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 —

Spot the difference — *any difference*

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