We defined integrals in order to compute areas,
but there are many other applications of integrals.
Integrals can be used to compute many other things.
In this video, I want to revisit the definition of Riemann integral,
and reinterpret the notation we use with a different geometrical flavour.
(And this interpretation will be very useful in other applications.)
Specifically, I will try to use the same idea to set up VOLUMES as integrals.
Let's recall the idea we use to define Riemann integral.
I have a region here whose area I want to compute.
It is the region surrounded by the graph of the function f,
and the x-axis, from x = a to x = b.
I want to compute its area.
To define Riemann integral,
first, we break the interval [a,b] into smaller pieces using a partition;
and this breaks the region into slices.
To make this process useful, I should have many many slices very very thin,
but to make the picture visible, I'm using only six.
And then, I'm going to approximate each one of the slices with a rectangle.
I'm being a bit vague about exactly which rectangle...
Now I'm going to write an equation for the area of all of those rectangles.
I'm going to look at one rectangle first: let's say this one!
I'm going to call the width of this rectangle Δx,
and then the height of this rectangle is f(x), for some value of x.
It's some value of the function f,
and I'm not being very precise about which value...
But therefore, the area of this rectangle is going to be f(x)⋅Δx.
And now if I add over all the pieces (if we put a sum here),
then I get the sum of the areas of the six rectangles.
And that's an approximation for the area of the region I'm interested in!
Now if my function is nice and integrable, perhaps continuous,
then when I take the limit as Δx → 0,
(as those rectangles become thinner and thinner)
this approximation becomes exact,
and that's what we call the value of the definite integral.
I have been very hand-wavy here,
but if we want to recall the precise formal way to do it,
we spent quite a few videos doing that.
What I want to do next is I want to reinterpret this notation,
in a way that is not very precise, but is geometrically very useful.
Since the integral is a limit of sums, I'm going to abuse notation,
and think of the integral as an actual sum of infinitely many pieces,
which are infinitely small.
Or more precisely, when I write "dx",
I'm going to think of dx as a Δx that is very very tiny;
(the technical term is INFINITESIMALLY SMALL)
and then of the integral, as the sum over infinitely many of those pieces.
What do I mean by this?
Here is the same region again.
This time, instead of breaking it into six slices,
I'm going to break it into infinitely many slices,
which are infinitesimally small.
I'm going to draw only one of them.
Now, I drew a piece that looks big, (that looks finite)
because it's the only way to see it,
but we should pretend that this is infinitesimally small.
I'm going to call the width of this piece "dx", to indicate that.
I drew only one, but I'm thinking of all the infinitely many pieces like this...
And I'm now looking at the area of this slice.
Now, since this is infinitesimally small, the area will be as well,
and therefore, I'm going to call its area "dA", because it's also an infinitesimal.
Now in this case, I'm going to say that when I pretend that this is a rectangle,
it's not an approximation, but it's exact... and here is why.
If I calculate the area of this slice by a rectangle,
then the error I'm making is just this piece.
This is the error I'm making if I replace it with a rectangle.
But if this dimension is infinitesimally small,
then the error is infinitesimally small SQUARED,
so by comparison, when this is very very tiny,
it's negligible (it's not important), and I can omit it in the limit.
So this area is equal to the height of the rectangle, which is f(x),
times the width, which is dx.
And now, the area of the full region,
is going to be the "sum" of all these pieces (infinitely many of them),
but that sum is what we call an integral.
Or in other words, it's ∫ₐᵇ(f(x))dx.
This was a bit hand-wavy,
it can be made rigorous (it can be formalized),
but I don't want to do that, for our purposes this is okay.
The advantage of thinking of it this way,
is that it helps with the geometric intuition,
and it helps us reuse the same ideas
for many other applications of the integral, apart from areas.
For example, it is useful to think of integrals this way
to figure out how to use integrals to compute volumes.
This is a carrot.
Let's say I want to compute its volume.
If I don't know what else to try,
maybe I can start by approximating it by a cylinder.
It is definitely not a cylinder, but it's close,
and that will give me a certain error.
But let's say I want something better.
Well, just like we did with areas,
(we took an arbitrary region and sliced it,
and approximated pieces (we used rectangles))
I can also take this volume and slice it, literally...
And now look at what we got:
if I take any one of the slices, well, this is pretty much a disc.
It is not an exact disc, but it's very close.
This is pretty much a disc, and I can approximate it by a disc,
and I know how to compute its volume.
And if it takes some kind of limit,
as the thickness of the slices goes to 0, and they become infinitesimally small,
this will no longer be an approximation,
but it will give me the exact value of the volume.
I'm going to make that specific with a complete algebraic example.
Here is a concrete example of a region whose volume I'm going to calculate.
I'm going to call R the region on the first quadrant,
between the graph of y = x² and the x-axis, from x = 0 to x = 1,
and I'm going to rotate this region around the x-axis.
This creates a solid, and I want to compute the volume of that solid.
At this moment, I invite you to pause the video, draw a picture,
and try to understand what this region looks like.
What does this volume look like?
Once you are comfortable with that,
keep watching and I will explain how to set this up as an integral.
This is the region R, between y = x² and the x-axis, from x = 0 to x = 1,
so this is the region.
And I'm going to rotate it around the x-axis.
When this is rotated around the x-axis, it creates a solid,
and it's going to look a little bit like this:
it's kind of like a cone that has been bent a little bit.
Now I want to compute the volume of this solid,
and to do that I'm going to a slice it into pieces:
infinitely many pieces, infinitesimally small.
I'm going to draw one of the pieces, and to do that,
let me cut a thin slice of the region R, here.
I'm going to say the width is dx,
because I'm pretending it's infinitesimally small.
I'm drawing it as a finite piece because it's the only way to visualize it,
but assume it's microscopic.
And then when you rotate it here, this will correspond to a slice like this one.
Now I want to figure out what the volume of this is slice is.
I'm going to call it "dV", because since the slice is infinitesimally thin,
the volume will also be infinitesimally small.
Now what is this piece? This is pretty much a disc.
It's not exactly a disc when this is finite,
but in the limit when this is infinitesimally small,
I can approximate it by a disc because the error I'm making is negligible.
So the volume of it will be the volume of the disc.
Now the volume of a disc is the surface area times the thickness
(or height in this case).
Now the height in this case is dx,
so the volume of this piece is dx times the surface area;
and if I call this radius R, then the surface area is π⋅R².
Now let's see what R corresponds to in the original picture.
Knowing where this comes from, R is this:
which is equal to y, which is equal to x².
(This is just in this specific case.)
So, I have a way to write the volume of an infinitesimally thin piece,
and then the volume of the whole thing,
will be the "sum" of all those infinitesimally thin pieces,
but that's actually an integral.
Since I am using x as the variable, the integral goes from 0 to 1.
And now the integral of the volume of this piece, which is this equation.
Now instead of R, I'm going to put x² because I'm using x as the variable,
and that's it!
I have succeeded at setting up this volume as an integral.
Now we can go ahead and use antiderivatives to compute it.
I will leave that part as an exercise,
but you can perform that computation and the final answer should be π/5.
To conclude this video, in case you want to practice this method,
here are four examples of solids whose volume you can compute
with exactly the same idea, that I have used in the one example in this video.
The calculations are more complicated but the idea is the same.
They are all carrots that you have to slice infinitesimally thin.