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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 37 "Finding areas by using tr
ansformation" }}{PARA 19 "" 0 "" {TEXT -1 16 "Dr. Wei-Chi Yang" }}
{PARA 19 "" 0 "" {TEXT -1 29 "http://www.radford.edu/~wyang" }}{PARA 
0 "" 0 "" {TEXT -1 71 "Example: Find the area bounded by the curve x=2
y-y^2, from y=0 to y=2. " }}{PARA 0 "" 0 "" {TEXT -1 44 "Method 1. We \+
can plot the graph and compute " }{TEXT 256 8 "directly" }{TEXT -1 1 "
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "implicitplot(x=2*y-y^2,x=-5.
.5,y=0..2,scaling=constrained);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "int(2*y-y^2,y=0..2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 176 "
Method 2. Or we can use the  transformation x=y to find the area (so t
his will be the area of y=2x-x^2 from x=1 to x=2). Note the areas thes
e two should be the same. See below:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "plot(2*x-x^2,x=0..2,y=0..
1,scaling=constrained);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "
int(2*x-x^2,x=0..2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 191 "Remark 1
: Notice that if we flip the graph of  y=2*x-x^2 across ' y-axis ', an
d find the area from x=-2 to x=0, this area should be the same as the \+
previous two, but how do we find function??" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 14 "f:=x->2*x-x^2;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "g:=x->f(-x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 5 "g(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot(g(x),x=-
2..0,y=0..1,scaling=constrained);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "int(g(x),x=-2..0);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
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