A significant amount of focus in statistics is on making inference about the averages or means of phenomena. For example, we might be interested in the average number of goals scored per game by a football team, or the average global temperature or the average cost of a house in a particular area.

The two types of averages that we usually focus on are the sample mean from a set of data and the expectation that comes from a probability distribution. For example if three men weigh 70kg, 80kg, and 90kg respectively then the sample mean of their weight is . Alternatively, we might say that the arrival times of trains are exponentially distributed with parameter we can use the properties of the exponential distribution to find the mean (or expectation). In this case the mean is .

It is this second kind of mean (which we will call the expectation from now on), along with the generalisation of taking the expectation of functions of random variables that we will focus on.…

• Look at our Riemann sums and think about taking a limit of them
• Define the definite integral
• Look at a couple of theorems about the definite integral
• Do an example
• Look at properties of definite integrals

That’s quite a lot, but we are more or less going to follow along with Stewart. Stewart just has a slightly different style to mine, so I recommend reading his for more detail, and mine for potentially a bit more intuition.

So, let’s begin…

We have seen in previous lectures/sections/semesters/lives that we can approximate the area under a curve by splitting it up into rectangular regions. Here are examples of splitting up one function into rectangles (and, in the last way trapezoids, but you don’t have to worry about this).…

We have now said that rather than taking pesky limits of Riemann sums to calculate areas under curves (ie. definite integrals), all we need is to find an antiderivative of the function that we are looking at.

As a reminder, to calculate the definite integral of a continuous function, we have:

where is any antiderivative of

Remember that to calculate the area under the curve of from, let’s say 2 to 5, we had to write:

And at that point we had barely even started because we still had to actually evaluate this sum, which is a hell of a calculation…then we have to calculate the limit. What a pain.

Now, we are told that all we have to do is to find any antiderivative of and we are basically done.

Can we find a function which, when we take its derivative gives us ?…

The Fundamental Theorem of Calculus, Part 2 (also known as the Evaluation Theorem)

If is continuous on then

where is any antiderivative of . Ie any function such that .

————-

This means that, very excitingly, now to calculate the area under the curve of a continuous function we no longer have to do any ghastly Riemann sums. We just have to find an antiderivative!

OK, let’s prove this one straight away.

We’ll define:

and we know from the FTC part 1 how to take derivatives of this. It’s just . This says that is an antiderivative of .…

for when is continuous on .

Proof:

we define and we want to find the derivative of . We will do this by using the fundamental definition of the derivative, so let’s look at calculating this function at and – ie. how much does it change when we change by a little bit?

But remember that the definite integral is just the area, so this difference is the area between a and x+h minus the area between a and x. Which is just the area between x and x+h. Using the properties of integrals, we can write this formally as:

and we can write, for :

Restated, we can think of this as the area between x and x+h divided by h.…

For now, let’s take some examples and see what the FTC is saying. I’ll restate it here:

The Fundamental Theorem of Calculus, part 1

If  is continuous on  then the function  defined by:

,     for

is continuous on  and differentiable on , and .

——

Let’s look at some examples. We’re going to take an example that we can calculate using a Riemann sum. Let’s choose .

If we integrate this from to some point – ie. calculate the area under the curve, we get:

.

Make sure that you can indeed get this by calculating the Riemann sum.

So, what does the FTC part 1 tell us? It says that if we take the derivative of this area, with respect to the upper limit, , then we get back .…

To some extent, this is what we have looked at so far (at least in terms of calculus, and building up to calculus):

However, we’re about to see some magic. We’re about to see the most important thing yet on this course, and indeed one of the most important moments in all of mathematical history.

We are going to see…actually, we are going to prove, that there is a relationship between rates of change and the area under a graph. This doesn’t sound that amazing, but its consequences have essentially allowed for the development of much of modern mathematics over the last 350 years.

The link that we are going to prove will allow us to find the area under graphs of functions for which taking the Riemann sum would be really hard.…

The absolute value function is just….a function. You give it a number, and it returns a number. In the same way that is a function. You give it a number and it returns that number multiplied by itself. So the absolute value function, which we write as takes a number and returns the same number if the number was positive, and the negative of the number if it was negative, thus returning always a positive number.

We can think of this as the function “how far away from the point 0 (the origin) on the real number line is x?”. It doesn’t care about what direction it is, only how far away it is.…

where a and b are the widths of the sections of the corridor. The question is to find the longest pipe that can be carried down this corridor. The word pipe here just means something long and with essentially 0 thickness. There is a huge simplification which is being assumed here, which is that the corridor is only 2 dimensional. Of course in a 3 dimensional corridor we have a lot more room to manoeuvre.

Let’s think about a pipe going round the corner. The longest pipe that can go through is the length of the shortest gap that it has to go through. So let’s think of a pipe at a particular angle with respect to the corner:

The pipe here is the blue line and is the angle that it makes with respect to the corner.…

This question screams proof by induction, so we start with the base case, which in this case is :

which is indeed divisible by 64.

Now, let’s assume that it holds true for some positive integer . ie:

for .

Now let’s see how we can use this to prove that the statement holds true for . For we have:

where we have manipulated the expression to contain the left hand side of the inductive hypothesis. Thereby, plugging in the inductive hypothesis, we get:

but clearly is an integer, so this is divisible by 64 and thus the statement holds true for , thus it holds true for all positive integers