categories: mathematics
Hello everyone,
THE TRAPEZOID METHOD
Today we'll show how to easily calculate an approximation of a given area using the trapezoid method, which allows for a better estimate of a given area's value than simply dividing that area into rectangles. So how does this method work? To answer that, we should recall how we last calculated the left, right, or midpoint Riemann sum. We took n rectangles with base \(\Delta x\), then calculated the areas of those rectangles and summed them up (that's the short version). The trapezoid method looks very similar. All you need to do is decide into how many parts we'll divide our area along the horizontal axis, then divide the required segment by that number of segments to get the final length of our \(\Delta x\), and then, based on that, determine both the left and right base of our trapezoid, since the height of our trapezoid is exactly that \(\Delta x\). The formula for the area of a trapezoid is given as: $$At = 0.5 * (a + b) * h$$ So we see that, having this data, we're able to calculate the area of that trapezoid. Knowing the starting point on the OX axis from which we calculate our area, we also know the length of the left base as well as the length of the right base, because if the function under which we want to calculate the area is given as f(x), then the right base will necessarily be equal to: $$f(x_0 + \Delta x)$$ where x0 is the starting point. You'll be able to see a few illustrations below, which I think will additionally help clarify this topic for you, but that's for its own time...
THE RIEMANN SUM IN SIGMA NOTATION
There are, however, two more topics I'd really like to cover, namely: sigma notation and writing the Riemann sum in sigma notation. Let me start in a somewhat unusual way — I want us to begin with an example that can illustrate why we can use the sigma (sum) symbol to calculate the area on a given interval of a function. Let's say we have a function given as: $$f(x) = \sqrt{x}$$. You can picture it as a somewhat truncated logarithmic function with a base greater than 1, since we're not considering it over negative numbers (not at this level). We want to calculate the area of this function using sigma notation on the interval from 1 to 5 using four rectangles of equal \(\Delta x\), so we calculate the total length of the interval as 4 (5 - 1), and then divide it by 4 to get our \(\Delta x\). So our \(\Delta x\) comes out to 1. Now the question arises: do we want our sum to be left- or right-sided? For illustration purposes, let's show what this looks like for the right-sided sum (we'll take the function values at the right edge of \(\Delta x\), i.e. the bottom side of the rectangle). So we start at 1 and then move by \(\Delta x\) = 1, and it's only the value at that argument that interests us for the first rectangle. So the formula will be: 1 + 1, for the next one (\(x_i\)) it will be 1 + 2, for the next one 1 + 3. So we see that in general we can define \(x_i\) as $$x_i = 1 + 1 * i$$, where i represents successive values. The height of the rectangle, on the other hand, is the value of the function at its right edge. $$f(x_i) = \sqrt{x_i} = \sqrt{1 + 1 * i}$$. So the general expression for the area will be of the form: $$1 * \sqrt{1 + 1 * i}$$. Now all that's left is to sum this expression under the sigma symbol, so the sum from i = 1 to 4 of the expression: \(1 * \sqrt{1 + 1 * i}\), i.e.: $$\sum\limits_{i=1}^4 (1 * \sqrt{1 + 1 * i})$$ and that's it — we've just defined a simple Riemann sum using sigma notation 🙂. Alright, let's now discuss the sigma symbol in more detail. The symbol itself, of course, denotes the summation operation in mathematics. Let's say, for example, that we have a sum from n = 1 to 3 of the expression $$2n - 1$$. We see that n is defined for the interval from 1 to 3, so we'll substitute n = 1, n = 2, n = 3 in turn, and then sum the resulting expressions: $$[2 - 1] + [4 - 1] + [6 - 1] = 9$$. Here we call n the summation index, over which we sum the given expressions. Let's represent the whole thing in notation as follows: $$\sum\limits_{i=1}^3 (2i - 1)$$ And that's essentially everything I wanted to show you.
Thanks for watching.
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