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?
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
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.
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
Until next time, enjoy calculus… muhahahaha! [vhb]
Cover image: Flickr/functoruser