du' = 2w \ dw $ and $ v' = \sin w $.

\vskp

$ = \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} -
\Big( w^{2} \sin w \Big\vert_{0}^{\sqrt{\pi / 2}} -
\int_{0}^{\sqrt{\pi / 2}} 2w \sin w \ dw \Big) } $

\vskp

$ = \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} -
w^{2} \sin w \Big\vert_{0}^{\sqrt{\pi / 2}} + \int_{0}^{\sqrt{\pi
/ 2}} 2w \sin w \ dw } $

\vskp

$ = \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} -
\Big( (\sqrt{\frac{\pi}{2}})^{2} \sin \sqrt{\frac{\pi}{2}} -
(0)^{2} \sin 0 \Big) + \int_{0}^{\sqrt{\pi / 2}} 2w \sin w \ dw }
$

\vskp

$ = \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} -
\frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} + \int_{0}^{\sqrt{\pi /
2}} 2w \sin w \ dw} $

\vskp

$ = \displaystyle { \int_{0}^{\sqrt{\pi / 2}} 2w \sin w \ dw } $

\vskp

\noindent Use integration by parts. Let $ u" = 2w $ and $ dv" =
\sin w \ dw $ such that $ du" = 2 \ dw
 $ and $ v" = - \cos w $

\vskp

$ = \displaystyle { -2 w \cos w \Big\vert_{0}^{\sqrt{\pi / 2}} +
\int_{0}^{\sqrt{\pi / 2}} 2 \cos w \ dw } $

\vskp

$ = \displaystyle { -2 \Big( (\sqrt{\frac{\pi}{2}}) \cos
\sqrt{\frac{\pi}{2}} - (0) \cos 0 \Big) + 2 \sin w
\Big\vert_{0}^{\sqrt{\frac{\pi}{2}}} } $

\vskp

$ = \displaystyle { -2(\sqrt{\frac{\pi}{2}}) \cos
\sqrt{\frac{\pi}{2}} + 2 \Big( \sin \sqrt{\frac{\pi}{2}} - \sin 0
\Big) } $

\vskp

$ = \displaystyle { -\sqrt{2 \pi} \cos \sqrt{\frac{\pi}{2}} + 2
\sin \sqrt{\frac{\pi}{2}} } $

\vskp \vskp \vskp


{\it PROBLEM 13:}  \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ x=y^2 $ and $ x=4 $ .

\vskp \vskp

\noindent {\it SOLUTION 13:} \ \ \ Compute the area of the
enclosed region bounded by the graphs of the equations $ x=y^{2} $
and $ x=4 $ . Begin by finding the points of intersection of the
two graphs.  From $ x=y^{2} $ and $ x=4 $  we get that

\vskp

$ y^{2} = 4 \ \ \ \ \longrightarrow $

\vskp

$ y = -2 $ or $ y = 2 $

\vskp

\noindent Using horizontal cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{-2}^{2} (right \ - \ left) \ dy }$

\vskp

$ = \displaystyle { \int_{-2}^{2} (4 - y^{2}) \ dy } $

\vskp

$ = \displaystyle { \Big( 4y - \frac{y^{3}}{3} \Big)
\Big\vert_{-2}^{2} } $

\vskp

$ = \displaystyle { \Big( 4 \cdot 2 - \frac{2^{3}}{3} \Big) -
\Big( 4 \cdot (-2) - \frac{(-2)^{3}}{3} \Big) } $

\vskp

$ = \displaystyle { \Big( 8 - \frac{8}{3} \Big) - \Big( -8 +
\frac{8}{3} \Big) } $

\vskp

$ = \displaystyle { 16 - \frac{16}{3} } $

\vskp

$ = \displaystyle { \frac{48}{3} - \frac{16}{3} } $

\vskp

$ = \displaystyle { \frac{32}{3} } $


\vskp \vskp \vskp


{\it PROBLEM 14:}  \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ x=y+3 $ and $ x=y^2-y $ .

\vskp \vskp

\noindent {\it SOLUTION 14:} \ \ \ Compute the area of the
enclosed region bounded by the graphs of the equations $ x=y+3 $
and $ x=y^{2}-y $ . Begin by finding the points of intersection of
the two graphs.  From $ x=y+3 $ and $ x=y^{2}-y $  we get that

\vskp

$ y+3 = y^{2}-y \ \ \ \ \longrightarrow $

\vskp

$ 0 = y^{2} - 2y - 3 \ \ \ \ \longrightarrow $

\vskp

$ 0 = (y-3)(y+1) \ \ \ \ \longrightarrow $

\vskp

$ y=3 $ or $ y=-1 $

\vskp

\noindent Using horizontal cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{-1}^{3} (right \ - \ left) \ dy }$

\vskp

$ = \displaystyle { \int_{-1}^{3} ((y+3)-(y^{2}-y)) \ dy } $

\vskp

$ = \displaystyle { \int_{-1}^{3} (-y^{2}+2y+3) \ dy } $

\vskp

$ = \displaystyle { \Big( -\frac{y^{3}}{3} + \frac{2y^{2}}{2} +3y
\Big) \Big\vert_{-1}^{3} } $

\vskp

$ = \displaystyle { \Big( -\frac{y^{3}}{3} + y^{2} +3y \Big)
\Big\vert_{-1}^{3} } $

\vskp

$ = \displaystyle { \Big( -\frac{3^{3}}{3} + 3^{2} + 3(3) \Big) -
\Big( -\frac{(-1)^{3}}{3} + (-1)^{2} + 3(-1) \Big) } $

\vskp

$ = \displaystyle { \Big( -9 + 9 + 9 \Big) - \Big( \frac{1}{3} + 1
- 3 \Big) } $

\vskp

$ = \displaystyle { \Big( -9 + 9 + 9 \Big) - \Big( \frac{1}{3} +
\frac{3}{3} - \frac{9}{3} \Big) } $

\vskp

$ = \displaystyle { \Big( 9 \Big) - \Big( -\frac{5}{3} \Big) } $

\vskp

$ = \displaystyle { \frac{27}{3} + \frac{5}{3} } $

\vskp

$ = \displaystyle { \frac{32}{3} } $

\vskp \vskp \vskp


{\it PROBLEM 15:} \ \ \ Compute the area of the enclosed region
bounded by the graphs of the equations $ x=y^3 $ and $ x=y^2+2y $.

\vskp \vskp

\noindent {\it SOLUTION 15:} \ \ \ Compute the area of the
enclosed region bounded by the graphs of the equations $ x=y^3 $
and $ x=y^{2}+2y $ . Begin by finding the points of intersection
of the two graphs.  From $ x=y^3 $ and $ x=y^2+2y $  we get that

\vskp

$ y^{3} = y^2 + 2y \ \ \ \ \longrightarrow $

\vskp

$ y^{3} - y^2 - 2y = 0  \ \ \ \ \longrightarrow $

\vskp

$ y ( y^{2} - y - 2 ) = 0 \ \ \ \ \longrightarrow $

\vskp

$ y ( y-2 ) ( y+1 ) = 0 \ \ \ \ \longrightarrow $

\vskp

$ y=0 $, $ y=2 $, or $ y=-1 $

\vskp

\noindent Using horizontal cross-sections we get that

\vskp

AREA $ = \displaystyle{ \int_{-1}^{0} (right \ - \ left) \ dy +
\int_{0}^{2} (right \ - \ left) \ dy }$


\vskp

$ = \displaystyle { \int_{-1}^{0} (y^{3} - (y^{2}+2y)) \ dy +
\int_{0}^{2} (y^{2}+2y - y^{3}) \ dy } $

\vskp

$ = \displaystyle { \int_{-1}^{0} (y^{3} - y^{2} - 2y) \ dy +
\int_{0}^{2} (y^{2}+2y - y^{3}) \ dy } $

\vskp

$ = \displaystyle { \Big( \frac{y^{4}}{4} - \frac{y^{3}}{3} -
\frac{2y^{2}}{2} \Big) \Big\vert_{-1}^{0} + \Big( \frac{y^{3}}{3}
+ \frac{2y^{2}}{2} - \frac{y^{4}}{4} \Big) \Big\vert_{0}^{2} } $

\vskp

$ = \displaystyle { \Big( \frac{y^{4}}{4} - \frac{y^{3}}{3} -
y^{2} \Big) \Big\vert_{-1}^{0} + \Big( \frac{y^{3}}{3} + y^{2} -
\frac{y^{4}}{4} \Big) \Big\vert_{0}^{2} } $

\vskp

$ = \displaystyle { \Big( \frac{0^{4}}{4} - \frac{0^{3}}{3} -
0^{2} \Big) - \Big( \frac{(-1)^{4}}{4} - \frac{(-1)^{3}}{3} -
(-1)^{2} \Big) + \Big( \frac{2^{3}}{3} + 2^{2} - \frac{2^{4}}{4}
\Big) - \Big( \frac{0^{3}}{3} + 0^{2} - \frac{0^{4}}{4} \Big) } $

\vskp

$ = \displaystyle { \Big( 0 \Big) - \Big( \frac{1}{4} +
\frac{1}{3} - 1 \Big) + \Big( \frac{8}{3} + 4 - 4 \Big) - \Big( 0
\Big) } $

\vskp

$ = \displaystyle { -\Big( -\frac{7}{12} \Big) + \Big( \frac{8}{3}
\Big) } $

\vskp

$ = \displaystyle { \frac{7}{12} + \frac{8}{3} } $

\vskp

$ = \displaystyle { \frac{7}{12} + \frac{32}{12} } $

\vskp

$ = \displaystyle { \frac{39}{12} } $

\end{document}
