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{SECT 0 {EXCHG {PARA 203 "" 0 "" {TEXT 204 36 "Rightend, Leftend and M
idpoint Rules" }{TEXT 204 0 "" }}{PARA 204 "" 0 "" {TEXT 205 12 "Wei-C
hi Yang" }{TEXT 205 0 "" }}{PARA 205 "" 0 "" {TEXT 206 23 "e-mail: wya
ng@runet.edu" }{TEXT 206 0 "" }}{PARA 205 "" 0 "" {TEXT 206 32 "URL: h
ttp://www.runet.edu/~wyang" }{TEXT 206 0 "" }}{PARA 206 "" 0 "" {TEXT 
207 0 "" }}{PARA 206 "" 0 "" {TEXT 207 160 "In this worksheet, we will
 experiment the rightend, leftend and midpoint rules for the Riemann s
um. Use this template for experimenting exercises from the text." }
{TEXT 207 0 "" }}{PARA 206 "" 0 "" {TEXT 207 0 "" }}{PARA 207 "> " 0 "
" {MPLTEXT 1 208 14 "with(student):" }{MPLTEXT 1 208 0 "" }}}{EXCHG 
{PARA 207 "> " 0 "" {MPLTEXT 1 208 13 "f:=x->tan(x);" }{MPLTEXT 1 208 
0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 39 "leftbox(f(x), x=
0..Pi/4, 10,color=RED);" }{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> "
 0 "" {MPLTEXT 1 208 29 "leftsum(f(x), x=0..Pi/4, 10);" }{MPLTEXT 1 
208 0 "" }}{PARA 206 "" 0 "" {TEXT 207 0 "" }}{PARA 206 "" 0 "" {TEXT 
207 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 9 "evalf(%);" }
{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 28 "r
ightsum(f(x),x=0..Pi/4,10);" }{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 
"> " 0 "" {MPLTEXT 1 208 35 "evalf(rightsum(f(x),x=0..Pi/4,10));" }
{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 27 "e
valf(int(f(x),x=0..Pi/4));" }{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "
> " 0 "" {MPLTEXT 1 209 29 "middlesum(f(x),x=0..Pi/4,10);" }{MPLTEXT 
1 210 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 209 9 "evalf(%);" 
}}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 210 0 "" }}}{EXCHG {PARA 208 
"" 0 "" {TEXT 211 93 "The followings are meant to show if we were to d
efine the rightend/leftend sums by ourselves:" }{TEXT 212 0 "" }}}
{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 37 "right:=proc(a,b,j,n) a+j
*(b-a)/n end;" }{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> " 0 "" 
{MPLTEXT 1 208 41 "left:=proc(a,b,j,n)  a+(j-1)*(b-a)/n end;" }
{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 206 "" 0 "" {TEXT 207 0 "" }}
{PARA 207 "> " 0 "" {MPLTEXT 1 208 62 "rsum:=proc(a,b,n) sum(((b-a)/n)
*f(right(a,b,j,n)),j=1..n) end;" }{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 
207 "> " 0 "" {MPLTEXT 1 208 16 "rsum(0,Pi/4,10);" }{MPLTEXT 1 208 0 "
" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 9 "evalf(%);" }
{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 61 "l
sum:=proc(a,b,n) sum(((b-a)/n)*f(left(a,b,j,n)),j=1..n) end;" }
{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 16 "l
sum(0,Pi/4,10);" }{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> " 0 "" 
{MPLTEXT 1 208 23 "evalf(lsum(0,Pi/4,10));" }{MPLTEXT 1 208 0 "" }}}
{EXCHG {PARA 206 "" 0 "" {TEXT 207 33 "Next we define the midpoint rul
e." }{TEXT 207 0 "" }}{PARA 206 "" 0 "" {TEXT 207 0 "" }}{PARA 207 "> \+
" 0 "" {MPLTEXT 1 208 61 "midpoint:=proc(a,b,j,n) (left(a,b,j,n)+right
(a,b,j,n))/2 end;" }{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> " 0 "" 
{MPLTEXT 1 208 65 "msum:=proc(a,b,n) sum(((b-a)/n)*f(midpoint(a,b,j,n)
),j=1..n) end;" }{MPLTEXT 1 208 0 "" }}}{EXCHG {PARA 207 "> " 0 "" 
{MPLTEXT 1 208 16 "msum(0,Pi/4,10);" }{MPLTEXT 1 208 0 "" }}}{EXCHG 
{PARA 207 "> " 0 "" {MPLTEXT 1 208 9 "evalf(%);" }{MPLTEXT 1 208 0 "" 
}}}{EXCHG {PARA 206 "" 0 "" {TEXT 207 110 "Remark: We observe that the
 midpoint rule is better in this case if we use the interval [0, Pi/4]
 and n = 10. " }{TEXT 207 0 "" }}{PARA 206 "" 0 "" {TEXT 207 144 "Ques
tion: If we fix the interval, and let 'n' get larger and larger, what \+
do you expect these three sums (right, left, and midpoint) will react?
" }{TEXT 207 0 "" }}}{PARA 209 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 
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