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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 10 "Calculus I" }}{PARA 257 "
" 0 "" {TEXT -1 46 "Lesson 26: The Fundamental Theorem of Calculus" }}
{PARA 256 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
255 "We are going to continue the connection between the area problem \+
and antidifferentiation. First we extend the area problem and the idea
 of using approximating rectangles for a continuous function which is \+
not necessarily positive over the interval [a,b]. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "restart; wi
th(student):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "f:=1+5*x;" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "plot(f,x=-2..1);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 407 "As we can see, the fucntion is no
t positive over the entire interval and if we \"owned the land between
 the graph and the x axis\" we would be thinking of the area of two tr
iangles one below the x axis from -2 to -.2; the other above the axis \+
from -.2 to 1. the first has area 1.8 *9*.5 and the other 1.2*6*.5 for
 a total of 11.7 square units. What happens when we use rightbox and s
um for n=100 in this case?" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "rightbox(f(x),x=-2..1,100); \+
rightsum(f(x),x=-2..1,100); evalf(%);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 599 "We see that the answer is nowhere close to 11.7 and is i
n fact negative. Looking at the approximating rectangles shows us that
 in the sum when we take the height to be f(x*) for some x* in the sub
interval , this value will be negative when x is between -2 and .2. So
 we guess that this process will be yielding the negative of the area \+
of the first triangle + the area of the second triangle. That is 1.8*(
-9)*.5 + 1.2*(4).5= -8.1 + 3.6=-4.5. This answer seems more consistent
 with the -4.275 above. This is considered a \"Net Area\" ie Area abov
e x axis - Area belowbelow x axis. We switch to n=1000." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "rightbox(f(x),x=-2..1,1000); rights
um(f(x),x=-2..1,1000); evalf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
38 "This seems to support the observation." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 256 179 "If g is conti
nuous (or monotone) on [a,b] then this approximating rectangle sum pro
cess will have a limit which we denote as the definite integral of g o
ver [a,b] the notation is " }{XPPEDIT 257 0 "Int(g,x = a .. b);" "6#-%
$IntG6$%\"gG/%\"xG;%\"aG%\"bG" }{TEXT 258 182 " and this definite inte
gral can be interpreted as the difference of areas where g is above th
e x axis and below the x axis. We say that g is an integrable function
 when this happens." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 29 "We try this for our function " }{XPPEDIT 18 0 "f (x)=1+
5*x" "6#/-%\"fG6#%\"xG,&\"\"\"F)*&\"\"&F)F'F)F)" }{TEXT -1 1 " " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "int(f,x=-2..1);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 186 "This is the answer -4.5 that we were exp
ecting. We also use the term ...\"integrating f from -2 to 1\" in desc
ribing this process of evaluating the definite integral. We can now de
fine an " }{TEXT 259 19 "indefinite integral" }{TEXT 260 0 "" }{TEXT 
-1 96 " which is a new function, based on this process of integrating \+
continuous functions g, such as :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 " G:=x-> Int( g(t),t=-2..x);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 124 " This is defining a new func
tion G for each x, which says to find its value you integrate g over t
he interval from -2 to x. " }{TEXT 261 124 "The interesting fact is th
at this G is itself an antiderivative for g. That is G is differentiab
le and its derivative is g. " }{TEXT 262 0 "" }{TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 125 "This is one part of the Fundamental theorem of
 Calculus. This yields a valuable tool in evaluating these definite in
tegrals. " }}{PARA 0 "" 0 "" {TEXT -1 39 " This says that over [a,b]  \+
G(b)-G(a) =" }{XPPEDIT 256 0 "Int(g,x = a .. b)" "6#-%$IntG6$%\"gG/%\"
xG;%\"aG%\"bG" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 321 
" This equation says that to find the definite integral, first we iden
tify an antiderivative of g over [a, b] then simply evaluate that anti
derivative at the two endpoints and subtract. Because of this theorem \+
we can identify the process of finding antiderivatives and finding ind
efinite integrals. We check this with our " }{XPPEDIT 18 0 "f (x)=1+5*
x" "6#/-%\"fG6#%\"xG,&\"\"\"F)*&\"\"&F)F'F)F)" }{TEXT -1 36 " we know \+
an antiderivative would be " }{XPPEDIT 18 0 "x+5*x^2/2;" "6#,&%\"xG\"
\"\"*(\"\"&F%*$F$\"\"#F%F)!\"\"F%" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 12 "F:=int(f,x);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 61 " You should note that here we have used the integral comm
and " }{TEXT 263 8 "int(f,x)" }{TEXT -1 89 " without specifying the in
terval to get this antiderivative and we also gave it a name F." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 51 "subs(x=-2,F); subs(x=1,F);subs(x=1,F)-subs(x=-2,F);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 121 "Here we see that we get the same answer \+
as we expected -4.5. in our previous worksheet we found the area under
 the curve " }{XPPEDIT 18 0 "x^2-4*x+5" "6#,(*$%\"xG\"\"#\"\"\"*&\"\"%
F'F%F'!\"\"\"\"&F'" }{TEXT -1 97 " over the interval [0,3] to be 6. We
 check the process by integrating this expression over [0,3]." }}
{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 22 "int(x^2-4*x+5,x=0..3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "
We further check it by finding an antiderivative for the expression an
d applying the fundamental theorem." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 70 "F:=int(x^2-4*x+5,x); subs(x=0,F); subs(x=3,F);subs(x=
3,F)-subs(x=0,F);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "This shows \+
the antiderivative and its values at 0 and 3 and then substracts givin
g the same answer of 6. " }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 
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