## UCT MAM1000 lecture notes part 8

Last time we looked at integrals which weren’t proper integrals because the limits of integration were infinite (either on one side, or both). This was one of the constraints we had on a well defined Riemann sum. The other constraint we had was that there were no infinite discontinuities in the integrand. Here we will show that sometimes we can indeed define an improper integral which does include such a discontinuity within the limits of integration.

Improper Integrals of the second kind: Infinite discontinuities in the integrand

We have seen what happens when we integrate $x$ from $\infty$ or to $-\infty$. Sometimes it gives a convergent value and we can find the integral, sometimes the improper integral does not converge and it gives us an answer of $\pm\infty$. Indeed sometimes it doesn’t blow up, but it just doesn’t converge – as in the integral of $\cos x$ over an infinite range.

Now we are going to look at what happens if the function itself has an infinite discontinuity in it and we want to integrate over this region, or include it at one of the limits of integration.…

## UCT MAM1000 lecture notes part 6 (part i)

Some thoughts on exam preparation

It feels like the exams are an age away, but believe me, the coming weeks will fly by and before you know it the exams will be upon you. There are things that you can do now, which are going to make the period before the exams, and the exams themselves be a lot less stressful than they would be otherwise.

Part of the issue with MAM1000 is that it’s a big course, which covers a lot of material. It will have been many months since you studied limits by the time you get to the exam, so keeping your hand in with all the topics covered will make life much easier for you.

Of course you need to do the tutorials each week, but I recommend, starting now, going through a couple of old tuts each week. It shouldn’t take too long as you have done them before, and the more times you do them, the faster they will go.…

The first rule of bootcamp is: You DO talk about bootcamp!

I know that the start of the semester is busy, but now is the time to consolidate what you learnt last semester if you feel that you are struggling. It will take a good number of hours each week for a couple of weeks, but with effort you can do it. It’s really useful to go through these with other people, but it’s also important to spend some time going through exercises on your own. I suggest finding a balance between the two.

I’m going to list some links below, some of which have explanations, some of which have videos and many of which have exercises. Here’s your task if you choose to accept it:

1. Put aside Whatsapp, Facebook, news feeds, anything distracting and make sure that you are in a quiet place – if possible take the material offline so you don’t have to have any internet connection when you are going through these.

## UCT MAM1000 lecture notes 1 (part iii) – integration by parts

Here we are going to come up with our second method for solving integrals which we can’t solve by inspection (noting that they are the antiderivative of some function which we know well).

We know the product rule for differentiation:

$\frac{d (f(x)g(x))}{dx}=f'(x)g(x)+f(x)g'(x)$

Integrating it gives us:

$f(x)g(x)+c=\int f'(x)g(x)dx+\int f(x)g'(x)dx$

We can then rearrange this to give:

$\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$

We have dropped the $c$ because this will come also from doing the second integral.

This is also sometimes written in an alternative form:

Let $f(x)=u$ and $g(x)=v$. Sometimes we will keep the functional dependence explicit, and sometimes we won’t – it is up to you to follow what things are functions of $x$ and what are constants. You will get more and more familiar with it over time.

Now we can write:

$f'(x)=\frac{du}{dx}\,\,\,\,\,\,\,\,g'(x)=\frac{dv}{dx}$

Now put this into the equation above for integration by parts.

$\int u \frac{dv}{dx} dx=uv-\int v \frac{du}{dx} dx$

cancelling the factors of $dx$:

$\int u dv=uv-\int v du$

(Actually the idea of cancelling the factors of $dx$ is rather sloppy, but here it goes through ok).…

## UCT MAM1000 lecture notes 1 (part ii) – integration by substitution – review.

This is just a quick reminder. If you find any of this confusing, there is a very important trick for making it easier – practice, practice, practice! It doesn’t take years to master this, it takes a few hours every week for a few weeks. You will become more and more familiar with the techniques and learn intuitively to know which technique to use in which situation.

If $F(x)=x^2$, what is its derivative? ie. how do we find a function $f(x)$ such that:

$f(x)=\frac{d (F(x))}{dx}$

The answer of course is $f(x)=2x$. We use our normal differentiation rules which you should now be very familiar with. How about if I told you that there was some function $F(x)$ whose derivative was $2x$ – ie. we reversed the question:

What is a function whose gradient at a point $x$ is $2x$?

How do you find what $F(x)$ is? We are trying to solve the equation

$2x=\frac{d F(x)}{dx}$

for $F(x)$.…

## UCT MAM1000 lecture notes 1 (part i) – preamble.

I will attempt to post notes for the coming sixty lectures on a daily basis. You can either ask questions about the topics which you don’t understand here, or email me directly, or of course come and chat in my office when I’m around.

• These notes are for the second semester of MAM1000. They are neither complete nor exact and no responsibility is held for the accuracy within. Mistakes are undoubtedly included. That being said, I hope that they can be a useful resource in addition to the course textbook (Stewart) and additional online materials.
• I am always very grateful when people find mistakes in these notes. These may be in the form of spelling, grammar, calculational errors, typos in formulae, typesetting errors and anything else which doesn’t seem to make sense. If an explanation is not clear, please contact me and I will do my best to explain it in another way.