good afternoon
so today we'll cover section 
5.2, definitive integral 
so we'll start from a very important definition 
we'll define the definitive integral
let's assume that we have a function F
defined on an interval 
AB, okay, then we'll take 
this interval and form a partition 
that is with sin points x_1, x_2
x_3 and so on 
x now coincides with A and x 
n coincides with b, okay 
then so make a partition x_0, x_1, 
x_n, then we'll pick 
sample points. Pick sampling points 
which we'll call x_1*, x_2*,  
x_n*. Basically 
inside each sub-interval 
the cycle x_1*, x_2*
and so on. remember these points 
they could be either the leftmost point of this subinterval, the rightmost point of this subinterval 
or somewhere in between. Okay, 
then we say that 
define 
an integral 
from a to b 
f(x) dx to be equal to 
the limit as n goes to infinity 
of the following sum. It goes from 1 to n 
f(x_1)* 
delta x. So here delta x 
normally if they're equal length it's b-a 
over n. That's the length of individual intervals. 
okay, so this 
definition is valid if
the limit exists
and is
the same for all
choices 
of sampling points. So for instance it could happen 
if I choose the sampling points to be 
the left side of each interval, the left hand rule, and I take the limit, I get a 
number
then I choose them to be the right side of each interval 
and if the limit is different, then this definition doesn't make sense 
so it only makes sense if the limit exists, and if it's the same for any 
choice
of all of the sampling points 
so I'm going to start by making an introduction 
so let's look at this new symbol and
talk about its components. First of all, 
this guy here is the integral sign 
 
and it looks like an elongated s 
why s? Because it's equal to a sum 
so it's the limit of a sum. So the symbol s comes from that word. 
and and b 
are the limits of integration 
limits of integration 
this heref
is called the integrand 
and finally 
this differential here 
tells me what is my independent variable
okay my horizontal axis 
is called X in this case and this
variable has to match, you know, the variable 
upon which the function f depends. 
We'll call this variable the silent variable of integration 
so let me show you. So 
the limit from a to b 
f(x)dx
actually can be written as the integral
of 
f(y)dy. Or I could insert any lette r
and the answer won't change. Why is that? 
well if you look look at the right hand
side of this definition
okay and if you take
f to be x^2
remember, we actually did this exercise last time 
we said let's consider 
the parabola f=x^2
and we actually calculated this sum. What was it equal to? 
1/3, right? So we got a number. Does this number depend on x? 
no, it's a number. It's not a function. So whatever you get as an answer here 
does not depend on x 
so x cannot appear in the final answer. So you can replace it by anything else 
the answer won't change. So we'll call it a silent 
variable
of integration 
so x or y or anything else 
questions? 
question? student: Yeah, I'm a little bit confused 
most of the time we plug in x as the independent variable and 
not y, so in the case of the right hand side 
are we plugging in a y value of a and a y value of b? Instructor: y is just a variable. 
I can call it alpha. I can call it... 
t. Absolutely, yeah. So you're right, usually the convention is that usually 
when you draw a graph, this is y, this is x 
so not to get confused, we don't often use y there 
but we might. We can. 
Okay. So finally 
this guy here also has a name 
this is called the Reinmann sum 
and we'll practice to evaluate such sums 
today. Okay? 
so what is the meaning of this new object? 
let me do it here 
so if
the function of x is non-negative 
on the interval [ab]
then the integral of 
f(x)dx what is the meaning of that? 
do you remember? Exactly 
because we already saw this right hand side. We only did it for non-negative 
functions
and we said, "that's the area."  Remember, we split it 
into a number of vertical 
rectangles. We calculated the area . So this is 
the area under 
the curve f
f(x) between 
a and b. So it has a very clear meaning
for non negative f
okay now what happens if f
changes sign so for instance we can
consider
this kind of function 
see this takes both positive and
negative values
so if you look at the definition you can
see we can do the Reinmann sum 
we can do these, so first we have these rectangles 
and we have these rectangles, okay, 
and if you look at the formula they all
have the same base
they all have different height, and the height of these is actually negative 
so we can do these separately. We can group these together, and we get the area 
area under this
part of the curve, and then this here 
gives me also an area, but it has a negative sign because 
each of the the components here has a
negative
height. So if we call this area 
A1 and this area A2
then the integral of 
f(x)dx is equal to A1 minus 
A2. This is called the net area 
or the signed area
net area
or signed
area
so normally we will assume that the function f
that appears under the integral is either continuous
or it has a finite number of
discontinuities
okay 
and next I'm going to show you an example 
 of using this definition to calculate
the integral. And this example 
is such that we will know the answer 
before we did any calculations
so in this example 
we said 
that f(x) is x-3
we'll say that 
x belongs on the interval from 1 to 4
 
evaluate 
f(x) 
dx by using 
the definition. Before I use the definition I'm going to give you an answer 
so this function
is a straight line
and 
this is what allows me to give you an
answer
or you are going to give me an answer, okay? 
so let's plot f(x) which is 
x-3 as a function of x. At 0,
we'll get -3. It crosses the horizontal axis 
at 3, okay, 
its value at 4 is equal to 1, it's value at 
1 is given at 2 
so
we're interested in 
the integral of this function
between 1 and 4. Okay? 
so we talked about that, it actually does change sign, 
so to calculate it simply, 
we can calculate this area and you can
calculate this area 
and remember that this area comes with a
negative sign, right? 
so let's call this A1
the positive area, let's call this A2
and let's calculate them from geometric considerations. This is easier 
what's the area of this object? It's a triangle 
1/2. It has a base of 1, a height of 1 
so it's 1x1 over 2
is 1/2. This one- what's the area of this? 
so
2. so this is 2
times 2 over 2 
is 2. So what's the answer? 
we have A1 minus
A2 is 1/2 minus 2
is -3/2. So whatever 
magic we're doing here, we're expecting to
get this answer
so let's keep that in mind 
and perform the calculations. We're going to use this definition
 in order to use the definitions
I have to know delta x, 
x_i*, and f of that amount. So let's do it step by step 
the easiest one is delta x 
we have to go b
minus A over n, so it's 4
minus 1 over n is 3
over n. Questions? 
now we have to create a partition so
A is 1, B 
is 4. What is
x1? so I'm going to have x1, x2
x3 and so on 
so each of these intervals have length 3/n
3/n
3/n. So what is the value
of x1? 
it's to the right of 1, the distance is 3/n, so it's one 
plus 3/n, right?
now what's the value of x2? 
it's one plus twice that thing, so 
2*(3/n)
this guy is 1 + 
3 times 3/n
So we can see the pattern. The pattern is that x_i
equals 1+ so for each i 
I have to step i times. So I have i
times delta x, 3
over n. Questions? 
so these are the values of x1 x2 x3
these are not quite what I need, I need sampling points 
and this is really my choice. I am going to use the right hand rule 
use the right
hand rule. What does it mean? 
that if I have  an interval xi-1
xi, I'm going to take my sampling point
to coincide with the rightmost border of that interval 
so xi*
is going to be equal to xi which is 1 plus 
plus i3/n
the last step before I can write down the definition is to evaluate the 
function
at this point f(xi*)
so I go here f(xi*) 
f(xi*) what's my function 
it's 3-- it's x-3. So I have to take 
the argument minus 3
and that's equal to 1+3i
over n minus 3, and that's equal to 
3i/n-2
okay so I evaluated my function
at all the sampling points 
and now I'm ready to use a definition
please ask me questions at any time
so we write down from 1 to 4
(x-3)dx
this is the integral we're trying to evaluate. This is the limit 
as n goes to infinity of the Reimann sum
and under the sum I have 
f(xi*), that's the expression I got. 3i/n
minus 2
times delta x. That's 3
over n. So again, this is f
f(xi*) and this is 
delta x
so now I have concrete numbers in the formulas 
and I have to evaluate first the sum
and then the limit, okay
so I continue to write limit
so to evaluate this sum 
this is a constant. It doesn't depend on i
so this guy can actually come out here
so I have 3/n and here I have a sum 
of 3i/n
minus 2
okay. The next step 
I'm going to split this in to two sums.
I'm going to have a sum of 3i
over n minus... like this 
next I notice that
in this sum I have i times a constant, 3/n. So 
this constant. This constant can come out
so I have
3/n
the sum of i, and this I can rewrite as 2
the sum of one 
questions? So I simplify that
as much as I could. And now I have 
two sums, and both of them we're going to 
evaluate. The first sum 
so let's do it here 
what does this stand for? This stands for 
one plus two plus three plus four, and so on
all the way to n, that is called an arithmatic progression 
and we have a formula for it, right? 
you don't have to remember this formula, but you must have seen it 
before in some class or another. Okay
and this one, I want you to tell me what this one is equal to 
So either one or n. So what do we do here? 
this is a machine. We have to take it and add it for each i 
i goes from 1 to n, so we are going to add this thing together 
n times. So we'll go 1 +1 +1 +1 +1
n times. So this is n
questions? So I evaluated both of these objects. So this 
n(n+1)
over 2, this is n 
so I continue here
limit as n goes to infinity 
I have to collect everything I have here
so 3/n
3 over
n, n(n+1) 
over 2 minus 2 
n. Okay. 
so something is going to simplify. I am going to 
divide through by n. One second
so 3
this n cancels, and I'm dividing by n 
so I have 1+1/n over 2
minus 2
okay, dividing through by n
all the terms. And now I can take the limit 
I do this in order to take the limit. The limit is easy 
because 1/n goes to zero as n becomes large 
so this is 0
so I have 3 times 3 halves 
-2 so this is the answer I get
and we'll go here. See? 
It's the same thing. So you can see two methods, right? This is 
done very quickly this is long and
painful
but we could only do this because these were triangles 
right? This new definition
eventually will give us a tool to evaluate the area under  
curves. Not just straight lines. Question? 
student: how did you divide everything by adding it? 
So here, let's write this 
I have 3, this n cancels
so I have 3, and I divide by n 
here. n+1 over n 
2, right? And 
n from here, 
I get this. 
more questions?
okay
very good 
so 
it is actually very rare when we can complete this calculation 
and get the exact numerical answer
again, this was possible because it's such  a simple function 
so sometimes it make sense to
approximate
let's not calculate this...the limit 
but we can get an approximate answer 
Okay. So in order to do this, we introduce 
the midpoint rule
so simply put 
this rule allows me to approximate this
integral
by a finite sum
both
of f of xi bar 
delta x, sorry, delta x. 
What's xi bar? So let's 
draw one of the intervals 
from xi-1 to xi. Remember the right point rule 
we take our sampling point here. Left point rule, we take it here. 
midpoint rule, we use the middle. So 
so we say that xi*
simply equals xi bar 
where it's the middle point. xi bar 
xi
okay
 
is it me? no okay
fine. 
So this is the midpoint, and thus the name 
was a question? okay. 
was that music? 
so this is the midpoint and let you
let me give you an example of how we can apply this rule to approximate 
a particular integral. So approximate the integral from 0 to 1
x^2 dx
with n = 4
by using
the midpoint 
rule. So let's draw 
every time they ask you to evaluate 
the Reinmann sum. The best thing to do is to draw your interval 
 actually right down the partition
find the sampling points. Do it systematically, and then it's very easy 
so we go from 0 to 1
they specify
n = 4. So this means that we divide it in to 4
subintervals. This is 1/2
one-quarter, three quarters. Four 
intervals. Now we have to find the sampling point 
Which has to be the middle point, because we are using the midpoint rule 
so this one here, x1 bar 
what is it equal to? 1 over 8
okay, x2 bar 
is 3 over 8
x3 bar 
5 over 8
and x4 bar is 7 over 8
questions? 
So we know our points, just put it here. 
we have to evaluate the function at these four points. And the function is x^2
so f(x) 
is x^2. So 
f(xi) is just xi^2
so we can say that the integral from 0
to 1
x^2dx is approximately equal to 
okay, each term here
will contain the multiplier delta X
so I'm going to take it out of the bracket 
and I forgot to calculate delta x. So 
what's delta x here? 
what is the distance between my 
points in the partition? One quarter, right? 
n=4, the total interval is one, so it's one quarter. 
so I have one quarter that comes out
and here I have the sum
f1-- oh, f
x1, x2
x3, 
x4. So one quarter 
of one
over eight squared, 3 over 8
squared, 5 over 8
squared, seven over 8
squared. So if you 
put it in your calculators, maybe one of you can do it here, if you have a 
calculator
we're going to evaluate this number and see how good 
this approximation is to this integral 
what is the exact value of this integral? 1/3.
We did it last time, right? So this is going to be 
something like 0.32
I believe something like this. So it's not going to be precisely 
the exact value, but it's going to approximate it 
and if we take a higher n, we will get a better answer
okay. Questions? Question? 
Student: "the 1/8, 3/8, 5/8, where do they come from?"
Where do they come from? So these are these values  
Okay. These are the sampling points for each interval 
I have to take the middle point and evaluate my function 
at that point so if I use different
color
inside each interval 
I calculated the middle 
of that interval. So my first interval goes from 0 to one quarter 
I only have four intervals, this, this, this and this. 
for the first interval, the middle corresponds to 1/8
I have to evaluate my function at that point 
the function is x^2, so the function evaluated at that point 
is 1/8 squared. That's the first 
term. Then I have to take the second inerval
its middle is given by 3/8, the function 
at that point is given by 3/8 squared
and so on. I add them up. More questions? 
So next let's go over some 
properties of definite integrals 
so the first property I write down here 
the integral from 
a to b f(x)
dx. The limits of integration, a to b
can be...can be swapped 
and we actually can go from b to a 
so if a is less than b, we can go this way 
or we can go the other way 
but the corresponding integral 
acquires a negative sign. Why is that? Because when we go
in the positive direction, the increment, delta x, 
is a positive quantity. When we go back, 
the value decreases .So the increment is negative. Okay? 
so if you look at the Reimann sum 
you end up multiplying by a negative value, delta x.  
so we have a minus sign here. Something that immediately follows from this 
is the following: the integral
of f(x)dx from a to a-- what's that equal to?
0. Two reasons for this: the first reason
comes from this formula if I take them
both to be a
then this is equal to minus the same thing, so this can only be zero  
the other reason
is that we're considering a the area
under a curve
of an infinitely thin 
base. So it has to be zero, right? 
okay
so then let's list
a bunch of properties that will help us evaluate
very many integrals and Riemann sums, too 
so property number 1 states
the integral
of a constant and it's equal to this 
so let's suppose that my function
is the constant, c
which can be negative or positive 
so then this object on the left
its either the area under this curve
if this is a positive constant or it's the area above that curve 
with a minus sign, so it's the same thing, and this is a rectangle  
so the area is equal to b-a
times C. and that's what we have here 
on the right hand side. Okay? So it's very easy 
to evaluate the definite integral of a
constant because it comes from this
geometric 
consideration. Questions? 
Okay. The next one talks about 
the integral of a sum 
of two functions. So here, both f and g
are functions of x. Question? Student: just to make sure, I can't see. Is that an equal sign? 
here yes. Student: So that only works
it works if-- oh, I see. nevermind. It works for both positive 
and negative values of C and also for C=0
So if I have a sum of two functions, then I can evaluate 
the integral separately. It's fdx
plus gdx
gdx. How do I know this? 
this is very easy to prove 
all I need to do is recall the definition
okay
so this is the definition of the integral f(x) 
so I can replace f with f+g
then I'll have f+g here 
so I can split this sum in to two sums 
one was f, the other one was g. Then I can evaluate  
the limits separately. The limit for the f sum plus the limit for the g sum
and that's equal to these two things 
Okay? So I'm sorry, I shouldn't forget
to put the limits of integration. Because what I just wrote actually as a very
different meaning. That's the antiderivative, that's not what I'm talking about here 
Okay. Property number 3: the integral of a constant
times a function
is given by the same constant
times the integral over the function
Proof? You go here again. If we have c times f here 
then we have c times f here. c comes out of the sum
c comes out of the limit, in order to have a c
times this which is given
by the integral times c
okay, and finally property number 
4 I'm not sure why it's listed as a
separate property in the book
but i's the integral
of the difference between two functions 
it actually follows from properties 2&3
but I'll write it down nonetheless
so the integral of a difference is the difference between the two integrals 
and next we apply all these rules 
to calculate an example
so let's look at the integral from 0 to 1
5+2
x^2dx
so one way to approach it will be to
do it by definition. Write down the limit of a sum
of this thing, create a partition 
calculate sampling points and then do the sums and then the limit 
it's very, very tedious. We're going to do the same thing 
by using the rules so we notice
that the integrand is a sum of two functions 
so by rule number two I'm going to
present it as the sum
of two integrals. 
So this is rule number 2. I can
evaluated separately. Okay, now 
the first one, we have to look at rule number 1
which is a constant, right? 
we have five here so
we have five
times 1 minus 0. And that's by rule number one 
so I dealt with this one nicely. 
Now this one we have to look at rule number 3.
we have a constant inside, 2
comes out. So we have 2 
so this
is rule number 3. So what's this equal to? 
I already asked you like five times today. 
1/3, right? 
so we're going to use this fact to write down the answer
Okay? Very easy. 
So if we can decompose it somehow
and follow different rules, we can 
evaluate integrals easily. Questions? 
now rule 
property number five 
let's suppose that I have are
an interval from a to b 
and I look
at an integral of fdx--
of x dx, and I 
define another point, C
which may or may not be between a and b 
actually, it could be anywhere else. Then I can rewrite this integral
as the sum of the integral from a to c
plus an integral from c
to b. 
I will illustrate this with a picture 
let's suppose for simplicity that f is positive 
and the quantity on the left is 
has the meaning of the integral under the curve. Question? Student:  "just want to make sure we're on the same page. C is
between a and b?" It doesn't have to be. So in this example it is  
but in general it doesn't have to be. It's just another constant. 
So... and you can prove it, of course. Here we illustrated it with an example that 
is more obvious
on the left we have the area under this curve. 
the whole thing. Now on the right, we have two integrals 
which is this area and this area
you add up the two, and 
you get the full thing. Okay? So this is the
the meaning of this property as an
example for instance we can say that
we can write this down 
as the integral from 0 to 1
1 to five. So this is an example 
Okay. Question? 
Student: can you draw an example where c is not between a and b? 
Yes. So let's consider the following geometry 
a, b, c
so
we have to assume
in this case that f is actually defined 
for the whole interval. Okay, so if f only exists here, 
we cannot do this. but if c is in the domain
of f, then we can potentially 
define c to not belong to this interval. 
so, so this is my function f
okay
so I have an integral from a to b 
fdx is equal to 
from a to c fdx
from c to b dx.
Now let's simplify this. So first of all, 
this first interval 
I'm going to use this rule and flip the limits 
so I get
abfdx
is minus from c
to a fdx
now the second integral spans 
the interval from c to b. okay? 
so I'm going to split this one in to this plus this 
so instead of this integral I'm going to write 
from c to a fdx
plus from a to b
fdx. Right? 
now you can see that this
the first and the second terms cancel
out and I get that identity
the left hand side is equal to the right hand side 
so it's a little bit more... less obvious
but it also works
more questions? Okay.
Finally, the last 
group of properties is called comparative properties 
they are
numbered from 6 to 8, So let's suppose that f 
is non-negative
then the integral from a to b f(x) 
dx is non-negative. This is obvious
because for non-negative fs
the integral is nothing but the area under the curve, and that's a non-
quantity. Okay? So we already know this. 
fine
number seven let's suppose that we have
two functions
and one of them is bigger than the other 
on the same, you know 
interval. Then we can compare 
the integral of f was the integral 
of g. The function is bigger 
and then the integral is bigger too. Let me illustrate this 
with an example
Let's suppose that both functions are 
non-negative, okay? For simplicity. So this is the larger of the functions, f
and this is g
okay? Between a and b
so the integral of f is this whole area 
and the integral of g is part of it 
right so you can see that the area under
the function g is smaller than the area
under the function f, 
and that's exactly what this says. Okay? 
this only works if f is greater than g. If they intersect somewhere, 
then we cannot make this conclusion
finally
the last property
property 8
Is like this. Let's suppose that we have two numbers 
m little and M capital 
Which work like this. so this is my function from a to b
and there's some value, m
and another one, M capital 
such that they enclose this function. One is always smaller 
than my function, and the other one is always larger 
for all values of x on the interval. Okay? Let's suppose I know 
that these two numbers exist.
make an estimate of the integral of f
okay? I know what it's bigger than and I know what it's smaller than 
what are these things? So the
integral of f is this area 
now look at this  box here
this is completely contained inside
this bigger area, and the area of this box 
is m times b-a. 
well I can see that the it
into love F youth gonna the smaller
done that this box on the other hand
but those are the books this on the
aerial the box
you buy em capital you minus a in a
completely contains
then the growth
so I can have there's bound the upper
bound on the bill
so a I will say this was a an example
and then I let you go
game
example like this
fifth find the %uh
so the integral
from 0 to 1/4
0 he to the Linus affects squared yes
so at this stage we have no idea
what this alias but got but I can read
it down a double in the quality
that defines what it is so
let's blow this function it some taking
functional X
day because you're a and one a
when ethical 0 itself is equal to 1
in 1x chemical 21 it's now the is
E to the -1 Pratt so for sure
I can write down the bowling double
inequality
he to the -1 is almost
less or equal to you to the minus
excluded and thus always less or equal
to what
so this function that lot
is bounded between if -1 which is
something like one quarter
and one someone to use proper 28
to a evaluate
I the bounds for this and I'll
on the left I'm gonna have you to the -1
times 110
and on the right they have one so
to conclude well I know
that this integral is between one
already
and one and this is much more
information
than I get by just looking at this rate
thank you much
