# What’s integration?

What’s the area of a rectangle? Easy! Width times height. Another one: what is the area of a circle? Yes! pi times the radius squared. What’s the area under a parabola? Easy!… well, not really (for now!).

This is a graph of the function f(x) = x². The region in red is the area we want to calculate, and as you can see it’s bound between x=0.5 and x=1. That’s not a shape we know. It is not a circle nor a rectangle nor a triangle. It does look like a trapezoid though, the only difference is the diagonal which is not straight but is curved.

So maybe we can approximate that area knowing that the area of a trapeziod is A = h·(B + b)/2. As you can see on the graph, h is 0.5, B is 1 (f(1) = 1² = 1) and b is f(0.5) = 0.5² = 0.25, so we know that the approximate area is 0.3125. At this point I can tell you that the exact area is 0.29166… Close enough! The approximated area is greater than the actual area because, as you can see on the graph, there is a space between the diagonal and the curve, which means that we’ve added another area!We can do better! Imagine you were to divide that shaded region on the graph in many rectangles, all with the same width but with different heights, in order to fit each of them in the shape.So, we have 5 rectangles, differing only by their height, which depends on the x value. The idea is to sum up the area of each rectangle to estimate the area under the curve. As you know, the area of a rectangle is width times height, where the width (x-axis) is always the same and the height (y-axis) is expressed in terms of the function, f(x) = x² in this case. f(x) = y, so to find the height of a rectangle at x = 1 we have to substitute the x in f(x) with 1. h = y = f(1) = 1² = 1. In fact, if you look at the graph, you can see that at x = 1 corresponds y = 1.

In this case, the width of each rectangle is 0.1, and you can verify this with the formula (ba)/n, where b is the value of x at which the area you want to calculate “ends”, a is the value of x at which the area “starts” and n is the number of rectangles. a and b are called “the bounds”.Here, a is 0.5, b is 1 and n is 5. (1-0.5)/5 = 0.5/5 = 0.1 . This value (which is, again, the width of each rectangle) is called Δx, which means “variation” or “difference”. So,

$A&space;\approx\sum_{i=1}^{n}f(x_{i})\Delta&space;x$

which means “sum all of the f(xᵢ)·Δx‘s with i going from 1 to n, which in this case is the number of rectangles. This way, the i increases by one everytime until it reaches n. This is exactly what happens: f(x₁)·Δx + f(x₂)·Δx + f(x₃)·Δx + f(x₄)·Δx + f(x₅)·Δx.Remember, the exact area is 0.29166 (I will explain you how I calculated it in the next lesson!) and you can see that the trapezoid area is closer to this number that it is the sum of the areas of the rectangles. So you might think: great! So I just need to use the first method for every curve! Less work but, what’s the point of this lesson? Well, give a look at this curve:

There is no way we can estimate the area with a trapezoid, square or anything else, and even if we could find a way to fit them, it would still be very inaccurate. We want the exact area! No estimation. Let’s go back to the rectangles method. If we could draw many rectangles, with the same technique as before, just with a much smaller Δx, they would fit perfectly under any curve. Let’s take the first example:

So much better! Now we have 12 rectangles within the same bound, which means Δx is 0.04166…; what we want is the lowest Δx possible, which means the highest number of rectangles. Now comes the concept of limit: Δx must be extremely small, or in other words, it must approach 0, but without ever reaching it. So we say that Δx tends to 0. This means that n tends to infinity. When this happens, we have the exact area under the curve.

$A&space;=&space;\lim_{n&space;\to&space;\infty}\sum_{i=1}^nf(x_i)\Delta&space;x$

This is called the Riemann sum, where:

$x_i&space;=&space;i&space;\Delta&space;x$ $\Delta&space;x&space;=&space;\frac{b-a}{n}$

Therefore:

$A&space;=&space;\lim_{n&space;\to&space;\infty}\sum_{i=1}^nf(i\frac{b-a}{n})\frac{b-a}{n}$

and

$A&space;=&space;\lim_{n&space;\to&space;\infty}\frac{b-a}{n}\sum_{i=1}^nf(i\frac{b-a}{n})$

which is the definition of definite integral;

$\int_a^bf(x)dx=\lim_{n&space;\to&space;\infty}\sum_{i=1}^nf(x_i)\Delta&space;x$

or

$\int_a^bf(x)dx=\lim_{n&space;\to&space;\infty}\frac{b-a}{n}\sum_{i=1}^nf(i\frac{b-a}{n})$

and it is read “integral from a to b of f of x with respect to x“. When you integrate, you integrate with respect to a variable, usually x, and the rest is treated like a constant.

If you notice, f(x)dx looks a lot like f(x)Δx. That’s in fact true! When Δx tends to 0, it becomes dx:

$\lim_{\Delta&space;x&space;\to&space;0}&space;\Delta&space;x&space;=&space;dx$

Now, I have talked about definite integral, which is used to calculate the numerical value of the area under a curve. But what if I want to know the formula of the area under a curve? Here comes the indefinite integral, called this way because the bounds of integration are not defined. In fact, a definite integral is treated the same way as an indefinite integral (except in some cases, which you will learn in the future). If you see a definite integral, first you calculate the result of the indefinite integral, then you plug in the values of a and b into the x. But this means that to calculate an indefinite integral you can’t use the Riemann sum, because it requires the bounds, which are absent in this kind of integral.

$F(x)=\int&space;f(x)&space;dx$

F(x) is the integral of f(x) and it is not the area under the graph of f(x), but the formula of the area. F(x) is called “primitive of f(x).

As an example, I will tell you that the formula for the area under the parabola of function f(x) = x² is x³/3, which means that

$F(x)=\int&space;x^{2}&space;dx=\frac{x^{3}}{3}+C$

The C is called additive constant, which I will talk about when I explain how to calculate an integral.

If I want to know the value of the area of this parabola from 0.5 (a) to 1 (b), I will just have to plug in the values of a and b into x³/3; but how?

$\int_a^b&space;f(x)dx=\left&space;[F(x)\right&space;]_{a}^{b}=F(b)-F(a)$

In this example, we have

$\int_{0.5}^1&space;x^{2}dx=\left&space;[\frac{x^{3}}{3}\right&space;]_{0.5}^{1}=\frac{1^{3}}{3}-\frac{0.5^{3}}{3}=0.29166...$

Now you should have understood what integration is and be ready for the next lesson, with which you will be able to actually calculate easy integrals and then move on to tougher ones!

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