Integration by Parts
Integration By Substitution
Last year I played the 2/3 numbers game, also called the Keynesian Beauty Contest with my first year maths class. The discussion can be found here: http://www.mathemafrica.org/?p=11143 I wanted to know if, telling my class the results from last year (including sketching for them the histogram of results), would change how they chose their numbers this year. Of course I can’t tell if it changed them, but what is fascinating is that either:
- Their guesses (if I didn’t tell them about the results from last year) would have been very different from those last year, or:
- They were completely unaffected by knowing what people did last year, which really means that they believed that the rest of the class would have been unaffected.
I plot here the results from the last three years and you can see how similar the results are, year on year.
One of the homework questions this week was the following:
a) Sketch the region R and find its area.
b) Find the volume of the solid obtained by rotating the region R around the y-axis.
The first thing to do is to sketch the graphs of and . Once you’ve done that, the other parts of the inequalities should be clear. It should look like the red region in the following plot:
Now we have to imagine bringing out a third axis, perpendicular to the picture above, ie. coming out towards us. We then want to rotate the red form here about the vertical axis. This we can imagine doing in the following animation:
Given this form we can then think about either taking horizontal cross-sections through it, which will give us thin annuli, or we can take vertical, circular slices to give us thin shells. Adding these together and integrating should give us the same answer whichever way we choose to slice it, but one way will be considerably easier.…
I gave a challenge question at the end of class a week or so ago. Here I will give the solution and show that it gives us something rather strange and surprisingly useful.
I wrote down the following, and asked you to prove it:
For . Now, N! can be thought of as the number of different orderings of pulling N objects out of a bag (without replacement) when they are all different. If you have N things in a bag, then there are N possible things that you can pull out first. There are then N-1 ways of pulling out the next object, N-2 ways of pulling out the next, etc. and finally, when you’ve pulled out N-1 objects there’s only a single possibility of pulling out the last. So:
And the number of ways of pulling no objects out of a bag is 1, because you just don’t pull anything out.…
Out of the blue I wrote down a rather confusing mass of indices and summations on the board a few days ago. Writing this down at the last minute was perhaps a bad idea, but I wanted to show what the general form for expanding a fraction into partial fractions was. Here I’m just motivating it a little more. It’s not something that you will need to use, but it’s often good to write things down in as general a form as you can.
Let’s say that we have an expression of the form:
Where P(x) is some polynomial of degree less than 3 (because the denominator is degree 3). We can write this as:
To find A, B and C, you cross-multiply, and then match coefficients of powers of x with those in P(x). If you have an irreducible quadratic in the denominator you will have terms of the form:
in your partial fraction, and of course if it’s an irreducible quadratic to an integer power greater than one, you will have multiple terms, just as you do for the (x-2) expression in the example above.…
If you want to understand maths, you really have to do it. I recommend going through these examples and using the substitutions given here as hints. Get a blank piece of paper, put your notes away and try to do these examples and see if you get the same answers as in class. If you don’t, write in the comments, and we can see where things may have gone astray.
I’ve been teaching integration by substitution, including by trig substitutions over the last few days, and a frequent question which a newbie substituter will ask is “how did you know to make that substitution?”. It’s a very reasonable question, and one that takes practice to build the correct intuition, but I’ll do my best to give some motivation now as to why we made some of the substitutions we made. We won’t solve the integrals, but we will motivate here why we make particular choices for substitutions.…
The following has been a rather interesting journey – from a test question which seemed fine, to a subtlety which seemed easy, to a discussion with a number of different mathematicians about the nature of distributions, measure theory and regularisation. I will try and make it as clear as possible in the post below. Note, as mentioned in the comments, we have actually only found the solution to this problem for a constrained range of x, and not x . I didn’t want to complicate things any more than necessary here for first year students, but the comments are very important too.
In a recent class test, there was a question, written by me, which was not quite the question that I wanted to ask. It turns out that it does have an answer, but it’s not an answer that can yet be found by the means at the class’s disposal.…
How would you go about finding the value of if you didn’t have a square root button on your calculator? Well, the most obvious thing might be to try some values, based on your knowledge of the square root function. You are being asked to find that x for which:
or, in other words, that x, which, when squared gives 3. We have to be a little careful here because we know that there will actually be two numbers which satisfy this (one positive, one negative), and we are interested in the positive one only.
So, we try some values, but we don’t do it randomly, we can see that because and that whatever number squared gives 3 must be between 1 and 2. We can try something called a binary intersection. This just means taking the values which we know bound the right answer (ie. we know that 1<x<2), and trying the number in the middle.…
The chain rule is a method for differentiating a function of a function, or differentiating composite functions.
Consider the expression . We notice that this is not a normal sine function. It has an as argument for the sine function. Therefore, we can consider the in the sine function as a whole different function. This can be broken into two functions, and .
If we consider , we can write .
In order to differentiate a composite function, , i.e; to find , we let
What this implies is that whatever is, u will be equal to that. Then, the process of differentiating is to find
- (as u will be a function of x).
Finally, we can write,
Back to our example, ; remember that and . We let
This leaves us with . We can simplify the equation by writing
Note that the u in the cosine function is replaced with .…