They say Lincoln had very little or no formal education. The former US president himself is known to have spoken few good things about his childhood schooling.

However, sometime back in June last year two mathematics professors in Illinois confirmed the discovery and authenticity of two papers believed to have been part of Linoln’s cyphering book — that is a 19th century term for “workbook”, in case you are wondering.

I recently came across an interesting animation that was shared with a rather poor choice of words: “Dots that are moving around in a circle! Connect them and you’ll get a moving octahedron…”

I find that this statement is faulty for reasons I will explain below; but first, the animation in question:

I will try to limit the mathematics to the necessities; generally, neither is this a two-dimensional figure, nor are those circles. I think the animation, in its present form, merely benefits from a rather tricky choice of perspective.

Before we actually talk about this, I hope to clear up a few things (read, prove a couple of lemma) regarding the construction and structure of an octahedron.

ONE OF THE BIGGEST PROBLEMS in contemporary theoretical physics is the dark matter-dark energy duo. Back when Einstein forced himself to tailor his theory of general relativity to a static universe, he had come up with what was termed the cosmological constant; it was, for all practical purposes, a sort of correction parameter which made sure his famous field equation worked in a static universe. Once turned off (i.e. the Lambda in the equation below (the upside-down V) once set to zero) one term is knocked off the equation once again returning it to a state of satisfying a changing universe.

Quantum fluctuations and the cosmological constant

The entire elaboration that was the cosmological constant has since died and been resurrected. Physicists are now toying with the idea that the value-equivalents of temporary fluctuations of energy in space (as a result of Heisenberg’s uncertainty principle) called quantum vacuum fluctuations is nothing but the cosmological constant.

A lot of hope was stacked on the Higgs Boson prior to its discovery, and if anything, that has all be trebled after the fact. Now, the latest attribution to the so-called God particle, is the mysterious behaviour of dark matter (no, not the comic,) and dark energy and, in turn, the cosmological constant itself.

Ride the Higgs boson seesaw?

In their PRL paper, “A Higgs seesaw mechanism as a source for dark energy,” physicists Lawrence Kraus and James Dent propose using the seesaw mechanism (which was originally used to understand neutrino masses which are a mere millionths of, say, particles in the Lepton family) to understand how the Higgs particle can create an energy density in the Grand Unified Scale (about 16 orders smaller than a proton) at which the three known energies are believed to tie to form a unification à la Einstein’s Grand Unification Theory — quite aptly shortened to GUT.

Promises of a minuscule order

The discovery of the Higgs boson has provided a much needed injection of fact into what was otherwise turning out to appear like a lot of wild conjecture.

Dark matter is what occupies a majority of the mass on any given galaxy, but can only be detected indirectly (by its gravitational influence) since it does not interact with light (things such as light shining off it do not occur.) Such an energy, called dark energy for obvious reasons, appears to have a magnitude of electroweak physics’ GUT order.

Kraus and Dent’s theory of Higgs coupling along the lines of the Neutrino seesaw explanation arrives at precisely this order of magnitude of energy contribution, as it were, to a quantum vacuum. The duo calls this an “extension of the standard model” — another promising, if not fully proven, foundation on which to explain their ideas.

The pith of this paper is that by operating on a factual entity (the Higgs boson,) using, analogously, a previously tested approach (the neutrino seesaw,) treated on the platform of a well-accepted idea (the standard model,) we end up with energies of exactly the same minuscule order of magnitude as the one we found hard to explain convincingly.

Perhaps this paper is more of a nudge in the right direction than a revelation itself, but sometimes in physics nudges are exactly what we need.

Cover image: Rosette nebula via Flickr/bobfamiliar

Laser eye surgery is without doubt an amazing procedure, giving people the opportunity to throw away their glasses and contact lenses. Laser eye surgery has actual been around a lot longer than most people realise and it was first carried out over 20 years ago, admittedly not with the same impressive results as we expect today.

In the early days, laser eye surgery was far more invasive than it is today, leaving people with extensive corneal scarring and variable results. Laser eye surgery risks were also far higher and they also tended to be more serious.

Today however, laser eye surgery is minimally invasive and the results are extremely impressive. Around 95% of people having the procedure will end up with 20:20 vision and almost 100% will have driving standard of vision or better. Laser eye surgery is the most common elective surgery available and is far more popular than even the most common cosmetic surgery procedures.

For those deciding whether or not to have laser eye surgery then one of the most commonly asked questions is just how long the visual results are likely to last. This article is going to outline the most important points you need to know about this aspect of laser eye surgery:

Laser re-treatments

Around 6% of all laser eye surgery procedures will have to be repeated with a ‘top up’ treatment. This is normally done for free by the clinic that carried out the original procedure and it may only be required for one eye. To illustrate how laser re-enhancements work, it is best to use an example.

Imagine your prescription was -7.00 Dioptres prior to your surgery and then after surgery your prescription was completely eliminated to zero, meaning you had perfect vision. Then over the course of the next year your prescription slowly regressed to -1.00.

This is obviously only a fraction of what your prescription was before you had the surgery but it is enough to affect your vision. The laser is then re-applied to re-correct this remaining prescription.

Long sighted prescriptions

Long sighted prescriptions (plus e.g. +5.00) are more likely to regress than short sighted (minus e.g. -3.00) prescriptions.

Higher prescriptions

The higher the prescription the more likely you are to need a laser re-enhancement. The reason for this is that the more something has changed, the more likely it is to return back to its original state.

Reading glasses as you get older

Regardless of whether you have had laser eye surgery or not, you are still going to need reading glasses as you reach your mid 40’s. This is related to the lens in your eye that slowly loses its power which is a condition called presbyopia.

Eye diseases

As we get older so the chances of us getting age related conditions such as cataracts increases. Cataracts typically start affecting us aged around 70 years old and they can cause a change in our prescription. So even though you may have had laser eye surgery previously the cataracts could mean you end up needing glasses as you get older.

So, as you can see, having laser eye surgery does not guarantee that you will be free from glasses/contact lenses for life but for most people they will at least have perfect distance vision for a prolonged period of time.

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Tim Harwood is an optometrist with over 8 years in practice with a specialist interest in both laser eye surgery and contact lenses. He has worked both in the United Kingdom as well as Australia and has worked for both multiple and independent opticians. Tim also provides useful information on his own website, TreatmentSaver.com.

Last session we saw two important aspects under integration: methods of solution and the steps to solve a definite integral. In this session, we shall be seeing why definite integrals can be solved this way. In other words, we’ll see why it works.

This is mathematics, so it isn’t like somebody decided you could subtract integrals of limit points and whatever you get is the right answer. And then we didn’t follow all of that for centuries to satisfy one man’s fancy. There is a reason it works (or proof, if you want to call it that,) and we will be understanding that today.

First of all, if you have not been here for sessions 1 and 2, my advice is that you head right back and go through them before continuing this session. You can read session 1 here and you’ll find session 2 here. Go on, I’ll wait.

The weird thing about areas

If you remember last session well, I asked you to find the area of a weird looking object. My point then was to show you basically that integration of something can tell you the area of that thing.

[sws_red_box box_size="100%"] Now, this is obviously not always true — a little common sense will tell us that. For instance, integration of velocity won’t give you the area of anything. What I intend to say is that if the area of something can be measured, such as an apple, integration will most likely give you its value. [/sws_red_box]

As always, to understand how this works, let’s reduce this real-world problem to a mathematical model. (That’s just a fancy way of saying let’s use a wooden plank and a toy car to understand the effects of toppling over instead of trying it with real cars, real mountain roads and real humans.

Here’s our mathematical model:

Graph 1

Ugh, not very interesting, is it? Well, what model were you expecting?

Now let’s zoom in to this part:

Graph 2

I hope you recognise where the region shown in graph 2 is present in graph 1. Got it? Good, now I’ve been a very kind soul and scribbled a few things on this graph:

1. I’ve marked two points: x = 2, y=0 and x = 4, y=4

2. I’ve marked another point, A, in red

3. I’ve marked the curve between (2,0) and (4,4) as C

4. I’ve shaded the area under curve C in light grey

5. I’ve pointed out to the graph’s equation in case you can’t find it anywhere

6. I’ve spoken to dolphins undersea

Now, once again, this is the equation for our graph:

Let’s start with the absolute basics. Firstly, whether you say y = something or f(x) = something, it doesn’t matter because y=f(x), they’re just two ways of saying the same thing. Secondly, the graph comes by plotting points by giving values to x, not y. You always give values to the thing on the right hand and see what happens to the thing on the left. Therefore, if I were to say x = 2, I get… y=0. Now, if x=3, y=1; and if x=4 then y=4. Notice the parts I’ve written in bold: these are the points I’ve marked on the graph, (2,0) and (4,4)

So what’s the fuss about all these points? Let’s do a little exercise. Let’s integrate eq. 1 with respect to x. I mean, that’s what we’re here for isn’t it? It’s been seven paragraphs into this third integration session and we still have not integrated anything? Well, try it yourself and see what you’ll get.

But wait! We’re geniuses, so let’s go one step further and try to make this thing definite. Sounds like fun.

Now how exactly do you do that? By finding the limits, of course. Yes, but how? Just ask yourself these simple questions:

1. What’s our superstar variable? Answer – x

2. What’s out curve? Answer – c

3. What’s the lowest value of our variable in our curve? Answer – 2

4. What’s the highest value of our variable in our curve? Answer – 4

So, there. For an equation (like eq. 1) given for curve C in variable x, the upper limit is 4 and lower limit is 2. How simple was that?

Good, now integrate eq. 1 with limits and you’ll get eq. 2 below. (I’ve shown the intermediate steps in case you’re lost.)

Aren’t we feeling creative? Now let’s do something else. Let’s join points (2,0) and (4,4) and make a triangle like I’ve shown in the figure below. I’ve now even given the points (vertices) names as M, N and O.

Graph 3

Great, now let’s find the area of that triangle. Yes, I see your stare. Do we have no better work? Right now, no, so let’s find the area of the triangle. We know it’s a right angled triangle, so the area is given by,

From graph 3, our base stretches from x=2 all the way to x=4 (i.e. the line MO) which is 4 – 2 = 2 units in length. Next, out height is the line NO, which stretches from y=4 to 4=0 putting its length at 4 – 0 = 4 units but you already knew all that.

So our area is,

Well, well, well, will you look at that! In both equations 2 and 3, our answers are roughly similar. Granted, eq. 3 is very slightly (2/3 times) greater, but that is because of the extra region as I’ve highlighted in grey in graph 3.

So what does that tell us? It tells us that integration of a curve gives you the area of graph under it. Beautiful!

[sws_red_box box_size="100%"] Please note: for the precise minded, if the slight difference arising due to the extra region in graph 3 does not satisfy you, there are more advanced proofs. We’ll discuss that once again in session 4.[/sws_red_box]
[sws_red_box box_size="100%"]Please note this too: the above integration-area equivalence is not always true. For instance, in a line integral, the integration restricts itself entirely to the line. In particular this integration-area equivalence arises when the lower limits of our region (the shaded part in graph 1) has a limit at the x-axis. What if it goes beyond? Or into the third dimension? [/sws_red_box]

Alright, here’s an exercise for you, and this is why I marked point A. Just like we’ve done above for curve C: MN, try to prove the same result for curve D: AMN. Hint: make two triangles with their only common vertex at M.

The fundamental theory of calculus

We’re here at last. Let us quickly sum up a couple of important things we have learnt since session 1. We know how these work, and the fundamental theorem of calculus will tell us why.

Firstly, we know integration can definitively be done over an a range of values with a fixed beginning and end called limits; secondly, we know this gives the area under graph in most general circumstances; thirdly, we know that integrating a function and then subtracting the integral at specific situations corresponding to its two limits will give us the definite value of that integral.

Alright, so here is what the fundamental theory of calculus says:

Huh? We’ve seen that before haven’t we? Let me explain the notation: the capital F stands for integral. While eq. * above is the standard notation, here is another way to write the same thing:

There, you have definitely seen eq. 4 before! We discussed it in session two when learning how to solve definite integrals, but we just learnt it like everything else then, we never called it with a fashionable name like, ‘the fundamental theorem of calculus.’

So now you might be wondering why this is held in such high regard. All it does is state what we already know, right, so what’s so fundamental about it?

In short, this theorem says our method of integration is correct. (Well, that’s reassuring.) But it also has other deep implications such as the area thing we saw above. A second fundamental theorem of calculus is even more fundamental (and rather obvious) and it simply says how differentiation and integration and interchangeable, like this below. (Note that x and t are variables, a is a constant.)

Since you probably are waiting for proof of this theorem, let me give you some good news and some bad news: the good news is, we’re not going to prove this theorem now. The bad news is, we’ll prove it after we learn yet another theorem called the mean value theorem in integration, or, as Doofenshmirtz would say, “Behold! The ‘for every definite integral there exists a rectangle with the same area and width -inator,’ or the FEDITERWSAW-inator.”

[sws_red_box box_size="100%"] If you don’t know who Dr Heinz Doofenshmirtz is, it’s time to abandon integration and turn on the Disney Channel. Seriously. [/sws_red_box]

It’s question time

After sessions 1 and 2, I received a few emails carrying doubts regarding stuff we have talked about so far. I’m going to take time today in session 3 to clarify a couple of those because it may help others too to be free and fully clarified when learning the ultra-important things coming up in sessions 4 and 5. (See contents at the bottom of this article.)

Josh Fraser asks, “What really is the exponent (e) in math? I’ve seen that in so many places.”

First of all, John, that’s not the exponent. If I say 2 is raised to the power 5, then 5 is the exponent, another name for power. The letter e you see used everywhere, which you’re referring to, is actually called Euler’s number after the Swiss mathematics superstar, Leonhard Euler (say leon-hard-oiler.) This e is just a mathematical constant like pi and is approximately equal to 2.7182. It is defined like this:

Why is it important? The Euler number (sometimes called Napier’s constant) is its own derivative and integral. In other words, almost always, no matter what you do with e — throw it off a cliff, run it over with a car, integrate it, drown it, differentiate it, light it on fire — it remains just as it is. Kind of adamant, eh? And, needless to say, such properties get us maths and physics nerds terribly interested.

Shania Hubbard asks, “How do I integrate the cube of cosine?”

How indeed! I was overjoyed when I saw this question because I remember being asked this in an examination and not being able to answer it till the very last moment when it struck me like a light bulb but it was too late. Anyway, here’s the question the way it’s supposed to look:

First try it yourself before looking at the answer below.

This one is solved by substitution; in fact, it is a good example of how substitution can sometimes get messy. Now if I directly substitute something for cos x, say u, then deriving it would go, 2 times the square of sin x, which is nowhere in the function. A little more effort will reveal this:

Now substitute u in place of sin x, so,

Substituting this in eq. 5 we get,

Now integrate eq. 6 as usual,

Now just substitute back u = sin x in eq. 7 and you have your answer:

Now try it for the cube of sin, tan and other trigonometric functions. Then try it for the fourth power and see how you can handle it.

Dave Moreno asks, “I’m going back to my college days with your Integration 101 series and I’m enjoying it more than I did in college. But I remember we were taught a formula method for substitution whereas you explained it in today’s session [session 2] without any formula. I actually understood and enjoyed it, but how different is it from the formula?”

Well, big question, Dave. First of all, congratulations on deciding to re-visit college calculus; there are not many who are brave enough to do that. But, you’re right, the substitution method can be taught with a formula. And it looks like this:

See how you made that “Ugh! What is that spelschtick? I’m gettin’ oatta heer!” look on your face? That’s why I didn’t mention this in session 2. In reality, you do not need to know this to perform substitution because this exists just to generalise the substitution rule and prove that it works. (What you see above is sufficient proof.)

When you’re working out mathematics, functions rarely show up in the convenient f’(g(x))g’(x) form as in eq. 8, so identifying them in this manner becomes simpler when you forget about this rule/formula hullabaloo and just do it the way we saw in session 2 (and even in Shania’s question above.) Trust me, forget this formula and you’ll fare better.

I hope that has covered most doubts. Other questions I received were in some way or other related to these three (especially the last two) so I figured if I answered these publicly, it should help others solve their own problems, if any.

It’s integration table time

We’ll look at ten hyperbolic functions today — yes, there are exactly ten; and no, we still have not let go of trigonometric functions. Here’s a look at the future: there are 32 more. Just the trigonometric ones, that is. And then there are about thrice as many left and then — I think I’ll stop now.

HYPERBOLIC TRIGONOMETRIC FORMS

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Note how the left hand sides of v16 – v25 are very similar to those of v6-v15 except that v16-v25 are hyperbolic.

Also note there is another way to represent hyperbolic functions: arc-. So sinh is arcsin, cosh is arccos, csch is arccsc and so on. I usually prefer the arcsec, arccot… representation but stuck to the more popular method in the table above.

So, in today’s session we’ve covered the fundamental theorem of calculus (which we will prove in session 4) and seen an introduction to the integration-area equivalence (which we’ll further understand after session 4.) In other words, sessions 3 and 4 are very closely linked and you might want to treat this as an introduction to session 4 which goes into the beginning of the dark depths of maths that have made students shiver for centuries.

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Here is the plan for session 4:

1. Reimann sums, mean value theorem in integrals, and hence proof of the fundamental theorem of calculus

2. Properties of definite integrals

3. Integration by parts

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Until next time, enjoy calculus… muhahahaha! [vhb]

Welcome to the second session of Integration 101, the place where you’re taught integration like you always hoped you’d be. Today we’ll talk about four things, two of which are carry-overs from the previous session, so if you have not read through session 1, do it now.

1. How do you solve the real world well problem I asked you last time

2. How to integrate a football

3. Two methods of integration plus an illegal short-cut

4. How definite integrals work

As promised, this session is inevitably going to have more maths than the last, although nothing to make your head spin. And we’ll make it fun in the process.

If you were actually a genius and solved the problems from our last session, you’ll know the first one is simply e raised to the power x; as for the second one, give yourself a pat if you got something even vaguely like this:

We’ll go through the third problem now.

How deep inside the well is the water?

Here is our question: A farmer goes to his well and wants to check the depth of the well till where the water has filled. So he picks up a stone and throws it inside and starts looking at his clock. He hears a splash sometime later when the stone hits the water and sees the time: it has been 20 seconds. How far below is the water?

The solution is a lot simpler than you would imagine: if velocity is the time derivative (i.e. differentiation with respect to time, t,) of position and acceleration is the time derivative of velocity, then by definition (integration is the reverse of differentiation) the integration of acceleration with respect to time is velocity and further integration of velocity should give us position.

Graphically, for differentiation –

and for integration –

We know the acceleration of the stone to be approximately 10m/s/s so integrate that with time t according to eq. 2 in session 1.

So the velocity of the stone is 10t and obviously depends on the time. For instance, at t=0 when it leaves the farmer’s hand its velocity is 0; at t=10, it’s velocity is 100m/s and so on. Now integrate the right hand side (velocity) further to get the position.

So far so good. Now that is the position of the particle with respect to the zeroth position (i.e. at the start of its travel, when it leaves the farmer’s hand.) All we need is it’s position at time t=20 because that is when it hit the water. Plugging in t=20s to our above equation we get the final position as 5 x 20 x 20 = 2000m.

Whopping! That’s how integration comes up in real life. Now I know you’re smart so you’re probably wondering what happened to that +C we have. Isn’t the answer actually (2000 + C) metres?

In short, it is, but to be precise, the C becomes equivalent to zero in this case and I’ll tell you why by the end of this session when we’ll solve this same problem slightly differently.

How to integrate a football?

That subheading is probably misleading, but it is valid. My intention is to explain to you why we integrate at all. Because if you don’t understand why we’re doing all this, you won’t feel like doing all this.

Once again, get into a time machine and travel back to grade 1. Remember how your teacher once gave you this threatening problem, “If Superman starts in his car from home and goes to his office in 20min and he travels at 50km/h, how far is his office from his house?”

You shuddered and shivered wondering if you had the right answer before standing up and saying, “About 16.5km.”

I say it isn’t. Not only do I know Superman personally, I also know mathematics enough to say you were wrong. That is merely his average speed: he never started at 50km/h and never drove into his office parking space at 50km/h and if the road to his office is 16.5km of straight, non-twisting glory, he’s one helluva lucky bloke.

Th point is, the guy’s speed definitely varied throughout. He slowed down at intersections, at turns, for maniac drivers and so on. Besides he started at 0km/h and ended at 0km/h which means there definitely was a lot of acceleration and deceleration going on. If you integrate these various speeds of Superman’s driving, you would get a more accurate idea of the distance between his home and office. That is one place where we use integration.

Now, we also spoke, in session 1, of how differentiation is taking apart and integration is pasting together. Why would a sane man want to spend his vacation doing that? There is a reason, and this is all it is: integration simplifies the summation process. You heard me, simplifies.

Think about this: you know how to find the area of a circle, you know how to find the area of a square, or the volumes of a cuboid or a pyramid. Now riddle me this — what is the area of this fine looking thing I have here:

Not so smart, eh?

Well, integration can help you here if you ever take up such a great quest as this. Coming down to earth, you can, for instance, find out the surface area of a more common shape such as a sphere, by finding the areas of each of it’s smaller, imaginary surface segments and integrating them. I always liked to compare this to a football:

What’s the difference, you ask? That’s the best part: you don’t have to know the area of each and every section to integrate it — just one. Find the area of one tiny, tiny, little section of any size and properties so long as it is on the football, and integrating it will magically give you the surface area! The same goes for volume etc.

Two (+1) methods of integration

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Do not confuse this section with the rules of integration. We will talk about that in future sessions. For now, understand the difference: methods of integration are few and fundamental and work on simple forms. The rules of integration will tell you how to break down a complex integral into simple forms so you can apply the methods of integration on those simple forms.

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Method #1 — The simple method, or, the look-at-the-table-for-heaven’s-sake method

In this method, your maths teacher gives you evil looking integrals and you quickly identify they look like something else. Kind of like looking at Lady Gaga and thinking, ‘sea anemone!’ This –

Now do the same thing with this:

See what happened there? We know this basic trigonometric identity:

Why did you do that? Because you recognised that the table (such as the one you learnt in session 1) has no formula for the integral on the left hand side, but the integral on the LHS looked very similar to the integral on the RHS and it struck you like a light bulb — isn’t this sec x for which I have a formula?

This decidedly takes some practice, so solve the following three questions:

In all cases, simply reduce the integral to something you can compare with the table. (All results used in the above problems have been covered in the table in session 1.)

[sws_yellow_box box_size="100%"] Fun fact: In eq. 1 you might recognise that 2(pi)r is the circumference of a circle with radius r. What do you think you would get when you integrate it? Try to answer this without integrating it first! [/sws_yellow_box]

Method #2 — The substitution method, or, the what-if-I-replace-the-earth-with-mars method

Perhaps the single most popular — and useful — method in integration and in most other branches of mathematics is the substitution method. This is where you try using a tennis ball instead of a football just to see how hilarious things can get.

Consider this:

First of all, we have nothing that looks even vaguely like this in out integrals list. Secondly, this has two functions while everything we integrated till now had just one!

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Weird Maths Tip: A function is something that does something to a variable. Like a doctor who operates on you or a physicist who cuts up an atom. Or a machine that totally transforms an object that goes through it:

Identifying functions can be a very useful skill in Calculus. For instance, e raised to the square of x has two functions: e and the power 2.

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In our eq. 2 we notice something interesting: sine and cosine are easily related to each other. In other words, sine and cosine can be converted into each other. One of several possibilities that come to our mind is this:

or,

But we don’t have cosx dx in eq. 2. How about it we try it the other way round?

or,

and if we throw the minus sign from RHS to LHS,

As it turns out, we do have sinx dx in eq. 2!

So far this is what we know:

1. The integral in eq. 2 has two functions, sine and cosine

2. Sine and cosine can be interrelated as in eq. 3

3. Eq. 3 has something (sinx dx) that looks exactly like something in eq. 2 (again, sinx dx)

4. So, if we substitute cosx as something, we can write sinx in terms of that something according to eq. 3

So, here’s what we do next: we’ll put cos x as some letter (go on, pick one of 25!) Let’s toss and settle for u. Put cos x = u, so, according to eq. 3, (relation highlighted in red)

Substituting eq. 4 in eq. 2,

becomes,

or,

which we can easily solve from the integral table in session 1 (for variable u, in this case,) as,

which, re-substituting for u = cos x, gives our answer:

Again, this takes some practice, so here are three problems:

Alright, now onto the bonus method.

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Maths tip nobody tells you until it’s too late

How would you solve this question?

The answer: you wouldn’t.

Remember, unlike algebra, you cannot solve integration for one value of x like you see above. Instead, integration is done over several values of the variable as we’ll see in the last section of this article. So the above question is basically meaningless.

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Method #3 — Guesswork, or, the method maths teachers don’t want you to know

Well, before I get you too excited, let me make this clear: this method does not work always; actually, it does, but it does not always simplify things. Sometimes — rarely, but there are occasions, when — it complicates your problem, so use it wisely.

The concept behind the guesswork approach is to guess at an integration method (because choosing the right method is often the hard part) and then blindly integrating it and then fixing the values later.

The best way to demonstrate this is using an example. So, consider,

It looks hard (or maybe I’m not in the mood to integrate elaborately) so I’ll guess: let’s try the power formula (v1.)

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So the blue part is what I think it should be. Let’s see how right I am. Common sense tells me that if I differentiate the blue part, I should end up with the question. Let’s see what happens:

Oh, almost! There’s that extra 5 in the numerator, so my guess was off by that much. Let’s balance it out by putting (i.e. multiplying by) a 5 in the denominator. So, let’s change our guess from,

to,

That was simple. Now let’s see if it’s any use. Why not try it the long way yourself: if you integrate the LHS, you do indeed end up with the RHS in red. So our guesswork method yielded the correct answer!

How definite integrals work

Now we come to the last section of our session today. How would you solve a definite integral?

Recall from session 1 that we had said a definite integral is one which has limits; that is what makes it definite. There are only two steps to solve a function of definite integration:

1. Solve the integration just as you would any indefinite integral, except, don’t write +C.

Spot the two limits? There is an upper limit (written above) and a lower limit (written below.)

2. a. Replace all x in your equation (or y or u or whatever your variable is) with the value of the upper limit and solve it.

2. b. Replace all x in your equation (or y or u or whatever your variable is) with the value of the lower limit and solve it.

3. Now subtract your answers in step 2. b. from your answer in step 2. a. to get your final answer.

Let’s try it with an actual problem now. I’ve labelled equations as steps 1, 2, 3… corresponding to those above.

Question:

Solution:

We have,

On solving the given integration, we get,

For upper limit 4, we have,

and for lower limit 2, we have,

Therefore, subtracting step 3. b. from 3. a. we arrive at our final answer: 682.66 – 10.66 = 672

Now try this out yourself for our first problem, that of the farmer and his well. You’ll notice that the answer in step 3. b. becomes zero since the lower limit is 0.

Here is how you might understand it: we are talking about the position of the stone. It starts at time t=0s, at the farmer’s hand so that becomes the lower limit, and travels up to when he hears the splash at t=20s, which becomes the upper limit. In other words, if we apply limits,

So the depth of the well till water is not 2000 + C but simply 2000m.

Also notice how integration is done over a range of values from 2 to 4 or 0 to 20 as in our previous examples. If, however, we try to integrate for one value of the variable (like we asked in the meaningless question previously) let’s see what happens.

basically means the upper limit and lower limit are both = 2, since no range of values exist. This give us,

Try it yourself and you’ll find that, in general,

That’s another important result worth remembering, although it appears seldom.

Here are just four simple and easy relations for this session:

FOUR REMAINING TRIGONOMETRIC INTEGRALS

I certainly hope you’re enjoying the Integration 101 series. Do email me with your thoughts, suggestions or requests and I’d be more than glad to get back to each one of you.

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Here’s a glimpse of what we’ll be discussing in the next session:

1. Definite integration and area: how the magic happens

2. The fundamental theorem of Calculus

Update:

3. Clarifying a few doubts and questions I received regarding sessions 1 and 2

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Until next time and after, differentiate your troubles and integrate your joys [vhb]
Cover image: Quadragonal symmetry axis – Flickr/John “Pathfinder” Lester

Calculus is divided into two parts: chopping wood and then pasting it back together. The chopping is what mathematicians call differentiation while the pasting together is called integration. You might be wondering where these fancy names come from — it does not matter, because so long as you understand this next section correctly, integration won’t be a problem.

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Addition, multiplication and integration

What many people fail to understand is that the deep roots of Calculus lie in grade 1 when they learnt addition. Things were simpler back then: 2 + 2 = 4 and it was always true. It was true at two o’clock and it was true at four o’clock.

Soon, you went to grade 2 and 3 where something intimidation came up: multiplication. That’s like adding and adding and adding and adding several times, simplified. You even remember your teacher telling you, “multiplication is repeated addition.” So, 2 + 2 + 2 + 2 + 2 + 2 + 2 = 14, which took you twenty minutes to solve in grade 1 could be solved in seconds by writing it as 7 x 2 = 14. Again, this was just as true at four o’clock as it was at two o’clock.

What many people don’t tell you is that integration is also repeated addition. But where you could add 2 seven times to get fourteen, imagine adding it 236,785,467,890,643,776,543,127 times. I bet you can’t even read that number, let alone add that many times. Besides, that number keeps changing whether it’s two o’clock or four!

Luckily for us, geniuses like Newton and Leibnitz (who absolutely hated each other) had a solution for this. This is what integration is all about. It tells you how to add 2 + 2 + 2 + 2 + 2 … 236,785,467,890,643,776,543,127 times without going crazy. So, if 2 + 2 = 4 is how we represent addition, and 7 x 2 = 14 is how we represent multiplication, this is how we represent integration:

See what happened there? The difference simply is that you’ll be adding figures where you used to add digits. And that integration symbol (which is really the violin clef symbol, because Newton probably ran out of symbols,) tells you what you’re doing with those two half-circles.

Get to the point

Alright, alright, I will. It’s just that whatever you read till now is terribly important. Start looking at integration that way, you’ll understand how easy, not to mention beautiful, all this is.

Now for some real mathematics. There are two types of integrals: definite and indefinite. This is what they basically are: indefinite integrals give you some value as an answer but also give you a + or - some value; definite integrals just give you the exact value.

Think of it this way: when you go to a dull pizza guy, he might give you a couple of dollars more or less in change. He’s indefinite. But go to your mathematics teacher and he’s sure to give you nothing more than a C. He’s definite. If only it were the other way round.

When is it indefinite?

Good question. How do you know if an integral is definite or not? Consider this: There’s a huge circular playground and you and your friend pick up a measuring tape and start measuring the length along its edge. You both start at your own points, and you don’t really know where to end, so each of your measurements overlap for some distance, like so:

See how, when you add it all up, you’re adding the red parts twice? That’s what’s indefinite about it. You know that the length around the circle is at least the length of the blue and green parts; but the answer to this indefinite integral will give you,

total length = (length of blue part + length of green part) +/- (length red parts)

Well, you know it’s obviously wrong, but you also know that, if you know the length of the red parts, you can get the correct answer by a simple addition/subtraction.

Now, if you and your friend are both smart and, before measuring, you decide on a starting and an ending point, you’ll measure it this way:

Now, notice how you get a definite answer,

total length = (length of blue part + length of green part)

What made the difference is the fact that there was a fixed starting point and a fixed ending point. In other words, your measurements would not start before a minimum limit and would not go beyond a maximum limit. Indeed, these fixed starting and ending points are called limits. And, if you have limits, you have a definite integral, and if you have no limits, you have an indefinite integral.

Mathematically, this is what an indefinite integral might look like:

And here’s a definite integral:

If you have no idea where all that came from or how, don’t worry. Calculus can be intimidating. But we’ll get to that in a matter of a session or two.

For now, keep in mind that the answer for a definite integral and an indefinite one are the same (the red highlighted part,) except that for an indefinite integral, you write +c after the answer to denote the indefinite part. This is a universal notation.

Some mathematics at last: the technique of integration

I know some of you must be writhing in your chairs because we have not done any real mathematics yet. We got to the mathematics faster when we discussed differentiation, but integration is a whole different ball game, so a little patience can go a long way.

Integration is a technique after all, like addition or multiplication, so how does this technique work? Here’s a glimpse. From differentiation we know,

So, if you were to take the left hand side and integrate it, you should get the right hand side, yes? It turns out you do get just that. (Again, see the red highlighted part.)

The reason why I compared integration with multiplication will now become clear to you. Just as we know 2 times 4 is 8, we know the relation in eq. 1 also; only, the multiplication symbol becomes the integration symbol. And, just as we know the proof that 2 times 4 is 8 (i.e. 4 + 4 = 8) we also know the proof of the relation in eq. 1. Again, just as we don’t care about the proof every time we say 2 x 4 = 8, we don’t care about the proof every time we use the relation in eq. 1 either. We know it is true.

So how exactly do you learn the various multiplication possibilities with 4? It goes 4 times 1 is 4, 4 times 2 is 8, 4 times 3 is 12 and so on. Similarly, there are set integration tables you will have to memorise. No bargaining here!

Integration table 1 of 5

Every day we’ll take a look at about ten integration relationships (or formulae, if you like,) over the next 5 sessions. That’s right, you genius, that’s 50 formulae in all! If you want to nitpick, there are 120 in all, but I’ll tell you why knowing 50 is sufficient.

First, here’s one all-important-never-forget-like-ever rule. (n is a constant, see how the dx becomes x on the right hand side?)

Since every session will be once in two to three days, you will have two to three days to learn ten formulae; isn’t that plenty? And you’ll know them all in less than a fortnight! So here is today’s set eq. v1 to v10.

NB 1 — I have highlighted some parts in red. These usually have a pattern and help in memorising easily.

NB 2 — x is the only variable; e is the exponent; anything else (n, a etc.) are constants.

NB 3 — All logarithms are natural, i.e. to the base e and not 10. This is why we write ln and not log.

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Something to keep you occupied

Here are a bunch of simple problems that you should be able to do following the above table. They’re as easy as Calculus gets. The first one is done for you as an example.

Example:

Solution:

substituting n = 7 in equation v1, we get,

That’s your answer in bold, right there. See how simple that was?

Now here are three problems for you to work out:

Problem 1: [Remember this result. It's a very popular relation for some reason!]

Problem 2:

[Hint: You can split integration between addition. So integration of (a+b) is integration of a + integration of b.]

Alright, I hear you, let’s make this last one even more interesting. So here is a real world problem (well, sort of,) that I came up with. It’s simple, so try it out using integration.

Problem 3: A farmer goes to his well and wants to check the depth of the well till where the water has filled. So he picks up a stone and throws it inside and starts looking at his clock. He hears a splash sometime later when the stone hits the water and sees the time: it has been 20 seconds. How far below is the water?

See how I just gave you a problem you never though could be solved with integration? Well, at least this is one problem that doesn’t begin, integration of something and something else…

Try it out. And in case you’re struggling, here are a couple of hints:

1. Take the starting time as t1 = 0 and position as x1 = 0. The finishing time is t2 = 20s and the finishing position is x2, which you must find. (That’s the depth of the well till it reaches water, right?)

2. Use this definition: velocity is the differentiation of position with respect to time t; and acceleration is the differentiation of velocity with respect to time t. And we know, for all practical purposes, that the acceleration of the stone due to gravity is 10m/s/s, don’t we?

3. Start by using eq. 2 above.

Now solve it!

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I hope you enjoyed today’s session. There wasn’t a lot of maths, but that’s OK. We’ll start slow. But there will be maths tomorrow, mind you. Here’s a glimpse of what we’ll be discussing in the next session:

1. Three methods of integration (including one devilish, illegal short cut maths teachers don’t want you to know)

2. How on earth does one solve this definite integral thingy?