Okay.
So we'll get started with integration.
So the problem that integration was
originally invented to
solve was, I have this function f of x.
So that's the blue curve here, and then I
imagine that there's a region underneath
my function, so it's above the x axis,
between a lower limit a an upper limit b.
And I want to
assign some sort of area to this region s.
So, often you hear
integration just as referred to as the
problem of finding the area under a curve.
And so the, the kind of the concept of
area, it's,
until there was calculus, the only way you
could the find
the area of something was figuring out how
to make it
into a rectangle and then taking the area
of that rectangle.
And so for instance,
if you had two triangles and they were
both right triangels,
you could sort of fit them together and
make a rectangle.
And then you would say, well, the area of
the triangle must be
half of that because there were two of
them that made it up.
Here, it's going to be a little bit
trickier, because it's, even the concept
of area is a little bit less clear when
you have a curved edge like this.
But the approach we’re going to take is
just take a lower bound for this area and
then an upper bound for this area.
So the lower bound is going to be these
red rectangles.
And so, I've, I've made this diagram a
little
bit simpler just by making the curve
upwards sloping everywhere.
So I'm going to evaluate the height of the
rectangle at the left side of each
rectangle.
And that means that each rectangle is
completely below the curve.
And then I'm going to do just the opposite
with the black rectangle.
I'm going to evaluate that at the right
side.
So each black rectangle is completely
above the curve, at
the top of each black rectangle is
completely above the curve.
And then my lower limit for my area
underneath
this curve is going to be the sum of the
areas of the red rectangles, and the upper
limit
is going to be the sum areas of the black
rectangles.
And I'm going to refer to this splitting
up of the region ab as a partition.
And what I want to do then is notice
I have n here as a subscript from my
partition.
So this would be p5, it's a partition into
five sub intervals.
And I can refine my partition just by
breaking this up at another point.
So, for instance, right in the middle
here,
I could add another point to my partition.
And then you're going to see that the, the
red sum
was going to get a little bit bigger,
because I could
fit a little bit more red rectangle in
here.
And the black sum is going to get a little
bit smaller,
because I could take a little bit away
from up here.
And then the idea is to look as I make my
partition finer and finer and finer.
So this p sub n.
And I have n going to infinity.
So my, my upper limit, my upper bound for
this area is coming down as n goes to
infinity.
And the red rectangles, my lower bound is
going up as n goes to infinity.
And so I'm going to
say that if those two limits are equal.
So,
if I, if I let my partition become finer
and finer, and finer in the limit.
If my lower bound is equal to my upper
bound, then I'm going to say
that the definite integral of f of x from
a to b, and I write it using.
So this symbol here that
I can't highlight, is integral from a to b
of f of x the x.
I'm just going to define that to be the
limit of, because these are equal, I can
pick either one, but I picked the, the
lower lower bound, and I'm going to call
that A.
And then the interpretation I get after
doing
all of this is that A can be interpreted
as the area of the region S.
And then I'm also going to say that my
function f of x is integrable.
And this, this condition at the end, or
this definition
means that both of these limits are equal
to one another.
If I were to continuously refine my
partition until and these limits.
If the upper limit stayed above the lower
limit, then f would not be an integrable
function.
So the, the notation, or the, the
definition of integrability
is that these two limits are going to
equal to one another.
And so, just to get used to working with
integrals.
So, I'll go over some basic properties.
So, if I look at the integral from b to a.
So now, I'll I've done is I've switched
the, the endpoint.
So, instead of my limit going from a to b,
I'm now going to go from b to a.
And let's pretend it's like the picture,
so a is less than b.
I'm just going to end up with minus the
integral, the
definite integral from a to b of f of x.
And so in particular, it's sort of like
area, but it's
somehow a little bit different.
Because the area of a rectangle doesn't
really change.
And if I look at it going this way, or if
I look at it going this way, it's the same
area.
But the, the value of an integral's going
to change depending
on if I'm going from left to right or
right to left.
If I integrate a function over an interval
of width
zero, so this is just a to a, the same
point,
the value of that integral is going to be
zero.
So here I let c be a real-valued constant.
If I look at the
integral from a to b of c, that's just
going to be c times b minus a.
And if you imagine what this is going to
look like, c says my
function is constant for all x, so it's
just a, a horizontal line.
And I'm integrating, I'm trying to find
the area underneath that between a and b.
So if b is greater than a, b minus a is
going to
be positive, that's just going to give me
the width of this region.
And then because it has a constant height,
that's just going to be a rectangle.
So all I've done here is just said width
times height is the area of a rectangle.
And then like for the derivative, we also
have a linearity property for integration.
So the integral from a to b of a function
that's equal to two simpler functions
added together, is just the integral of
the
first function plus the integral of the
second function.
Similarly, I can scale it.
So, if I have a integral of a function
f of x times a real valued constant c,
then the answer
to that is just c times the value of the
integral.
And now I'm going to choose a point c that
is
strictly between a and b, so I'm splitting
up the integral ab into two pieces.
And
if I integrate the function f of x over
the first piece and add that
to the integral of f of x over the second
piece, that's going to
be equal to the value of the integral over
the whole integral.
So as you just integral from a to c of f
of x plus the integral from c to
b of of x, is the same thing as the
integral from a to b of f of x.
And then if f of x is greater than or
equal to zero at every point in between a
and b, then the integral of that function
over that
interval has to be greater than or equal
to zero.
And also if f of x is greater than or
equal
to g of x at every point in the interval,
then
the integral of f of x over the interval
is greater
than the integral of g of x over that same
interval.
Oops.
And then this number 9 is essentially the
same thing I did
with my partition, except now I just have
a partition of size one.
All I'm saying is that if f of x is
greater than some constant m over the
entire interval and less than some
constant capital M over that interval.
Then I can make a lower bound and an upper
bound
for the integral by just saying that this
integral, or sorry,
this rectangle that fits entirely
underneath the curve that's going to
have an area smaller than the value of
the, the integral.
And this rectangle where the curve fits
entirely inside the rectangle is
going to have a value larger than the
value of the integral.
Okay.
And so, just a, a quick example of, of how
we can use this are principle to evaluate
an integral.
Suppose I want to evaluate the integral
from 0 to 4 of 4x minus x squared.
So, it's not really clear how you could do
that right from the start.
but I think if I change this function
around a little bit most of you will
recognize it.
And if you don't, I'll tell you what it is
at the end so it makes sense, hopefully.
So the first thing I'm going to do is just
write it backwards.
I want to have the x squared first.
So I'll say, minus x squared minus 4x.
And then I'm going to add and subtract 4.
So really, what I'm going to do is
underneath the
square root, I'm going to add 0 to this
quantity.
But I'm going to pick a very special value
of 0 that I can write the 4 minus 4.
And so, I'm going to put the positive 4
here And then because everything in that
quantity is going to be negative, if I put
the positive four here, that's my 4 minus
4.
But now,
we can recognize this as x minus 2
squared.
So I can rewrite, oops, what have I done?
Maybe I have to keep hitting that.
so now I can recognize this as x minus
2 squared, the integral from 0 to 4.
And I think I have my picture here.
So this is what this function looks like.
And in fact, if you get rid of the square
root here, you think of this
function as y equals f of x, then what
you're going to have is y squared plus
x minus 2 squared equals 2 squared.
Which is just the area of, or the equation
for a circle with center at 2,0.
And so that means that I can use my
area interpretation, because I know how to
compute the
area of half a circle to say that the
value of this integral is equal to 2 pi.
And just in case you forgot, at the bottom
there
I wrote the equation of a circle with
center at x0,y0.