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{SECT 0 {PARA 256 "" 0 "" {TEXT 256 23 "Laboratory Exercise 3.3" }
{TEXT 257 25 "\nCaculating Riemann Sums\n" }{TEXT 258 14 "(CCH Text 2.
3)" }{MPLTEXT 1 0 2 "\n\n" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 19 "The \+
function is ..." }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }{XPPEDIT 18 0 "f
(x)=(1/4)(6+7x-x^3)" "/-%\"fG6#%\"xG-*&\"\"\"\"\"\"\"\"%!\"\"6#,(\"\"'
F**&\"\"(F*F&F*F**$F&\"\"$F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "f:=x->(1/4)*(6+7*x-x^3);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG:6#%\"xG6\"6$%)operatorG%&ar
rowGF(,(#\"\"$\"\"#\"\"\"9$#\"\"(\"\"%*$F1F.#!\"\"F4F(F(" }}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 9 "Problem 1" }}{PARA 0 "" 0 "" {TEXT -1 30 "
The left hand Riemann sum for " }{XPPEDIT 18 0 "f(x)" "-%\"fG6#%\"xG" 
}{TEXT -1 99 " on [-1,1] using 30 subintervals can be calculated as fo
llows.\nThe width of the entire interval is " }{XPPEDIT 18 0 "(1-(-1))
=2" "/,&\"\"\"\"\"\",$\"\"\"!\"\"F(\"\"#" }{TEXT -1 24 ", so each box \+
should be " }{XPPEDIT 18 0 "2/30=1/15" "/*&\"\"#\"\"\"\"#I!\"\"*&\"\"
\"F%\"#:F'" }{TEXT -1 12 " wide.  The " }}{PARA 0 "" 0 "" {TEXT -1 69 
"boxes have left hand endpoints that start at -1 and progress through \+
" }{XPPEDIT 18 0 "-1+1/15" ",&\"\"\"!\"\"*&\"\"\"\"\"\"\"#:F$F'" }
{TEXT -1 2 ", " }{XPPEDIT 18 0 "-1+2/15" ",&\"\"\"!\"\"*&\"\"#\"\"\"\"
#:F$F'" }{TEXT -1 5 ", ..." }{XPPEDIT 18 0 "-1+29/15" ",&\"\"\"!\"\"*&
\"#H\"\"\"\"#:F$F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 23 "The
 first box has area " }{XPPEDIT 18 0 "f(-1)(1/15)" "--%\"fG6#,$\"\"\"!
\"\"6#*&\"\"\"\"\"\"\"#:F(" }{TEXT -1 27 " and the next one has area \+
" }{XPPEDIT 18 0 "f(-1+1/15)(1/15)" "--%\"fG6#,&\"\"\"!\"\"*&\"\"\"\"
\"\"\"#:F(F+6#*&\"\"\"F+\"#:F(" }{TEXT -1 53 ".  If we continue this i
t \nbecomes the following sum." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "s
um(f(-1+i/15)*(1/15),i=0..29);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"#
H\"#5" }}{PARA 0 "" 0 "" {TEXT -1 104 "\nThat of course does not show \+
the structure very well, so let's turn to the \"student\" package of M
aple.\n" }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(
student):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "leftsum(f(x),x=-1..1,3
0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%$SumG6$,(#!\"\"\"\"%\"\"\"%
\"iG#\"\"(\"#g*$,&F)F+F,#F+\"#:\"\"$F(/F,;\"\"!\"#HF2" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 9 "evalf(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"
+++++H!\"*" }}{PARA 0 "" 0 "" {TEXT -1 22 "This is of course the " }
{XPPEDIT 18 0 "29/10" "*&\"#H\"\"\"\"#5!\"\"" }{TEXT -1 13 " given abo
ve." }{MPLTEXT 1 0 1 "\n" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Probl
em 2" }}{PARA 0 "" 0 "" {TEXT -1 30 "The right hand sum is similar." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ri
ghtsum(f(x),x=-1..1,30);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%$SumG6
$,(#!\"\"\"\"%\"\"\"%\"iG#\"\"(\"#g*$,&F)F+F,#F+\"#:\"\"$F(/F,;F+\"#IF
2" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(\");" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+++++J!\"*" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }
{MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Problem 3" }}
{PARA 257 "" 1 "" {TEXT -1 15 "To approximate " }{XPPEDIT 18 0 "int(f(
x),x=-1..1)" "-%$intG6$-%\"fG6#%\"xG/F(;,$\"\"\"!\"\"\"\"\"" }{TEXT 
-1 114 " to within one decimal place with a right hand sum \nmeans tha
t we must set the error to be less than 0.05.  Since " }{XPPEDIT 18 0 
"Delta x= (1-(-1))/n" "/*&%&DeltaG\"\"\"%\"xGF%*&,&\"\"\"F%,$\"\"\"!\"
\"F,F%%\"nGF," }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2/n" "*&\"\"#\"\"\"%
\"nG!\"\"" }{TEXT -1 15 " \nwe must have " }{XPPEDIT 18 0 "[f(1)-f(-1)
]" "7#,&-%\"fG6#\"\"\"\"\"\"-F%6#,$\"\"\"!\"\"F-" }{XPPEDIT 18 0 "(Del
ta x)" "*&%&DeltaG\"\"\"%\"xGF$" }{TEXT -1 16 " less than 0.05\n" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "(f(1)-f(-1))*2/n;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#,$*$%\"nG!\"\"\"\"'" }}{PARA 0 "" 0 "" {TEXT -1 17 "
So we must have  " }{XPPEDIT 18 0 "6/n<0.05" "2*&\"\"'\"\"\"%\"nG!\"\"
$\"\"&!\"#" }{TEXT -1 1 "." }{MPLTEXT 1 0 1 "\n" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "solve(\"<0.05,n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$
-%*RealRangeG6$,$%)infinityG!\"\"-%%OpenG6#\"\"!-F$6$-F*6#$\"$?\"F,F'
" }}{PARA 0 "" 0 "" {TEXT -1 61 "\nWell, that was not too helpful, so \+
let's solve it directly.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "6/0.05
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+++++7!\"(" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 27 "rightsum(f(x),x=-1..1,120);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$-%$SumG6$,(#!\"\"\"\"%\"\"\"%\"iG#\"\"(\"$S#*$,&F)F+F
,#F+\"#g\"\"$F(/F,;F+\"$?\"F2" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "eva
lf(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+,++DI!\"*" }}{PARA 0 "
" 0 "" {TEXT -1 1 "\n" }{MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "" 
{TEXT -1 9 "Problem 4" }}{PARA 258 "" 1 "" {TEXT -1 15 "To approximate
 " }{XPPEDIT 18 0 "int(f(x),x=-1..1)" "-%$intG6$-%\"fG6#%\"xG/F(;,$\"
\"\"!\"\"\"\"\"" }{TEXT -1 183 " to within one decimal place with the \+
average of the left and right hand sums\nmeans that we must set the er
ror to be less than 0.1.  In other words, we can have twice the error \+
if we " }}{PARA 0 "" 0 "" {TEXT -1 48 "take the average of the over an
d underestimates." }}{PARA 258 "" 0 "" {TEXT -1 6 "Since " }{XPPEDIT 
18 0 "Delta x= (1-(-1))/n" "/*&%&DeltaG\"\"\"%\"xGF%*&,&\"\"\"F%,$\"\"
\"!\"\"F,F%%\"nGF," }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2/n" "*&\"\"#\"
\"\"%\"nG!\"\"" }{TEXT -1 14 " we must have " }{XPPEDIT 18 0 "[f(1)-f(
-1)]" "7#,&-%\"fG6#\"\"\"\"\"\"-F%6#,$\"\"\"!\"\"F-" }{XPPEDIT 18 0 "(
Delta x)" "*&%&DeltaG\"\"\"%\"xGF$" }{TEXT -1 14 " less than 0.1" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "(f(1)-f(-1))*2/n;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#,$*$%\"nG!\"\"\"\"'" }}{PARA 0 "" 0 "" {TEXT -1 16 "
So we must have " }{XPPEDIT 18 0 "6/n<0.1" "2*&\"\"'\"\"\"%\"nG!\"\"$
\"\"\"!\"\"" }{TEXT -1 1 "." }{MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "6/0.1;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+++++g!\")" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
57 "(rightsum(f(x),x=-1..1,60) + leftsum(f(x),x=-1..1,60))/2;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%$SumG6$,(#!\"\"\"\"%\"\"\"%\"iG#\"
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0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 16 "Problems 5 \+
and 6" }}{PARA 0 "" 0 "" {TEXT -1 48 "The picture of the boxes for 10 \+
subintervals on " }{XPPEDIT 18 0 "int(f(x),x=-1..3)" "-%$intG6$-%\"fG6
#%\"xG/F(;,$\"\"\"!\"\"\"\"$" }{TEXT -1 16 " is given below." }}{PARA 
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6$7&FU7$FS$\"++++'\\#FG7$$\"\"\"F*FZ7$FgnF*F/-F$6$7&Fin7$Fgn$\"\"$F*7$
$\"+++++9FGF^o7$FaoF*F/-F$6$7&Fco7$Fao$\"++++kKFG7$$\"+++++=FGFho7$F[p
F*F/-F$6$7&F]p7$F[p$\"++++#>$FG7$$\"+++++AFGFbp7$FepF*F/-F$6$7&Fgp7$Fe
p$\"++++)o#FG7$$\"+++++EFGF\\q7$F_qF*F/-F$6$7&Faq7$F_q$\"++++c;FG7$F^o
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Far$\"1Z6MI=7(z\"F]u7$$\"1nmm\"fBIY#Far$\"1s?\\aOHF>F]u7$$\"1LLLLO[kLF
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" 0 "" {TEXT -1 239 "The reason that the boxes did not stay under the \+
graph is that the function is not increasing over the entire interval.
\nIf it is monotonic, then the boxes are all over or under the functio
n.  The same problem occurs with a right hand\nsum.\n" }{MPLTEXT 1 0 
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