I realise now, in all the excitement of the FTC that I hadn’t written a post about the definite integral…that’s shocking! ok, here we go…the plan for this post:
- 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).…
OK, get ready for some Calculus-Fu!
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 ?…
OK, now we come onto the part of the FTC that you are going to use most. We are finally going to show the direct link between the definite integral and the antiderivative. I know that you’ve been holding your breaths until this moment. Get ready to breath a sign of relief:
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.
and we know from the FTC part 1 how to take derivatives of this. It’s just . This says that is an antiderivative of .…
So, we are now ready to prove the FTC part 1. We’re going to follow the proof in Stewart and add in some discussion as we go along to motivate what we are doing. What we are going to prove is that:
for when is continuous on .
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.…
OK, so up to now we can’t actually use the FTC (Fundamental Theorem of Calculus) to calculate any areas. That will come from the FTC part 2.
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:
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 .…
We’ve seen some intriguing things in this course so far, and we’ve developed some clever tricks, from how to find the gradient of just about any function we can throw at you, to proving statements to be true for an infinite number of cases.
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.…
I’ve been getting a lot of questions about absolute values, and so I thought I would try and clarify things here as much as possible. I’ll give some basic definitions and intuition, and then go through some examples, from easier to harder.
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.…
You have a corridor which has an L-shape in it. The corridor looks like this:
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.…
Prove that for every positive integer , is divisible by 64.
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:
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