Area. A(2) = sin(0) π 2 + sin(π/2)π 2 = π For 3 subintervals we will find

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1 Area In order to quantify the size of a -dimensional object, we use area. Since we measure area in square units, we can think of the area of an object as the number of such squares it fills up. Using this idea we can derive formulas for the area of a square, rectangle, triangle, etc. With a little bit of ingenuity we can also figure out the area of a circle. What about more complicated shapes, such as the area between a sine wave and the x-axis? In order to answer this question, we will need to use the power of calculus. A general technique for finding the area of a complicated shape is to break it up into smaller pieces which have known areas. Unfortunately, it is not possible to do this for most - dimensional objects, such as the sine wave in question. Instead, we can think about approximating the complicated shape with shapes that we can find the area of. The simplest and most practical -dimensional figure to use here is a rectangle, because it is much more flexible than using a square, but it is still very easy to calculate its area. If we only use a few rectangles, they will overlap our function of interest in many places, so it will be a rather crude approximation. However, if we use more rectangles, which are more refined, then we can get a better approximation. Just like with Euler s method, we can think of taking a limit as the width of these rectangles approaches 0, in order to find the exact area of the function or object of interest. The only real limitation we need to be concerned about is being able to compute the area of these rectangles. If we have billions of rectangles, then clearly we will need to use a computer to do the work, but still, computers have their own limitations. Let s begin with a simple illustration. Let s try and approximate the area between a sine wave and the x-axis, over the interval [0, π] (so we don t need to worry about what happens when we cross the x-axis). For simplicity (and scalability), let s divide the interval into equal subintervals, of length x. For convenience, we will use the value of the function at the left endpoint for the height of each rectangle. Finally, we must decide how many subintervals we want. We ll begin with just two, in order to illustrate the process. If we denote the approximated area function as A(n), where n is the number of subintervals, we will find For 3 subintervals we will find A() = sin(0) π + sin(π/)π = π 1.57 A(3) = sin(0) π 3 + sin(π/3)π 3 + sin(π/3)π π(sin(π/3) + sin(π/3)) = = π We could continue this way to find more accurate approximations, but more interesting than the specific case is a slightly more general problem. Let s think about finding the area underneath an arbitrary function f(x) over the interval [a, b]. If we want n subintervals, then each will be of length x = b a n and the approximated area underneath the curve will be A(n) = f(a) x + f(a + x) x f(a + (n 1) x) x + f(a + n x) x

2 This is a very interesting result. When written in this way, the problem of approximating the area underneath a curve looks very familiar to something we ve already done - approximating the solution to a differential equation. If we consider the differential equation df dx = f(x) where the value of F (a) is known, we find that F (b) = F (a) + f(a) (x) + f(a + x) x f(a + (n 1) x) x + f(a + n x) x and rearranging the terms we have the result F (b) F (a) = f(a) (x) + f(a + x) x f(a + (n 1) x) x + f(a + n x) x Above is only an approximation to F (b), but we can it as accurate as we like, simply by using small enough intervals. Rather than focusing on semantics, let us shift our attention to this extremely amazing, yet surprising result: the process of approximating a solution to a differential equation is the same as approximating the area underneath a curve. Since we can make this approximation as accurate as we like, this means that the process of solving a differential equation is the same as finding the area underneath a curve. Because we can solve differential equations through integration, this means that the process of integration is the same as finding the area underneath a curve. In order to integrate (or antidifferentiate) a function, we simply need to find the area beneath the curve. In order to find the area beneath a curve, we simply need to antidifferentiate it. These two seemingly unrelated ideas are actually one in the same! In order to find the area underneath a function over an interval, we simply evaluate the antiderivative of the function at the endpoints and subtract the difference. Noting that sin(x)dx = cos(x) + c we find that the area underneath the sine curve from [0, π] is cos(π) ( cos(0)) = = which was the value we began to approximate with rectangles. We should note that when we find the area undearneath a curve in this way, we are really finding a signed area. In places where the function is above the x-axis we have positive area, and in places where the function is below the x-axis, we have negative area. Using this reasoning, if we calculate the area under a sine wave over [0, π], we get 0, because on [0, π] the sine function is positive, and on [π, π] the sine function is negative, and the negative portion is the mirror image of the positive portion of the function. It follows that the integral is 0 because there is exactly as much positive and negative area. To find the conventional area between a curve and the x-axis, we need to look at the integral of the magnitude of a function, so we don t have anymore negative area. Not only can we use integration to find the area underneath curves, but if we can become more proficient at finding the area underneath curves, we can think of numerically finding

3 antiderivatives by finding the area underneath a curve. In order to become more proficient, we will need to become familiar with summation. First, we note that the reason for the notation of the indefinite integral should now be clear - in the process of integration, we want to sum up rectangles to approximate the area underneath a function, and look in the limit as the length of each rectangle approaches 0, so that they in a sense become infintessimal in length. We use the Greek letter Σ to represent summation in a succinct form. We can write a sample sum in the form We call i the index of the this sum, and each x i is a single term in the sum. There is nothing unique about the choice of i, and in general we can use whatever variable we like for the index of the sum. For every different value of the index, we have a corresponding term. We evaluate this sum by adding each of the terms together. Thus, this sum would be evaluated x i x i = x 0 + x 1 + x + x 3 Where each x i is some value. If we let x 0 = 1, x 1 = 4, x = 3, and x 3 = 3 then we would find x i = x 0 + x 1 + x + x 3 = = 11 It is noteworthy that for sums with a finite number of terms, it does not matter in which order the terms are added. If one is considering infinite sums however, the order does matter. It is also possible for the index to appear in the term of a sum as follows 1 + i = (1 + 0) + (1 + 1) + (1 + 4) = 8 In general we can have some arbitrary mix of both the index and other factors in each term. Finally, if we have a constant factor in every term of a sum, we can factor the constant from the sum, evaluate the sum, and multiply the result by that factor in the end. Thus + i = (1 + i ) = 1 + i = 8 = 16 Using this notation we can better define the problem of finding the area underneath a curve. To find the area beneath a general function f over an interval [a, b], using n subintervals, we have I l = f(x i ) x

4 and I r = n f(x i ) x i=1 where x = b a n and the x i are the endpoints of our subintervals. Looking at the way x is defined, we see that as the number of intervals n increases, the width of the intervals x correspondingly decreases. We call I l the left-hand sum, and I r the right-hand sum, which are both special cases of Riemann sums, which we will discuss in more detail soon. As we increase n, both the right-hand and left-hand sums become closer to the area underneath the curve of interest, and if we look in the limit that n, it turns out the sums take on the same value, which is exactly the area underneath the curve. Let s try and apply this machinery to finding the area underneath the curve f(x) = x, over [0, b]. We know that it should be b because the resulting figure is simply a triangle. We also know because the antiderivative of x is x /, and evaluating the antiderivative at the endpoints of the interval and subtracting yields b / 0 = b /. We should also be able to find the same result by approximating the area using rectangles, and looking in the limit as the length of the rectangles approaches 0. For n subintervals we find x = b/n, and the area is I l = f(0 + i x) x = ib n b n 1 n = ib n = b n i = b n(n 1) n = b (1 1 n ) where we use the fact that for the sum of the first n integers, n i = n(n + 1) In order to find the exact area underneath the curve, we look in the limit as n of the left-hand sum. Doing so lim I b l = lim n n (1 1 n ) = b which is the familiar result. Using right-hand sums we would find the exact same thing. We stated above that I l and I r are special cases of Riemann sums. This leads us to the question of what a general Riemann sum is. For a general Riemann sum the length of the subintervals need not all be the same length. Rather than considering the limit as n, we look in the limit as the widths of the subintervals approach 0. Also, we look at both upper and lower sums, where the upper sums overestimate the area with each rectangle, and the lower sums underestimate the area of with each rectangle. Finally, we impose the

5 condition that the upper sums and lower sums both converge to the same limit. When this condition is met, we say a function is Riemann integrable, and define the definite integral as the limit of Riemann sums. The definite integral of a function is written b a f(x)dx and the definite integral represents the area underneath the function f on the interval [a, b]. We call a the lower limit of integration, and b the upper limit of integration. If the upper and lower Riemann sums don t converge, then we say a function is not Riemann integrable, and in a sense we cannot define the area underneath the curve in this way. In truth, one must continue far into the study of mathematics and science to begin to see the shortcomings of the Riemann integral. For now we must suffice ourselves with the fact that all bounded functions with finitely many discontinuities are Riemann integrable (in fact, even more functions that this are Riemann integrable). If so many functions are Riemann integrable, one might ask the question what is not Riemann integrable? One of the most simple examples is given as follows. Imagine a function that is 0 for rational numbers, and 1 for irrational numbers. Let us integrate over the interval [0, 1]. Since between every two rational numbers is an irrational, and between every two irrationals a rational, over every single interval we will an upper sum of 1, and a lower sum of 0, so the Riemann integral will not exist. In truth though, there are many many more irrationals than rationals in this interval, so we would except (using intuition from higher-level mathematics) that the integral should be 1. In order to understand these issues better, one must focus on studying analysis, and eventually measure theory, in which a more powerful version of the integral can be constructed (the Lebesgue integral).