## Math 100: Week 4 Saturday, Sep 28 2013

This is a post for my math 100 calculus class of fall 2013. In this post, I give the 4th week’s recitation worksheet (no solutions yet – I’m still writing them up). More pertinently, we will also go over the most recent quiz and common mistakes. Trig substitution, it turns out, is not so easy.

Before we hop into the details, I’d like to encourage you all to avail of each other, your professor, your ta, and the MRC in preparation for the first midterm (next week!).

1. The quiz

There were two versions of the quiz this week, but they were very similar. Both asked about a particular trig substitution

$\displaystyle \int_3^6 \sqrt{36 - x^2} \mathrm{d} x$

And the other was

$\displaystyle \int_{2\sqrt 2}^4 \sqrt{16 - x^2} \mathrm{d}x.$

They are very similar, so I’m only going to go over one of them. I’ll go over the first one. We know we are to use trig substitution. I see two ways to proceed: either draw a reference triangle (which I recommend), or think through the Pythagorean trig identities until you find the one that works here (which I don’t recommend).

We see a ${\sqrt{36 - x^2}}$, and this is hard to deal with. Let’s draw a right triangle that has ${\sqrt{36 - x^2}}$ as a side. I’ve drawn one below. (Not fancy, but I need a better light).

In this picture, note that ${\sin \theta = \frac{x}{6}}$, or that ${x = 6 \sin \theta}$, and that ${\sqrt{36 - x^2} = 6 \cos \theta}$. If we substitute ${x = 6 \sin \theta}$ in our integral, this means that we can replace our ${\sqrt{36 - x^2}}$ with ${6 \cos \theta}$. But this is a substitution, so we need to think about ${\mathrm{d} x}$ too. Here, ${x = 6 \sin \theta}$ means that ${\mathrm{d}x = 6 \cos \theta}$.

Some people used the wrong trig substitution, meaning they used ${x = \tan \theta}$ or ${x = \sec \theta}$, and got stuck. It’s okay to get stuck, but if you notice that something isn’t working, it’s better to try something else than to stare at the paper for 10 minutes. Other people use ${x = 6 \cos \theta}$, which is perfectly doable and parallel to what I write below.

Another common error was people forgetting about the ${\mathrm{d}x}$ term entirely. But it’s important!.

Substituting these into our integral gives

$\displaystyle \int_{?}^{??} 36 \cos^2 (\theta) \mathrm{d}\theta,$

where I have included question marks for the limits because, as after most substitutions, they are different. You have a choice: you might go on and put everything back in terms of ${x}$ before you give your numerical answer; or you might find the new limits now.

It’s not correct to continue writing down the old limits. The variable has changed, and we really don’t want ${\theta}$ to go from ${3}$ to ${6}$.

If you were to find the new limits, then you need to consider: if ${x=3}$ and ${\frac{x}{6} = \sin \theta}$, then we want a ${\theta}$ such that ${\sin \theta = \frac{3}{6}= \frac{1}{2}}$, so we might use ${\theta = \pi/6}$. Similarly, when ${x = 6}$, we want ${\theta}$ such that ${\sin \theta = 1}$, like ${\theta = \pi/2}$. Note that these were two arcsine calculations, which we would have to do even if we waited until after we put everything back in terms of ${x}$ to evaluate.

Some people left their answers in terms of these arcsines. As far as mistakes go, this isn’t a very serious one. But this is the sort of simplification that is expected of you on exams, quizzes, and homeworks. In particular, if something can be written in a much simpler way through the unit circle, then you should do it if you have the time.

So we could rewrite our integral as

$\displaystyle \int_{\pi/6}^{\pi/2} 36 \cos^2 (\theta) \mathrm{d}\theta.$

How do we integrate ${\cos^2 \theta}$? We need to make use of the identity ${\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}}$. You should know this identity for this midterm. Now we have

$\displaystyle 36 \int_{\pi/6}^{\pi/2}\left(\frac{1}{2} + \frac{\cos 2 \theta}{2}\right) \mathrm{d}\theta = 18 \int_{\pi/6}^{\pi/2}\mathrm{d}\theta + 18 \int_{\pi/6}^{\pi/2}\cos 2\theta \mathrm{d}\theta.$

The first integral is extremely simple and yields ${6\pi}$ The second integral has antiderivative ${\dfrac{\sin 2 \theta}{2}}$ (Don’t forget the ${2}$ on bottom!), and we have to evaluate ${\big[9 \sin 2 \theta \big]_{\pi/6}^{\pi/2}}$, which gives ${-\dfrac{9 \sqrt 3}{2}}$. You should know the unit circle sufficiently well to evaluate this for your midterm.

And so the final answer is ${6 \pi - \dfrac{9 \sqrt 2}{2} \approx 11.0553}$. (You don’t need to be able to do that approximation).

Let’s go back a moment and suppose you didn’t re\”{e}valuate the limits once you substituted in ${\theta}$. Then, following the same steps as above, you’d be left with

$\displaystyle 18 \int_{?}^{??}\mathrm{d}\theta + 18 \int_{?}^{??}\cos 2\theta \mathrm{d}\theta = \left[ 18 \theta \right]_?^{??} + \left[ 9 \sin 2 \theta \right]_?^{??}.$

Since ${\frac{x}{6} = \sin \theta}$, we know that ${\theta = \arcsin (x/6)}$. This is how we evaluate the left integral, and we are left with ${[18 \arcsin(x/6)]_3^6}$. This means we need to know the arcsine of ${1}$ and ${\frac 12}$. These are exactly the same two arcsine computations that I referenced above! Following them again, we get ${6\pi}$ as the answer.

We could do the same for the second part, since ${\sin ( 2 \arcsin (x/6))}$ when ${x = 3}$ is ${\sin (2 \arcsin \frac{1}{2} ) = \sin (2 \cdot \frac{\pi}{6} ) = \frac{\sqrt 3}{2}}$; and when ${x = 6}$ we get ${\sin (2 \arcsin 1) = \sin (2 \cdot \frac{\pi}{2}) = \sin (\pi) = 0}$.

Putting these together, we see that the answer is again ${6\pi - \frac{9\sqrt 3}{2}}$.

Or, throwing yet another option out there, we could do something else (a little bit wittier, maybe?). We have this ${\sin 2\theta}$ term to deal with. You might recall that ${\sin 2 \theta = 2 \sin \theta \cos \theta}$, the so-called double-angle identity.

Then ${9 \sin 2\theta = 18 \sin \theta \cos \theta}$. Going back to our reference triangle, we know that ${\cos \theta = \dfrac{\sqrt{36 - x^2}}{6}}$ and that ${\sin \theta = \dfrac{x}{6}}$. Putting these together,

$\displaystyle 9 \sin 2 \theta = \dfrac{ x\sqrt{36 - x^2} }{2}.$

When ${x=6}$, this is ${0}$. When ${x = 3}$, we have ${\dfrac{ 3\sqrt {27}}{2} = \dfrac{9\sqrt 3}{2}}$.

And fortunately, we get the same answer again at the end of the day. (phew).

2. The worksheet

Finally, here is the worksheet for the day. I’m working on their solutions, and I’ll have that up by late this evening (sorry for the delay).

Ending tidbits – when I was last a TA, I tried to see what were the good predictors of final grade. Some things weren’t very surprising – there is a large correlation between exam scores and final grade. Some things were a bit surprising – low homework scores correlated well with low final grade, but high homework scores didn’t really have a strong correlation with final grade at all; attendance also correlated weakly. But one thing that really stuck with me was the first midterm grade vs final grade in class: it was really strong. For a bit more on that, I refer you to my final post from my Math 90 posts.

## An intuitive introduction to calculus Saturday, Sep 7 2013

This is a post written for my fall 2013 Math 100 class but largely intended for anyone with knowledge of what a function is and a desire to know what calculus is all about. Calculus is made out to be the pinnacle of the high school math curriculum, and correspondingly is thought to be very hard. But the difficulty is bloated, blown out of proportion. In fact, the ideas behind calculus are approachable and even intuitive if thought about in the right way.

Many people managed to stumble across the page before I’d finished all the graphics. I’m sorry, but they’re all done now! I was having trouble interpreting how WordPress was going to handle my gif files – it turns out that they automagically resize them if you don’t make them of the correct size, which makes them not display. It took me a bit to realize this. I’d like to mention that this actually started as a 90 minute talk I had with my wife over coffee, so perhaps an alternate title would be “Learning calculus in 2 hours over a cup of coffee.”

So read on if you would like to understand what calculus is, or if you’re looking for a refresher of the concepts from a first semester in calculus (like for Math 100 students at Brown), or if you’re looking for a bird’s eye view of AP Calc AB subject material.

1. An intuitive and semicomplete introduction to calculus

We will think of a function ${f(\cdot)}$ as something that takes an input ${x}$ and gives out another number, which we’ll denote by ${f(x)}$. We know functions like ${f(x) = x^2 + 1}$, which means that if I give in a number ${x}$ then the function returns the number ${f(x) = x^2 + 1}$. So I put in ${1}$, I get ${1^2 + 1 = 2}$, i.e. ${f(1) = 2}$. Primary and secondary school overly conditions students to think of functions in terms of a formula or equation. The important thing to remember is that a function is really just something that gives an output when given an input, and if the same input is given later then the function spits the same output out. As an aside, I should mention that the most common problem I’ve seen in my teaching and tutoring is a fundamental misunderstanding of functions and their graphs

For a function that takes in and spits out numbers, we can associate a graph. A graph is a two-dimensional representation of our function, where by convention the input is put on the horizontal axis and the output is put on the vertical axis. Each axis is numbered, and in this way we can identify any point in the graph by its coordinates, i.e. its horizontal and vertical position. A graph of a function ${f(x)}$ includes a point ${(x,y)}$ if ${y = f(x)}$.

The graph of the function $x^2 + 1$ is in blue. The emphasized point appears on the graph because it is of the form $(x, f(x))$. In particular, this point is $(1, 2)$.

Thus each point on the graph is really of the form ${(x, f(x))}$. A large portion of algebra I and II is devoted to being able to draw graphs for a variety of functions. And if you think about it, graphs contain a huge amount of information. Graphing ${f(x)= x^2 + 1}$ involves drawing an upwards-facing parabola, which really represents an infinite number of points. That’s pretty intense, but it’s not what I want to focus on here.

1.1. Generalizing slope – introducing the derivative

You might recall the idea of the ‘slope’ of a line. A line has a constant ratio of how much the ${y}$ value changes for a specific change in ${x}$, which we call the slope (people always seem to remember rise over run). In particular, if a line passes through the points ${(x_1, y_1)}$ and ${(x_2, y_2)}$, then its slope will be the vertical change ${y_2 - y_1}$ divided by the horizontal change ${x_2 - x_1}$, or ${\dfrac{y_2 - y_1}{x_2 - x_1}}$.

The graph of a line appears in blue. The two points $(0,1)$ and $(1,3)$ are shown on the line. The horizontal red line shows the horizontal change. The vertical red line shows the vertical change. The ‘slope’ of the blue line is the length of the vertical red line divided by the length of the horizontal red line.

So if the line is given by an equation ${f(x) = \text{something}}$, then the slope from two inputs ${x_1}$ and ${x_2}$ is ${\dfrac{f(x_2) - f(x_1)}{x_2 - x_1}}$. As an aside, for those that remember things like the ‘standard equation’ ${y = mx + b}$ or ‘point-slope’ ${(y - y_0) = m(x - x_0)}$ but who have never thought or been taught where these come from: the claim that lines are the curves of constant slope is saying that for any choice of ${(x_1, y_1)}$ on the line, we expect ${\dfrac{y_2 - y_1}{x_2 - x_1} = m}$ a constant, which I denote by ${m}$ for no particularly good reason other than the fact that some textbook author long ago did such a thing. Since we’re allowing ourselves to choose any ${(x_1, y_1)}$, we might drop the subscripts – since they usually mean a constant – and rearrange our equation to give ${y_2 - y = m(x_2 - x)}$, which is what has been so unkindly drilled into students’ heads as the ‘point-slope form.’ This is why lines have a point-slope form, and a reason that it comes up so much is that it comes so naturally from the defining characteristic of a line, i.e. constant slope.

But one cannot speak of the ‘slope’ of a parabola.

The parabola $f(x) = x^2 + 1$ is shows in blue. Slope is a measure of how much the function $f(x)$ changes when $x$ is changed. Some tangent lines to the parabola are shown in red. The slope of each line seems like it should be the ‘slope’ of the parabola when the line touches the parabola, but these slopes are different.

Intuitively, we look at our parabola ${x^2 + 1}$ and see that the ‘slope,’ or an estimate of how much the function ${f(x)}$ changes with a change in ${x}$, seems to be changing depending on what ${x}$ values we choose. (This should make sense – if it didn’t change, and had constant slope, then it would be a line). The first major goal of calculus is to come up with an idea of a ‘slope’ for non-linear functions. I should add that we already know a sort of ‘instantaneous rate of change’ of a nonlinear function. When we’re in a car and we’re driving somewhere, we’re usually speeding up or slowing down, and our pace isn’t usually linear. Yet our speedometer still manages to say how fast we’re going, which is an immediate rate of change. So if we had a function ${p(t)}$ that gave us our position at a time ${t}$, then the slope would give us our velocity (change in position per change in time) at a moment. So without knowing it, we’re familiar with a generalized slope already. Now in our parabola, we don’t expect a constant slope, so we want to associate a ‘slope’ to each input ${x}$. In other words, we want to be able to understand how rapidly the function ${f(x)}$ is changing at each ${x}$, analogous to how the slope ${m}$ of a line ${g(x) = mx + b}$ tells us that if we change our input by an amount ${h}$ then our output value will change by ${mh}$.

How does calculus do that? The idea is to get closer and closer approximations. Suppose we want to find the ‘slope’ of our parabola at the point ${x = 1}$. Let’s get an approximate answer. The slope of the line coming from inputs ${x = 1}$ and ${x = 2}$ is a (poor) approximation. In particular, since we’re working with ${f(x) = x^2 + 1}$, we have that ${f(2) = 5}$ and ${f(1) = 2}$, so that the ‘approximate slope’ from ${x = 1}$ and ${x = 2}$ is ${\frac{5 - 2}{2 - 1} = 3}$. But looking at the graph,

The parabola $x^2 + 1$ is shown in blue, and the line going through the points $(1,2)$ and $(2,5)$ is shown. The line immediately goes above and crosses the parabola, so it seems like this line is rising faster (changing faster) than the parabola. It’s too steep, and the slope is too high to reflect the ‘slope’ of the parabola at the indicated point.

we see that it feels like this slope is too large. So let’s get closer. Suppose we use inputs ${x = 1}$ and ${x = 1.5}$. We get that the approximate slope is ${\frac{3.25 - 2}{1.5 - 1} = 2.5}$. If we were to graph it, this would also feel too large. So we can keep choosing smaller and smaller changes, like using ${x = 1}$ and ${x = 1.1}$, or ${x = 1}$ and ${x = 1.01}$, and so on. This next graphic contains these approximations, with chosen points getting closer and closer to ${1}$.

The parabola $x^2 + 1$ is shown in blue. Two points are chosen on the parabola and the line between them is drawn in red. As the points get closer to each other, the red line indicates the rate of growth of the parabola at the point $(1,2)$ better and better. So the slope of the red lines seems to be getting closer to the ‘slope’ of the parabola at $(1,2)$.

Let’s look a little closer at the values we’re getting for our slopes when we use ${1}$ and ${2, 1.5, 1.1, 1.01, 1.001}$ as our inputs. We get

$\displaystyle \begin{array}{c|c} \text{second input} & \text{approx. slope} \\ \hline 2 & 3 \\ 1.5 & 2.5 \\ 1.1 & 2.1 \\ 1.01 & 2.01 \\ 1.001 & 2.001 \end{array}$

It looks like the approximate slopes are approaching ${2}$. What if we plot the graph with a line of slope ${2}$ going through the point ${(1,2)}$?

The parabola $x^2 + 1$ is shown in blue. The line in red has slope $2$ and goes through the point $(1,2)$. We got this line by continuing the successive approximations done above. It looks like it accurately indicates the ‘slope’ of the parabola at $(1,2)$.

It looks great! Let’s zoom in a whole lot.

When we zoom in, the blue parabola looks almost like a line, and the red line looks almost like the parabola! This is why we are measuring the ‘slope’ of the parabola in this fashion – when we zoom in, it looks more and more like a line, and we are getting the slope of that line.

That looks really close! In fact, what I’ve been allowing as the natural feeling slope, or local rate of change, is really the line tangent to the graph of our function at the point ${(1, f(1))}$. In a calculus class, you’ll spend a bit of time making sense of what it means for the approximate slopes to ‘approach’ ${2}$. This is called a ‘limit,’ and the details are not important to us right now. The important thing is that this let us get an idea of a ‘slope’ at a point on a parabola. It’s not really a slope, because a parabola isn’t a line. So we’ve given it a different name – we call this ‘the derivative.’ So the derivative of ${f(x) = x^2 + 1}$ at ${x = 1}$ is ${2}$, i.e. right around ${x = 1}$ we expect a rate of change of ${2}$, so that we expect ${f(1 + h) - f(1) \approx 2h}$. If you think about it, we’re saying that we can approximate ${f(x) = x^2 + 1}$ near the point ${(1, 2)}$ by the line shown in the graph above: this line passes through ${(1,2)}$ and it’s slope is ${2}$, what we’re calling the slope of ${f(x) = x^2 + 1}$ at ${x = 1}$.

Let’s generalize. We were able to speak of the derivative at one point, but how about other points? The rest of this post is below the ‘more’ tag below.

## A proof from the first sheet (SummerNT) Monday, Jun 24 2013

In class today, we were asked to explain what was wrong with the following proof:

Claim: As $x$ increases, the function

$\displaystyle f(x)=\frac{100x^2+x^2\sin(1/x)+50000}{100x^2}$

approaches (gets arbitrarily close to) 1.

Proof: Look at values of $f(x)$ as $x$ gets larger and larger.

$f(5) \approx 21.002$
$f(10)\approx 6.0010$
$f(25)\approx 1.8004$
$f(50)\approx 1.2002$
$f(100) \approx 1.0501$
$f(500) \approx 1.0020$

These values are clearly getting closer to 1. QED

Of course, this is incorrect. Choosing a couple of numbers and thinking there might be a pattern does not constitute a proof.

But on a related note, these sorts of questions (where you observe a pattern and seek to prove it) can sometimes lead to strongly suspected conjectures, which may or may not be true. Here’s an interesting one (with a good picture over at SpikedMath):

Draw $2$ points on the circumference of a circle, and connect them with a line. How many regions is the circle divided into? (two). Draw another point, and connect it to the previous points with a line. How many regions are there now? Draw another point, connecting to the previous points with lines. How many regions now? Do this once more. Do you see the pattern? You might even begin to formulate a belief as to why it’s true.

But then draw one more point and its lines, and carefully count the number of regions formed in the circle. How many circles now? (It doesn’t fit the obvious pattern).

So we know that the presented proof is incorrect. But lets say we want to know if the statement is true. How can we prove it? Further, we want to prove it without calculus – we are interested in an elementary proof. How should we proceed?

Firstly, we should say something about radians. Recall that at an angle $\theta$ (in radians) on the unit circle, the arc-length subtended by the angle $\theta$ is exactly $\theta$ (in fact, this is the defining attribute of radians). And the value $\sin \theta$ is exactly the height, or rather the $y$ value, of the part of the unit circle at angle $\theta$. It’s annoying to phrase, so we look for clarification at the hastily drawn math below:

The arc length subtended by theta has length theta. The value of sin theta is the length of the vertical line in black.

Note in particular that the arc length is longer than the value of $\sin \theta$, so that $\sin \theta < \theta$. (This relies critically on the fact that the angle is positive). Further, we see that this is always true for small, positive $\theta$. So it will be true that for large, positive $x$, we’ll have $\sin \frac{1}{x} < \frac{1}{x}$. For those of you who know a bit more calculus, you might know that in fact, $\sin(\frac{1}{x}) = \frac{1}{x} - \frac{1}{x^33!} + O(\frac{1}{t^5})$, which is a more precise statement.

What do we do with this? Well, I say that this allow us to finish the proof.

$\dfrac{100x^2 + x^2 \sin(1/x) + 50000}{100x^2} \leq \dfrac{100x^2 + x + 50000}{100x^2} = 1 + \dfrac{1}{100x} + \dfrac{50000}{100x^2}$

, and it is clear that the last two terms go to zero as $x$ increases. $\spadesuit$

Finally, I’d like to remind you about the class webpage at the left – I’ll see you tomorrow in class.

## Are the calculus MOOCs any good: After week 1 Saturday, Jan 12 2013

This is a continuation of a previous post.

I’ve been following the two Coursera calculus MOOCs: the elementary introductory to calculus being taught by Dr. Fowler of Ohio State University, and a course designed around Taylor expansions taught by Dr. Ghrist of UPenn, meant to be taken after an introductory calculus course. I’ve completed the ‘first week’ of Dr. Fowler’s course (there are 15 total), and the ‘first unit’ of Dr. Ghrist’s course (there are 5 total), and I have a few things to say – after the fold.

## Are the calculus MOOCs any good? Tuesday, Jan 8 2013

I like the idea of massive online collaboration in math. For example, I am a big supporter of the ideas of the polymath projects. I contribute to wikis and to Sage (which I highly recommend to everyone as an alternative to the M’s: Maple, Mathematica, MatLab, Magma). Now, there are MOOCs (Massice open online courses) in many subjects, but in particular there are a growing number of math MOOCs (a more or less complete list of MOOCs can be found here). The idea of a MOOC is to give people all over the world the opportunity to a good, diverse, and free education.

I’ve looked at a few MOOCs in the past. I’ve taken a few Coursera and Udacity courses, and I have mixed reviews. Actually, I’ve been very impressed with the Udacity courses I’ve taken. They have a good polish. But there are only a couple dozen – it takes time to get quality. There are hundreds of Coursera courses, though there is some overlap. But I’ve been pretty unimpressed with most of them.

But there are two calculus courses being offered this semester (right now) through Coursera. I’ve been a teaching assistant for calculus many times, and there are things that I like and others that I don’t like about my past experiences. Perhaps the different perspective from a MOOC will lead to a better form of calculus instruction?

There will be no teaching assistant led recitation sections, as the ‘standard university model’ might suggest. Will there be textbooks? In both, there are textbooks, or at least lecture notes (I’m not certain of their format yet). And there will be lectures. But due to the sheer size of the class, it’s much more challenging for the instructors to answer individual students’ questions. There is a discussion forum which essentially means that students get to help each other (I suppose that people like me, who know calculus, can also help people through the discussion forums too). So in a few ways, this turns what I have come to think of as the traditional model of calculus instruction on its head.

And this might be a good thing! (Or it might not!) Intro calculus instruction has not really changed much in decades, since before the advent of computers and handheld calculators. It would make sense that new tools might mean that teaching methods should change. But I don’t know yet.

So I’ll be looking at the two courses this semester. The first is being offered by Dr. Jim Fowler and is associated with Ohio State University. It’s an introductory-calculus course. The second is being offered by Dr. Robert Ghrist and is associated with the University of Pennsylvania. It’s sort of a funny class – it’s designed for people who already know some calculus. In particular, students should know what derivatives and integrals are. There is a diagnostic test that involves taking a limit, computing some derivatives, and computing an integral (and some precalculus problems as well). Dr. Ghrist says that his course assumes that students have taken a high school AP Calculus AB course or the equivalent. So it’s not quite fair to compare the two classes, as they’re not on equal footing.

But I can certainly see what I think of the MOOC model for Calculus instruction.

## Math 90: Week 8 Wednesday, Oct 24 2012

Today, we had a set of problems as usual, and a quiz! (And I didn’t tell you about the quiz, even though others did, so I’m going to pretend that it was a pop quiz)!. Below, you’ll find the three problems, their solutions, and a worked-out quiz.

## Math 90: Week 7 Wednesday, Oct 17 2012

I haven’t quite yet finished writing up the solutions to the problems we did in class yesterday. But I wanted to go ahead an make the solutions to the test available. If you ask me for them, I can send you a link to them.

But please note that there is an error in the key! In particular, on problem 7(b), I forgot that we only care about $t \geq 0$. So the final answer should not include $t = 1/2$.
The notes for the day are after the fold: (more…)

## Math 90: Week 5 Wednesday, Oct 3 2012

A few administrative notes before we review the day’s material: I will not be holding office hours this Wednesday. And there are no classes next Monday, when my usual set of office hours are. But I’ve decided to do a sort of experiment: I don’t plan on reviewing for the exam specifically next week, but a large portion of the class has said that they would come to office hours on Monday if I were to have them. So I’m going to hold them to that – I’ll be in Kassar House 105 (the MRC room) from 7-8:30 (or so, later perhaps if there are a lot of questions), and this will dually function as my office hours and a sort of review session.

But this comes with a few strings attached: firstly, I’ll be willing to answer any question, but I’m not going to prepare a review; secondly, if there is poor turnout, then this won’t happen again. Alrighty!

The rest is after the fold –

## Math 90: Week 4 Friday, Sep 28 2012

It was quiz-day!

The class did pretty well on the quiz. I wrote the quiz, and I’m pleased with the skill-level demonstrated. The average was about a 77%, and the median was an 80%. (For stat-witty folk, this means that the lower scores were somehow ‘lower’ than the average scores).

Anyhow, the solutions to the day’s problems and the quiz are below the fold:

## A MSE Collection: A list of basic integrals Monday, Jun 11 2012

The Math.Stackexchange (MSE) is an extraordinary source of great quality responses on almost any non-research level math question. There was a recent question by the user belgi, called A list of basic integrals, that got me thinking a bit. It is not in the general habit of MSE to allow such big-list or soft questions. But it is an unfortunate habit that many very good tidbits get lost in the sea of questions (over 55000 questions now).

So I decided to begin a post containing some of the gems on integration techniques that I come across. I don’t mean this to be a catchall reference (For a generic integration reference, I again recommend Paul’s Online Math Notes and his Calculus Cheat Sheet). And I hope not to cross anyone, nor do I claim that mixedmath is to be the blog of MSE. But there are some really clever things done to which I, for one, would like a quick reference.

Please note that this is one of those posts-in-progress. If you know of another really slick bit that I missed, please let me know. And as I come across more, I’ll update this page accordingly.

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