## UCT MAM1000 lecture notes part 10

### Volumes

Having understood how to calculate the areas between two curves, or simply the area under a curve as a limit of the Riemann sum we can start to think about how to approximate not areas but volumes. If we can calculate an area by adding together small rectangles, perhaps we can calculate volumes by adding together small boxes, or other shapes. In fact we will use not boxes but cylinders, though perhaps not cylinders as we normally think of them.

You  probably think of a cylinder as a tube with a circular cross-section, but in fact that is a particular type of cylinder called a circular cylinder. The general idea of a cylinder is a three dimensional object with a constant cross section.…

## On shifting the integration region in an improper integral

I gave a rather unclear explanation of the following in a lecture a couple of days ago and promised to (at least attempt to) clarify a bit.

You know that: $\int_0^{10}\frac{1}{x} dx$

is divergent and gives an answer of $\infty$. If you have an integral of the form: $\int_3^{10}\frac{1}{x-3}$

as we had in class, you don’t need to calculate this separately if you already know about the behaviour of $1/x$ about $x=0$ as they are exactly the same thing, just shifted (at least their convergence properties are the same).

The take-home message is that if you have: $\int_a^b \frac{c}{(x-a)^p}dx$

and you already know the behaviour of: $\int_0^d \frac{1}{x^p}dx$

then the integral you are interested in will have the same convergence or divergence behaviour.

 How clear is this post?

## UCT MAM1000 Lecture notes part 9

### Areas between curves

We know how to calculate the area between a curve and the horizontal ( $x$) axis of a graph. We have learnt a number of very sophisticated techniques which allow us to get an analytic answer.

It is a fairly simple extension to study not the area between a curve and the horizontal axis, but between two curves. You can see here the link between the two: We can see in the figure below how we can approximate the area by a sum of rectangles. In this case we have used the left point approximation. Each rectangle here has $\Delta x=0.02$ and the heights are $f(x_i)-g(x_i)$ where the $x_i$ are the left points of each rectangle, in this case $=0.02(i-1)$. So $x_1=0$, $x_2=0.02$, etc. We started off studying integration by taking the limit of a Riemann sum. We can do exactly the same thing here, taking the simplest possible Riemann sum, with rectangular regions which stretch between the two curves (see the diagrams on page 448 in Stewart) and then taking the limit to find an expression in terms of integrals.…

## Paradoxes of infinity and some more bonus material on prime numbers

The bonus material from yesterday:

Perfect numbers are those numbers which are equal to the sum of their divisors (other than themselves). For instance, 6 is divisible by 1, 2 and 3 (and 6, but we don’t include this in what follows). If you add these together, you get 6. 6 is a perfect number. 8 is divisible by 1, 2 and 4. Add these together and you get 7 – so 8 is not a perfect number (actually it is called deficient, the other alternative is an abundant number).

It was shown by Euclid that $2^{p-1}(2^p-1)$ is an even perfect number whenever $2^p-1$ is a prime (actually a prime of this form is called a Mersenne prime). In fact, it was shown by Euler two millennia later that every even perfect number could be written in this form! This is known as the Euclid-Euler theorem. So far there are 48 Mersenne primes known, and therefore 48 even perfect numbers known.…

## 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 7

Today we are going to look at improper integrals. This will give us access to a whole host of integrals which, naively, looking at them in terms of Riemann sums don’t make obvious sense.

When we defined the definite integral we gave some constraints. We can now integrate (either approximately or exactly): $\int_a^bf(x)dx$

as long as $[a,b]$ is finite and as long as there are no infinite discontinuities in $a\le x\le b$. An infinite discontinuity means that $f(x)$ is not bounded at some point in $[a,b]$ (intuitively this means that the function goes to $\pm \infty$ at some value of $x$ in $[a,b]$.

If we have an integral which does not abide by these constraints, we may still be able to calculate an answer for the area under the curve, but it will now be called an improper integral. The reason that these are defined as improper is because they will not themselves be well defined as Riemann sums, however, they will be limits of Riemann sums as we will soon see.…

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

OK, so the last post was a bit abstract, so it’s good to run through some examples here. Here are a few examples which illustrate some of the theory we discussed last time.

Example 1: A quadratic function in the denominator which can be written as the product of two linear terms $\frac{3x-6}{x^2+5x+6}=\frac{3x-6}{(x+2)(x+3)}$

Now split this up into the sum of two terms: $\frac{3x-6}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}$

multiply out the right hand side: $\frac{3x-6}{(x+2)(x+3)}=\frac{A(x+3)+B(x+2)}{(x+2)(x+3)}=\frac{x(A+B)+3A+2B}{(x+2)(x+3)}$

and match the coefficient of $x$ and the constant part to find $A$ and $B$: $A+B=3$ and $3A+2B=-6$. The solution to this is $A=-12$ and $B=15$. Our final expression is: $\frac{-12}{x+2}+\frac{15}{x+3}$

and if we need to integrate this we can use the normal techniques we have learnt up to now.

The quadratic in the following can’t be factored into linear terms in $x$ (well, not using real numbers, anyway). We can complete the square in the quadratic term though: $\frac{3x-6}{(x^2+6x+12)(x-3)(x+2)}=\frac{3x-6}{((x+3)^2+3)(x-3)(x+2)}$

we now write this as: $\frac{3x-6}{((x+3)^2+3)(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}+\frac{Cx+D}{((x+3)^2+3)}$

Note that the quadratic term has not just a constant in the numerator, but a term $Cx+D$.…

## 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.…

## UCT MAM1000 lecture notes part 5

Note that the last example I gave in class today can be done in a much simpler way which doesn’t involve a trig substitution. I will include both the way we did it in class as well as the simpler way below. Thank you to the student who came and pointed this out!

Today we looked at integrals of non-trigonometric functions for which substituting in some trig expression makes the undoable doable. Let’s start with the example: $\int\frac{1}{\sqrt{1-x^2}}dx$

I drew a quick sketch of this function, just because it’s good exercise to do that. You see that the denominator is the expression for a unit upper half semi-circle centred at the origin. Inverting this (ie. $\frac{1}{circle}$ gives a graph which passes through the point $(0,1)$ and asymptotes to infinity at $\pm 1$.

Whenever we see something of the form $1-x^2$ we should be reminded of the trig identity $1-\sin^2\theta=\cos^2\theta$ because this will simplify things.…

## Trig integrals

Trigonometric functions generally tell us about behaviour which happens in a periodic way – the motion of the earth around the sun, the position of a pendulum. This might be related to the (AC) current flowing down an electrical circuit, the change in risk of certain weather patterns throughout the year or the position of a planet in the sky.

Here we’re going to use a combination of simple integrals that we should all know by now and integration by substitution and parts along with trigonometric identities to solve certain integrals.

Note that some of the tricks we use are going to appear to have been pulled out of a hat, and until you become familiar with these methods, that’s unfortunately the case for some time. But spend time looking through them and seeing the structure of the methods and they will become clearer and clearer.

Integrating the $\tan$ function $\int\tan\theta d\theta$

Often it will be best to try and put our integral in the form of $\sin$ and $\cos$.…