## The Fundamental Theory of Calculus part 2 (part i)

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 $f$ is continuous on $[a,b]$ then

$\int_a^b f(x) dx=F(b)-F(a)$

where $F$ is any antiderivative of $f$. Ie any function such that $F'=f$.

————-

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:

$g(x)=\int_a^x f(t)dt$

and we know from the FTC part 1 how to take derivatives of this. It’s just $g'(x)=f(x)$. This says that $g$ is an antiderivative of $f$.…

## The Fundamental Theorem of Calculus part 1 (part iii)

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:

$\frac{d}{dx} \int_a^x f(t) dt=f(x)$

for $x\in [a,b]$ when $f$ is continuous on $[a,b]$.

Proof:

we define $g(x)=\int_a^x f(t)dt$ and we want to find the derivative of $g$. We will do this by using the fundamental definition of the derivative, so let’s look at calculating this function at $x$ and $x+h$ – ie. how much does it change when we change $x$ by a little bit?

$g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt$

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:

$g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt=\left(\int_a^{x}f(t)+\int_x^{x+h}f(t)\right)-\int_a^{x}f(t)=\int_x^{x+h}f(t)dt$

and we can write, for $h\ne 0$:

$\frac{g(x+h)-g(x)}{h}=\frac{1}{h}\int_x^{x+h}f(t)dt$

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

## The Fundamental Theorem of Calculus part 1 (part ii)

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 $f$ is continuous on $[a,b]$ then the function $g$ defined by:

$g(x)=\int_a^x f(t) dt$,     for $a\le x\le b$

is continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'(x)=f(x)$.

——

Let’s look at some examples. We’re going to take an example that we can calculate using a Riemann sum. Let’s choose $f(x)=x^2$.

If we integrate this from $0$ to some point $x$ – ie. calculate the area under the curve, we get:

$\int_0^x t^2 dt=\frac{x^3}{3}$.

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, $x$, then we get back $f(x)$.…

## The Fundamental Theorem of Calculus, part 1 (part i)

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.…

## All you’ve ever wanted to know about absolute values (and weren’t afraid to ask)

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 $f(x)=x^2$ 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 $f(x)=|x|$ 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.…

## Taking a pipe round a corner corridor optimisation question

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 $\theta$ is the angle that it makes with respect to the corner.…

## Prove that for every positive integer n, 9^n – 8n -1 is divisible by 64.

Prove that for every positive integer $n$, $9^n-8n-1$ is divisible by 64.

This question screams proof by induction, so we start with the base case, which in this case is $n=1$:

$9^1-8-1$ which is indeed divisible by 64.

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

$9^k-8k-1=64p$ for $p\in\mathbb{Z}$.

Now let’s see how we can use this to prove that the statement holds true for $n=k+1$. For $n=k+1$ we have:

$9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(9^k-8k-1)+64k$

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

$9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(64p)+64k=64(9p+k)$

but clearly $9p+k$ is an integer, so this is divisible by 64 and thus the statement holds true for $n=k+1$, thus it holds true for all positive integers $k$

 How clear is this post?

## A tricky complex numbers problem

The question is as follows:

If $\frac{\pi}{6}\in arg(z+a)$ and $\frac{2\pi}{3}\in arg(z-a)$ and $a\in \mathbb{R}$, find $z$.

So, let’s think about the information given and what we are trying to find. We want to find the complex number $z$ which satisfies this slightly strange set of constraints, and the constraints are given in terms of $z$ and $a$. So, by the looks of things, the answer will depend on $a$ and so the final expression should be a function of $a$.

Now let’s explore the constraints. In fact, let’s simply take $z+a$ and $z-a$ as two complex numbers, but importantly two numbers which differ only by a real number $2a$, so wherever they lie in the complex plane, they have the same imaginary part and differ only by an real part.

Now, the constraints are about the arguments of the two complex numbers. It doesn’t tell us anything about the magnitude of the numbers, so all the information tells us is the direction are in relation to the origin.…

## Why did we choose that range for theta when doing trig substitutions?

Remember when we are doing a trig substitution, for instance for an integral with:

$\sqrt{a^2-x^2}$

We said that we should choose $x=a\sin\theta$, which seemed reasonable, but we also said that $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. Where did this last bit come from?

Well, we want a couple of things to hold true. The first is that any substitution that we make, we have to be able to undo. That is, we will substitute $x$ for a function of $\theta$ but in the end we need to convert back to $x$ and so to do that we have to be able to write the inverse function of, in this case $x=a\sin\theta$. The $\sin$ function is itself not invertible because it’s not one to one, so we have to choose a range over which it is one to one. We could choose $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$ or we could choose $\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}$ (amongst an infinite set of possibilities). That would also be invertible. However, remember that we are going to end up with a term of the form:

$\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}$

So if we want this to simplify, we had better choose our range of $\theta$ such that $\cos\theta$ is positive, so that we can write $\sqrt{\cos^2\theta}=\cos\theta$.…

## Integrals with sec and tan when the power of tan is odd

We went through an example in class today which was

$\int tan^6\theta \sec^4\theta d\theta$

In this case we took out two powers of sec and then converted all the other $\sec$ into $latex\ tan$, which left a function of tan times $sec^2\theta d\theta$. We wanted to do this because the derivative of $\tan$ is $\sec^2$ and so we can do a simple substitution. If we have an odd power of $\tan$, we can employ a different trick. Let’s look at:

$I=\int \tan^5\theta\sec^7\theta d\theta$.

Here, sec is an odd power and so we can’t employ the same trick as before. Now we want to convert everything to a function of $\sec$ and have only a factor which is the derivative of $\sec$ left over. The derivative of $\sec$ is $\sec\tan$, so let’s try and take this out:

$I=\int \tan^5\theta\sec^7\theta d\theta=\int \tan^4\theta\sec^6\theta (\sec\theta\tan\theta)d\theta$.

Now convert the $\tan$ into $\sec$ by $\tan^2\theta=\sec^2\theta-1$:

$I=\int (\sec^2\theta-1)^2\sec^6\theta (\sec\theta\tan\theta)d\theta=\int (\sec^{10}\theta-2\sec^8\theta+\sec^6\theta) (\sec\theta\tan\theta)d\theta$

where here we have just expanded out the bracket and multiplied everything out.…