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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 257 
"" 0 "" {TEXT -1 30 "Lesson 1:  Area Between Curves" }}}{SECT 0 {PARA 
3 "" 0 "" {TEXT -1 11 " Exercise 1" }}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 
18 0 "g(x)=x^2+2" "6#/-%\"gG6#%\"xG,&*$F'\"\"#\"\"\"F*F+" }{TEXT -1 0 
"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "f(x) = 2*x + 5" "6#/-%\"fG6#%\"
xG,&*&\"\"#\"\"\"F'F+F+\"\"&F+" }}{PARA 0 "" 0 "" {TEXT -1 123 "(a) Pl
ot both functions on the same axes. (b) Find the area of the region en
closed between the curves from x = 0 to x = 6. " }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 9 "restart: " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 17 "g:= x -> x^2 + 2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "
f:= x -> 2*x + 5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "a:= pl
ot(g(x), x = -1..7, thickness=2, color = red):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 54 "b:= plot(f(x), x = -1..7, thickness=2, color =
 brown):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "display(\{a,b\}
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "We need to find the interse
ction points for these two plots. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "solve(x^2 + 2 - 2*x - 5, x);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 57 "The area enclosed by these curves from x = 0 to x = 5 \+
is " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "int(2*x+5-x^2+2,
x = 0 .. 3);" "6#-%$intG6$,**&\"\"#\"\"\"%\"xGF)F)\"\"&F)*$F*F(!\"\"F(
F)/F*;\"\"!\"\"$" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "int(x^2+2-2*x+5,x \+
= 3 .. 6);" "6#-%$intG6$,**$%\"xG\"\"#\"\"\"F)F**&F)F*F(F*!\"\"\"\"&F*
/F(;\"\"$\"\"'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "int(2*x + 5 - x^2 - 2, x = 0
..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "int(x^2 + 2 - 2*x \+
+ 5, x = 3..6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Thus the total
 are enclosed is 57 + 9 = 66 sq units. " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 58 "Let's redo the plot where we shade the areas in question.
 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "m:= plot(f(x), x = 0..
7, thickness=2, color = magenta, axes=boxed):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 63 "n:= plot(g(x), x = 0..7, thickness=2, color = bl
ue,axes=boxed):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "p:= seq
( plot([0 + i * (3/100) , t, t = g(0 + i*(3/100))..f(0 + i * (3/100))]
, thickness=2, color=red), i = 0..100):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 119 "q:= seq( plot([3 + i * (3/100), t, t = f(3 + i *(3/1
00))..g(3 + i *(3/100))], thickness=2, color = brown), i = 0..100):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "r:= textplot([6,45,`g`], a
lign=RIGHT):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "s:= textplo
t([6.5,7,`f`],align=RIGHT):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "display(\{m,n,p,q,r,s\});" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 
11 " Exercise 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "The parabola " 
}{XPPEDIT 18 0 "y = a*x^2+b*x+c" "6#/%\"yG,(*&%\"aG\"\"\"*$%\"xG\"\"#F
(F(*&%\"bGF(F*F(F(%\"cGF(" }{TEXT -1 28 " is tangent to the graph of \+
" }{XPPEDIT 18 0 "y = 2+abs(x-3)" "6#/%\"yG,&\"\"#\"\"\"-%$absG6#,&%\"
xGF'\"\"$!\"\"F'" }{TEXT -1 105 " at two points and the area of the re
gion bounded by their graphs is 10. Find a, b, and c. Make a sketch." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 8 "Solutio
n" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 28 "The axis of the para
bola is " }{XPPEDIT 18 0 "x = - b/(2*a) " "6#/%\"xG,$*&%\"bG\"\"\"*&\"
\"#F(%\"aGF(!\"\"F," }{TEXT -1 27 ". That is also the axis of " }
{XPPEDIT 18 0 "y = 2 +abs(x-3)" "6#/%\"yG,&\"\"#\"\"\"-%$absG6#,&%\"xG
F'\"\"$!\"\"F'" }{TEXT -1 6 " , so " }{XPPEDIT 18 0 " -b/(2*a) = 3 " "
6#/,$*&%\"bG\"\"\"*&\"\"#F'%\"aGF'!\"\"F+\"\"$" }{TEXT -1 5 ", or " }
{XPPEDIT 18 0 "b = -6*a" "6#/%\"bG,$*&\"\"'\"\"\"%\"aGF(!\"\"" }{TEXT 
-1 85 " . The point where the slope of the parabola is 1 is on both gr
aphs. Call the point [" }{XPPEDIT 18 0 "x[0],y[0]" "6$&%\"xG6#\"\"!&%
\"yG6#F&" }{TEXT -1 8 "]. Then " }{XPPEDIT 18 0 "x[0]-1 = a*x[0]^2-6*a
*x[0]+c;" "6#/,&&%\"xG6#\"\"!\"\"\"F)!\"\",(*&%\"aGF)*$&F&6#F(\"\"#F)F
)*(\"\"'F)F-F)&F&6#F(F)F*%\"cGF)" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "
1 = 2*a*x[0]-6*a;" "6#/\"\"\",&*(\"\"#F$%\"aGF$&%\"xG6#\"\"!F$F$*&\"\"
'F$F(F$!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 33 "eq1 := x0-1 = a*x0^2 -6*a*x0 + c;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eq2 := 1 = 2*a*x0 - 6*a;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "ac := solve(\{eq1,eq2\},\{a
,c\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "Finally, the area betwe
en the curves is 100, so the righthand half is 50." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 46 "eq3 := Int (a*x^2-6*a*x+c-(2+x-3),x=3..x0)=
50;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "eq3 := int (a*x^2-6*
a*x+c-(2+x-3),x=3..x0)=50;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "sol :=solve(subs(ac,eq3),x0);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "assign(\{x0=sol[1]\}); assign(ac);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 98 "plot([2+abs(x-3),a*x^2-6*a*x+c],x=sol[2]-
2..sol[1]+2,thickness=2, color=[red,green], thickness=2);" }}}}{SECT 
0 {PARA 3 "" 0 "" {TEXT -1 11 " Exercise 3" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 80 " Sketch the region bounded by the given curves and find t
he area of the region. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 34 " x = 3y, x + y = 0, 7x + 3y = 24. " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plot
s):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "f:= x -> x/3;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "g:= x -> -x;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "h:= x -> (24 - 7*x) / 3;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "a:= plot(f(x), x = -.5..7, t
hickness=2, color = brown):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
54 "b:= plot(g(x), x = -.5..7, thickness=2, color = blue):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "c:= plot(h(x), x = -.5..7, thicknes
s=2, color = magenta):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "d
:= textplot([6,4,`f`], thickness=2, color = brown):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 56 "e:= textplot([1.5,7,`h`], thickness=2, co
lor = magenta):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "k:= text
plot([2,-4,`g`], thickness=2, color = blue):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 112 "p:= seq( plot([0 + i * (3/50) , t, t = g(0 + i*
(3/50))..f(0 + i * (3/50))], thickness=2, color=red), i = 0..50):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "q:= seq( plot([3 + i * (3/5
0), t, t = g(3 + i *(3/50))..h(3 + i *(3/50))], thickness=2, color = b
rown), i = 0..50):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "displ
ay(\{a,b,c,d,e,k,p,q\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "We ne
ed to find the points of intersection. " }}{PARA 0 "" 0 "" {TEXT -1 
32 "We set f = g, h = g, and h = f. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "solve(f(x) = g(x), x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "solve(h(x) = g(x), x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "solve(h(x) = f(x), x);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 35 "Hence the desired area is given by " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "int(x/3
--x,x = 0 .. 3);" "6#-%$intG6$,(*&%\"xG\"\"\"\"\"$!\"\"F)%#%?GF+F(F+/F
(;\"\"!F*" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "int(8-7*x/3--x,x = 3 .. 6
);" "6#-%$intG6$,*\"\")\"\"\"*(\"\"(F(%\"xGF(\"\"$!\"\"F-%#%?GF-F+F-/F
+;F,\"\"'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "int(x/3 +
 x, x= 0..3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "int(8 - (7
*x)/3 + x, x = 3..6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Thus tot
al area is 12 sq units. " }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 11 " Ex
ercise 4" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 " For what values of m \+
do the line y = mx and the curve y = " }{XPPEDIT 18 0 "x/(x^2+1);" "6#
*&%\"xG\"\"\",&*$F$\"\"#F%F%F%!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "
" {TEXT -1 47 "enclose a region? Find the area of the region. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart: " }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 30 "First lets plot the curve y = " }{XPPEDIT 18 0 "x/(x^2+1)
;" "6#*&%\"xG\"\"\",&*$F$\"\"#F%F%F%!\"\"" }{TEXT -1 1 " " }}{PARA 0 "
" 0 "" {TEXT -1 47 "along with an example of y = mx, where say m = " }
{XPPEDIT 18 0 "1/2" "6#*&\"\"\"F$\"\"#!\"\"" }{TEXT -1 1 "." }}{PARA 
0 "" 0 "" {TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "
plot(\{x/2, x/(x^2 + 1)\}, x = -.5...5, thickness=2);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 68 "We need to determine how large the slope \+
m can be and still enclose " }}{PARA 0 "" 0 "" {TEXT -1 64 "a region. \+
From the above plot, the magnitude of m is determined " }}{PARA 0 "" 
0 "" {TEXT -1 35 "by the derivative of the curve y = " }{XPPEDIT 18 0 
"x/(x^2+1);" "6#*&%\"xG\"\"\",&*$F$\"\"#F%F%F%!\"\"" }{TEXT -1 11 " at
 x = 0. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "f:= x -> x / (x
^2 + 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(f);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "%(0);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 65 "Thus f ' (0) = 1. Hence a region is enclosed provide
d 0 < m < 1. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "Let's replot the
 curve and shade the region enclosed for an example m, say m = 1/4. " 
}}{PARA 0 "" 0 "" {TEXT -1 56 "Then we calculate the area for an arbit
rary value of m. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "g:= x \+
-> (1/4) * x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "solve(f(x)
 = g(x), x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "a:= plot(f(
x), x = 0..2, thickness=2, color = blue):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 53 "b:= plot(g(x), x = 0..2, thickness=2, color = brown
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "p:= seq( plot([0 + i
 * (sqrt(3)/50) , t, t = g(0 + i*(sqrt(3)/50))..f(0 + i * (sqrt(3)/50)
)], thickness=2, color=red), i = 0..50):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "display(\{a,b,p\});" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "solve( x / (x^2 + 1) = m * x,x);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 63 " Thus the intersection of the two curves is precis
ely when x = " }{XPPEDIT 18 0 "sqrt(-m(-1+m))/m;" "6#*&-%%sqrtG6#,$-%
\"mG6#,&\"\"\"!\"\"F)F,F-F,F)F-" }{TEXT -1 3 " . " }}{PARA 0 "" 0 "" 
{TEXT -1 41 " Hence the area is enclosed is given by: " }}{PARA 0 "" 
0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "i
nt(x/(x^2+1)-m*x,x = 0 .. sqrt(-m(-1+m))/m);" "6#-%$intG6$,&*&%\"xG\"
\"\",&*$F(\"\"#F)F)F)!\"\"F)*&%\"mGF)F(F)F-/F(;\"\"!*&-%%sqrtG6#,$-F/6
#,&F)F-F/F)F-F)F/F-" }{TEXT -1 3 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "Let's
 have maple do the integration. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 51 "int(x / (x^2 + 1) - m*x, x = 0..sqrt(-m*(-1+m))/m);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 " Thus the area enclosed, for 0 \+
< m < 1, is precisely: " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "1/2*ln(1/m)-1/2+1/2*m;" "6#,(*(\"\"\"F%\"\"#!\"\"-%#lnG6#*&F%F%%
\"mGF'F%F%*&F%F%F&F'F'*(F%F%F&F'F,F%F%" }{TEXT -1 2 ". " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 11 " Exercise \+
5" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Find the value of d such that
 the area of the region bounded by the " }}{PARA 0 "" 0 "" {TEXT -1 
14 "parabolas y = " }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 
3 " - " }{XPPEDIT 18 0 "d^2;" "6#*$%\"dG\"\"#" }{TEXT -1 9 " and y = \+
" }{XPPEDIT 18 0 "d^2;" "6#*$%\"dG\"\"#" }{TEXT -1 3 " - " }{XPPEDIT 
18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 8 "is 576. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart: \+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Let's get a sample picture. Say d \+
= 3. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "f:= x -> x^2 - 9;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "g:= x -> 9 - x^2;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "solve(f(x) = g(x), x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "a:= plot(f(x), x = -4..4, th
ickness=2, color = brown):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
53 "b:= plot(g(x), x = -4..4, thickness=2, color = blue):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "c:= seq( plot([-3 + i * (6/50) , t
, t = f(-3 + i*(6/50))..g(-3 + i * (6/50))], thickness=2, color=red), \+
i = 0..50):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(\{a,
b,c\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "We can see from the ab
ove plot that in the general case the functions f and g intersect " }}
{PARA 0 "" 0 "" {TEXT -1 13 "at d and -d. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "The enclosed area is given by \+
" }{XPPEDIT 18 0 "int(-x^2+d^2+d^2-x^2,x = -d .. d);" "6#-%$intG6$,**$
%\"xG\"\"#!\"\"*$%\"dGF)\"\"\"*$F,F)F-*$F(F)F*/F(;,$F,F*F," }{TEXT -1 
1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "int(2* d^2 - 2* x^2
, x = -d..d );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "solve((8/
3)*d^3 = 576, d)[1];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Thus when
 d = 6 or -6, the region enclosed has area 576." }}}}}{MARK "0 0 0" 3 
}{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }