categories: mathematics
Hello everyone,
In the last piece we managed to present the Riemann sum in sigma notation. However, there remained a certain gap related, of course, to the question: "What is it even for?".
Do you remember when in the first article we touched on the topic of what an integral actually is? Well, we can define an integral in general as the area under the graph of a given function, determined by the sum of the areas of n shapes that we can define under that graph on a given interval (with respect to the definite integral). So imagine we have some region that we managed to express in sigma notation. We then have n rectangles (if that's the shape we chose) on a specific interval of the OX axis. And now here comes the magic: if we assume that this n tends to infinity (a limit), then the limit of that Riemann sum, defined as a function of n, will give us the exact value of the area under the graph.
Alright, time to move on to concrete examples instead of wandering in the clouds. Let's say we're looking for the exact area under the graph of the function: $$\frac{1}{5} * (x^2)$$, on the interval from 2 to 6. As always, let's first determine the length of our interval on the OX axis, which will be 6 - 2 = 4 = L, so our \(\Delta x\), since we're dividing this region into n equal parts, will be $$\frac{4}{n} = \Delta x$$. Now we need to decide whether we'll compute the right, left, or midpoint Riemann sum... I've become quite skilled with the right-hand one, so I'll choose that here. We know that the height of each rectangle will give us the value \(f(x_i)\), where \(x_i\) is of course the value of the function's argument on the OX axis at the bottom-right corner of that rectangle. So we have: $$x_i = 2 + \Delta x * i = 2 + \frac{4}{n}*i$$. So we see that the value of our function will be given as follows: $$f(x_i) = \frac{1}{5} * (x_i)^2 = \frac{1}{5} * (2 + \frac{4}{n} * i)^2$$. And now, since we established that we want the exact value of the area, we need n rectangles, so we'll sum from i = 1 to n the expression $$(2 + \frac{4}{n}) * i)^2 * \frac{4}{5n}$$. Then we just add the limit symbol $$\lim_{n \to \infty}$$ to this sum. And that's essentially everything on this topic :)
Since I want to pack as much content into as little time as possible, and keep the theory to a minimum while giving you as many examples as possible, there's still one more thing we need to cover in this article. I decided to also tell you about the FUNDAMENTAL THEOREM OF CALCULUS (a moment of silence...). I capitalized that for good reason. On the scale of importance for understanding integrals, this theorem deserves a solid 10/10. It allows for incredibly simple solving of definite integrals on ANY interval. I know, sounds great ;) So let's talk about how to apply this beast in practice. I want to point out right away that I won't be giving you formulas to memorize by rote — memorize it however you like, as long as you're able to apply it correctly :) So let's say you have a definite integral on the interval from 0 to 10, say of the function $$f(x) = 1 * x$$. I deliberately chose such an easy function to show you the idea. To calculate such an integral (the area under the graph of this function on the interval from 0 to 10), you first need to calculate the antiderivative of the function 1 * x. This operation of finding the antiderivative of a given function is exactly what's called finding the indefinite integral of that function. This operation is the inverse of finding the derivative of a given function, if that helps you understand it (that's why I told you it's best to memorize those derivative formulas by heart, because they really are VERY USEFUL!). So what will the antiderivative of the function x be? Let's think... What function, after being differentiated, gives us x as a result? Of course, it's $$\frac{x^2}{2}$$, because if you now subtract one from the exponent of this function and move the number from the exponent in front of x (according to the definition of computing a derivative), the twos will cancel out and you'll get just 1 * x. And there you go, you've just determined the indefinite integral of the function x! Now, having knowledge of the indefinite integral of this function, you need to perform a subtraction over the given interval, in this case from 0 to 10, so: $$\frac{10^2}{2} - \frac{0^2}{2} = 50$$. So the area under the graph of the function x on the interval from 0 to 10 (the definite integral) equals 50. Let's now recap how to apply this theorem in practice (fine... ;) I'll give you the "statement" of this theorem at the end of the article, read it if you feel like it ;) ).
1) Find the indefinite integral (antiderivative) of the given function.
2) Compute the difference of this antiderivative over the initial interval
3) Write down the result and voila!
So let's now talk about the formal definition of what we applied in practice: "If we define a function as the definite integral of another function f, then the new function is an antiderivative of f. So to find the definite integral of f on the interval from a to b, you need to find an antiderivative of f, denoted F, and calculate F(b) - F(a)". However, as I already said, this is just another clever little rule, of no use to anyone unless you understand the rule itself and how to apply it. Thanks for reading.
Cheers ;)
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