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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 256 
"" 0 "" {TEXT -1 69 "Lesson 4:  Applications of Integration 2: Average
 Value of a Function" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Let's exa
mine why we should define the " }{TEXT 256 13 "average value" }{TEXT 
-1 15 " of a function " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 19 " o
ver the interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT 
-1 8 " to be  " }{XPPEDIT 18 0 "f_a = int(f(x),x = a .. b)/(b-a);" "6#
/%$f_aG*&-%$intG6$-%\"fG6#%\"xG/F,;%\"aG%\"bG\"\"\",&F0F1F/!\"\"F3" }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 44 "We could start by partitioning the interval " }{XPPEDIT 18 0 "[
a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 6 " into " }{XPPEDIT 18 0 "n;" "6#%
\"nG" }{TEXT -1 29 " equal sub-intervals.  Since " }{XPPEDIT 18 0 "f;
" "6#%\"fG" }{TEXT -1 68 " should be approximately constant on each su
b-interval (at least if " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 54 "
 is large), we could approximate the average value of " }{XPPEDIT 18 
0 "f;" "6#%\"fG" }{TEXT -1 48 " by computing the average value of the \+
function " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 21 " we get by repl
acing " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 117 " on each subinter
val by its value at the right-hand endpoint of the sub-interval.  This
 is illustrated below, taking " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT 
-1 20 " to be the function " }{XPPEDIT 18 0 "cos(x);" "6#-%$cosG6#%\"x
G" }{TEXT -1 4 " on " }{XPPEDIT 18 0 "[0, Pi/2];" "6#7$\"\"!*&%#PiG\"
\"\"\"\"#!\"\"" }{TEXT -1 52 ", with 6 sub-intervals.  (The graph of t
he function " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 88 " consists of
 the 6 line segments that make up the tops of the rectangles in the fi
gure.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "student[rightbox]
(cos(x), x=0..Pi/2,6);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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AULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve \+
1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" }}}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "The average value of " }{XPPEDIT 
18 0 "g;" "6#%\"gG" }{TEXT -1 32 " should just be the mean of the " }
{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 17 " values taken by " }
{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 4 ".  (" }{XPPEDIT 18 0 "n = 6;
" "6#/%\"nG\"\"'" }{TEXT -1 37 " in the example above.)  This mean is
" }{XPPEDIT 18 0 "sum(f(a+k*(b-a)/n),k = 1 .. n)/n;" "6#*&-%$sumG6$-%
\"fG6#,&%\"aG\"\"\"*(%\"kGF,,&%\"bGF,F+!\"\"F,%\"nGF1F,/F.;F,F2F,F2F1
" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "1/(b-a);" "6#*&\"\"\"F$,&%\"bGF$%
\"aG!\"\"F(" }{XPPEDIT 18 0 "sum(f(a+k*(b-a)/n),k = 1 .. n);" "6#-%$su
mG6$-%\"fG6#,&%\"aG\"\"\"*(%\"kGF+,&%\"bGF+F*!\"\"F+%\"nGF0F+/F-;F+F1
" }{XPPEDIT 18 0 "(b-a)/n;" "6#*&,&%\"bG\"\"\"%\"aG!\"\"F&%\"nGF(" }
{TEXT -1 35 " .  Apart from the first factor of " }{XPPEDIT 18 0 "1/(b
-a);" "6#*&\"\"\"F$,&%\"bGF$%\"aG!\"\"F(" }{TEXT -1 41 ", we recognise
 this as a Riemann sum for " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 
5 "; as " }{XPPEDIT 18 0 "proc (n) options operator, arrow; infinity e
nd;" "6#R6#%\"nG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }{TEXT 
-1 49 ", our approximation should therefore converge to " }{XPPEDIT 
18 0 "1/(b-a);" "6#*&\"\"\"F$,&%\"bGF$%\"aG!\"\"F(" }{XPPEDIT 18 0 "in
t(f(x),x = a .. b);" "6#-%$intG6$-%\"fG6#%\"xG/F);%\"aG%\"bG" }{TEXT 
-1 65 ", so it seems reasonable to define the (exact) average  value o
f " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 28 " to be this last expre
ssion." }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 7 "Example" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 77 "If a freely falling body starts from rest
, then its displacement is given by " }{XPPEDIT 18 0 "s = gt^2/2;" "6#
/%\"sG*&%#gtG\"\"#F'!\"\"" }{TEXT -1 14 ".  Fix a time " }{XPPEDIT 18 
0 "T;" "6#%\"TG" }{TEXT -1 34 ", and let the velocity after time " }
{XPPEDIT 18 0 "T;" "6#%\"TG" }{TEXT -1 6 " be V." }}{PARA 0 "" 0 "" 
{TEXT -1 53 "(a). Show that the average velocity, with respect to " }
{XPPEDIT 18 0 "t;" "6#%\"tG" }{TEXT -1 20 ", over the interval " }
{XPPEDIT 18 0 "[0, T];" "6#7$\"\"!%\"TG" }{TEXT -1 5 ", is " }
{XPPEDIT 18 0 "V/2;" "6#*&%\"VG\"\"\"\"\"#!\"\"" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
257 9 "Solution." }{TEXT -1 63 "  First, notice that we are asked to f
ind the average value of " }{TEXT 258 8 "velocity" }{TEXT -1 2 ", " }
{XPPEDIT 18 0 "v = gt;" "6#/%\"vG%#gtG" }{TEXT -1 9 ".  Since " }
{XPPEDIT 18 0 "V;" "6#%\"VG" }{TEXT -1 25 " is the velocity at time " 
}{XPPEDIT 18 0 "T;" "6#%\"TG" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" 
{XPPEDIT 18 0 "V = gT;" "6#/%\"VG%#gTG" }{TEXT -1 65 ".  The average v
elocity, with respect to time, over the interval " }{XPPEDIT 18 0 "[0,
 T];" "6#7$\"\"!%\"TG" }{TEXT -1 4 ", is" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 
"(1/(T - 0))*int(g*t, t=0..T);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&
%\"TG\"\"\"%\"gGF&#F&\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "and
 this is indeed " }{XPPEDIT 18 0 "V/2;" "6#*&%\"VG\"\"\"\"\"#!\"\"" }
{TEXT -1 3 ".  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "(b). Define " 
}{XPPEDIT 18 0 "S = g*T^2/2;" "6#/%\"SG*(%\"gG\"\"\"*$%\"TG\"\"#F'F*!
\"\"" }{TEXT -1 27 ", the displacement at time " }{XPPEDIT 18 0 "T;" "
6#%\"TG" }{TEXT -1 53 ".  Write the velocity as a function of displace
ment, " }{XPPEDIT 18 0 "s;" "6#%\"sG" }{TEXT -1 75 ", and show that th
e average velocity, with respect to s, over the interval " }{XPPEDIT 
18 0 "[0, S];" "6#7$\"\"!%\"SG" }{TEXT -1 5 ", is " }{XPPEDIT 18 0 "2*
V/3;" "6#*(\"\"#\"\"\"%\"VGF%\"\"$!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 259 9 "Solut
ion." }{TEXT -1 59 "  In part (a), all quantities were written as func
tions of " }{XPPEDIT 18 0 "t;" "6#%\"tG" }{TEXT -1 47 ".  Now we have \+
to write everything in terms of " }{XPPEDIT 18 0 "s;" "6#%\"sG" }
{TEXT -1 23 ", through the relation " }{XPPEDIT 18 0 "s = gt^2/2;" "6#
/%\"sG*&%#gtG\"\"#F'!\"\"" }{TEXT -1 15 ".  Solving for " }{XPPEDIT 
18 0 "t;" "6#%\"tG" }{TEXT -1 9 ", we get " }{XPPEDIT 18 0 "t = sqrt(2
*s/g);" "6#/%\"tG-%%sqrtG6#*(\"\"#\"\"\"%\"sGF*%\"gG!\"\"" }{TEXT -1 
19 ", so (for example) " }{XPPEDIT 18 0 "v = gt;" "6#/%\"vG%#gtG" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "sqrt(2*g*s);" "6#-%%sqrtG6#*(\"\"#\"
\"\"%\"gGF(%\"sGF(" }{TEXT -1 36 ".  Now let's compute the average of \+
" }{XPPEDIT 18 0 "v;" "6#%\"vG" }{TEXT -1 30 ", considered as a functi
on of " }{XPPEDIT 18 0 "s;" "6#%\"sG" }{TEXT -1 20 ", over the interva
l " }{XPPEDIT 18 0 "[0, S];" "6#7$\"\"!%\"SG" }{TEXT -1 1 ":" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "S := g*T^2/2;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"SG,$*&%\"gG\"\"\")%\"TG\"\"#F(#F(F+" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "w := (1/(S - 0))*int(sqrt(2*
g*s), s=0..S);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"wG,$*$-%%sqrtG6#
*&)%\"gG\"\"#\"\"\")%\"TGF-F.F.#F-\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "assume(g>0); assume(T>0);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 12 "simplify(w);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*
&%#g|irG\"\"\"%#T|irGF&#\"\"#\"\"$" }}}}}{MARK "2 1 0" 0 }{VIEWOPTS 1 
1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }