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Okay.
So we'll get started with integration.

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So the problem that integration was
originally invented to

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solve was, I have this function f of x.

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So that's the blue curve here, and then I
imagine that there's a region underneath

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my function, so it's above the x axis,
between a lower limit a an upper limit b.

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And I want to

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assign some sort of area to this region s.
So, often you hear

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integration just as referred to as the
problem of finding the area under a curve.

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And so the, the kind of the concept of
area, it's,

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until there was calculus, the only way you
could the find

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the area of something was figuring out how
to make it

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into a rectangle and then taking the area
of that rectangle.

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And so for instance,

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if you had two triangles and they were
both right triangels,

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you could sort of fit them together and
make a rectangle.

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And then you would say, well, the area of
the triangle must be

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half of that because there were two of
them that made it up.

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Here, it's going to be a little bit
trickier, because it's, even the concept

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of area is a little bit less clear when
you have a curved edge like this.

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But the approach we’re going to take is
just take a lower bound for this area and

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then an upper bound for this area.

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So the lower bound is going to be these
red rectangles.

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And so, I've, I've made this diagram a
little

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bit simpler just by making the curve
upwards sloping everywhere.

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So I'm going to evaluate the height of the

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rectangle at the left side of each
rectangle.

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And that means that each rectangle is
completely below the curve.

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And then I'm going to do just the opposite

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with the black rectangle.

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I'm going to evaluate that at the right
side.

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So each black rectangle is completely
above the curve, at

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the top of each black rectangle is
completely above the curve.

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And then my lower limit for my area
underneath

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this curve is going to be the sum of the

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areas of the red rectangles, and the upper
limit

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is going to be the sum areas of the black
rectangles.

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And I'm going to refer to this splitting
up of the region ab as a partition.

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And what I want to do then is notice

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I have n here as a subscript from my
partition.

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So this would be p5, it's a partition into
five sub intervals.

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And I can refine my partition just by
breaking this up at another point.

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So, for instance, right in the middle
here,

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I could add another point to my partition.

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And then you're going to see that the, the
red sum

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was going to get a little bit bigger,
because I could

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fit a little bit more red rectangle in
here.

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And the black sum is going to get a little
bit smaller,

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because I could take a little bit away
from up here.

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And then the idea is to look as I make my
partition finer and finer and finer.

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So this p sub n.

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And I have n going to infinity.

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So my, my upper limit, my upper bound for

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this area is coming down as n goes to
infinity.

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And the red rectangles, my lower bound is
going up as n goes to infinity.

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And so I'm going to

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say that if those two limits are equal.
So,

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if I, if I let my partition become finer
and finer, and finer in the limit.

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If my lower bound is equal to my upper
bound, then I'm going to say

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that the definite integral of f of x from
a to b, and I write it using.

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So this symbol here that

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I can't highlight, is integral from a to b
of f of x the x.

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I'm just going to define that to be the
limit of, because these are equal, I can

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pick either one, but I picked the, the

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lower lower bound, and I'm going to call
that A.

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And then the interpretation I get after
doing

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all of this is that A can be interpreted

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as the area of the region S.

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And then I'm also going to say that my
function f of x is integrable.

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And this, this condition at the end, or
this definition

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means that both of these limits are equal
to one another.

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If I were to continuously refine my
partition until and these limits.

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If the upper limit stayed above the lower

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limit, then f would not be an integrable
function.

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So the, the notation, or the, the
definition of integrability

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is that these two limits are going to
equal to one another.

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And so, just to get used to working with
integrals.

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So, I'll go over some basic properties.

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So, if I look at the integral from b to a.

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So now, I'll I've done is I've switched
the, the endpoint.

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So, instead of my limit going from a to b,
I'm now going to go from b to a.

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And let's pretend it's like the picture,
so a is less than b.

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I'm just going to end up with minus the
integral, the

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definite integral from a to b of f of x.

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And so in particular, it's sort of like
area, but it's

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somehow a little bit different.

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Because the area of a rectangle doesn't
really change.

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And if I look at it going this way, or if

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I look at it going this way, it's the same
area.

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But the, the value of an integral's going
to change depending

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on if I'm going from left to right or
right to left.

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If I integrate a function over an interval
of width

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zero, so this is just a to a, the same
point,

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the value of that integral is going to be
zero.

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So here I let c be a real-valued constant.
If I look at the

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integral from a to b of c, that's just
going to be c times b minus a.

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And if you imagine what this is going to
look like, c says my

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function is constant for all x, so it's
just a, a horizontal line.

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And I'm integrating, I'm trying to find
the area underneath that between a and b.

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So if b is greater than a, b minus a is
going to

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be positive, that's just going to give me
the width of this region.

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And then because it has a constant height,
that's just going to be a rectangle.

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So all I've done here is just said width
times height is the area of a rectangle.

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And then like for the derivative, we also
have a linearity property for integration.

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So the integral from a to b of a function
that's equal to two simpler functions

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added together, is just the integral of
the

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first function plus the integral of the
second function.

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Similarly, I can scale it.
So, if I have a integral of a function

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f of x times a real valued constant c,
then the answer

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to that is just c times the value of the
integral.

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And now I'm going to choose a point c that
is

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strictly between a and b, so I'm splitting

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up the integral ab into two pieces.
And

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if I integrate the function f of x over
the first piece and add that

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to the integral of f of x over the second
piece, that's going to

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be equal to the value of the integral over
the whole integral.

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So as you just integral from a to c of f
of x plus the integral from c to

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b of of x, is the same thing as the
integral from a to b of f of x.

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And then if f of x is greater than or
equal to zero at every point in between a

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and b, then the integral of that function
over that

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interval has to be greater than or equal
to zero.

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And also if f of x is greater than or
equal

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to g of x at every point in the interval,
then

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the integral of f of x over the interval
is greater

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than the integral of g of x over that same
interval.

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Oops.

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And then this number 9 is essentially the
same thing I did

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with my partition, except now I just have
a partition of size one.

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All I'm saying is that if f of x is
greater than some constant m over the

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entire interval and less than some
constant capital M over that interval.

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Then I can make a lower bound and an upper
bound

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for the integral by just saying that this
integral, or sorry,

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this rectangle that fits entirely
underneath the curve that's going to

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have an area smaller than the value of
the, the integral.

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And this rectangle where the curve fits
entirely inside the rectangle is

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going to have a value larger than the
value of the integral.

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Okay.

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And so, just a, a quick example of, of how

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we can use this are principle to evaluate
an integral.

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Suppose I want to evaluate the integral
from 0 to 4 of 4x minus x squared.

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So, it's not really clear how you could do
that right from the start.

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but I think if I change this function
around a little bit most of you will

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recognize it.

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And if you don't, I'll tell you what it is
at the end so it makes sense, hopefully.

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So the first thing I'm going to do is just
write it backwards.

145
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I want to have the x squared first.
So I'll say, minus x squared minus 4x.

146
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And then I'm going to add and subtract 4.

147
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So really, what I'm going to do is
underneath the

148
00:10:16,870 --> 00:10:20,048
square root, I'm going to add 0 to this
quantity.

149
00:10:20,048 --> 00:10:24,300
But I'm going to pick a very special value
of 0 that I can write the 4 minus 4.

150
00:10:24,300 --> 00:10:31,000
And so, I'm going to put the positive 4
here And then because everything in that

151
00:10:31,000 --> 00:10:33,376
quantity is going to be negative, if I put

152
00:10:33,376 --> 00:10:36,370
the positive four here, that's my 4 minus
4.

153
00:10:36,370 --> 00:10:37,394
But now,

154
00:10:37,394 --> 00:10:42,130
we can recognize this as x minus 2
squared.

155
00:10:44,300 --> 00:10:46,530
So I can rewrite, oops, what have I done?

156
00:10:48,940 --> 00:10:55,528
Maybe I have to keep hitting that.
so now I can recognize this as x minus

157
00:10:55,528 --> 00:11:01,880
2 squared, the integral from 0 to 4.
And I think I have my picture here.

158
00:11:01,880 --> 00:11:04,980
So this is what this function looks like.

159
00:11:04,980 --> 00:11:09,255
And in fact, if you get rid of the square
root here, you think of this

160
00:11:09,255 --> 00:11:13,980
function as y equals f of x, then what
you're going to have is y squared plus

161
00:11:13,980 --> 00:11:16,530
x minus 2 squared equals 2 squared.

162
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Which is just the area of, or the equation
for a circle with center at 2,0.

163
00:11:26,810 --> 00:11:28,727
And so that means that I can use my

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area interpretation, because I know how to
compute the

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area of half a circle to say that the
value of this integral is equal to 2 pi.

166
00:11:40,528 --> 00:11:44,268
And just in case you forgot, at the bottom
there

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I wrote the equation of a circle with
center at x0,y0.