Riemann Sums When we first introduced the integral we stated that there is a link between integration and area. We will further investigate this connection by using Riemann sums. We will see that Riemann sums lead us to the definite integral, which allows us to calculate the area beneath an arbitrary curve, given some interval. We will begin by reviewing summation notation. We use the Greek letter Σ to represent summation in a succinct form. We can write a sample sum in the form 3 X xi i=0 We call i the index of the this sum, and each xi 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 3 X xi = x0 + x1 + x2 + x3 i=0 Where each xi is some value. If we let x0 = 1, x1 = 4, x2 = 3, and x3 = 3 then we would find 3 X xi = x0 + x1 + x2 + x3 = 1 + 4 + 3 + 3 = 11 i=0 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 2 X 1 + i2 = (1 + 0) + (1 + 1) + (1 + 4) = 8 i=0 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 2 X i=0 2 2 + 2i = 2 X i=0 2 2(1 + i ) = 2 2 X 1 + i2 = 2 · 8 = 16 i=0 Our goal to use sums to find the area underneath an arbitrary curve. Suppose that we have the curve given by f (x) = x on the interval [0, 2]. If we draw a line down from f (2) to the x-axis we see that the area between the curve and the x-axis is a triangle. We can easily calculate this area as 1 1 b·h= 2·2=2 2 2 What if we didn’t know this formula for the area under a triangle? We could approximate the area using a number of rectangles underneath the curve. Let us divide the interval [0, 2] into four subintervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2]. The length of each subinterval is ∆x = 0.5. We’ll denote x0 = 0, x1 = 0.5 and so on until x4 = 2, as the points at the boundaries of our subintervals. Since our function is f (x) = x, we find that f (xi ) = xi . Now we’ll assocate a rectangle to each of these subintervals. In one case, we’ll use the function value at the left endpoint for the height of the rectangle, and denote the sum by Il . In the other case, we’ll use the function value at the right endpoint for the height of the rectangle, denoted by Ir . We find that Il = 3 X f (xi ) · ∆x = i=0 3 X xi · 0.5 = 0 · 0.5 + 0.5 · 0.5 + 1 · 0.5 + 1.5 · 0.5 = 1.5 i=0 We can also consider the right-endpoint sum Ir = 4 X i=1 f (xi ) · ∆x = 4 X xi · 0.5 = 0.5 · 0.5 + 1 · 0.5 + 1.5 · 0.5 + 2 · 0.5 = 2.5 i=1 We see that the left approximation is too low, and the right approximation is too high. Looking graphically we see that much of the curve is above the rectangles in the left estimate, and much of it is below the rectangles in the right estimate. What if we decrease the size of our intervals, and correspondingly sum the area of more, smaller rectangles? This should decrease the amount of error, as the rectangles will fill up the area beneath the triangle more accurately. To find the area beneath a general function f over an interval [a, b], using n subintervals, we have n−1 X f (xi )∆x Il = i=0 and Ir = n X f (xi )∆x i=1 where b−a n and the xi 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 the sums Il and Ir defined in this way Riemann sums. 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. We define this as the definite integral. ∆x = Definition: Definite Integral Let f be a bounded function on the interval [a, b]. The definite integral of f from a to b is defined as the limit of the Riemann sums Z b f (x)dx = lim n→∞ a n−1 X f (xi )∆x = lim n→∞ i=0 n X f (xi )∆x i=1 where x0 , . . . , xn are the endpoints of n subintervals dividing [a, b], each of length ∆x = b−a n It turns out that almost all practical functions are Riemann integrable. Z 1√ 1 − x2 dx Example 1 Estimate 0 Solution Since the definite integral is the limit of Riemann sums in the limit that the length of the subintervals go to zero, we can approximate the definite integral using finite sums. For ease of calculation, let us consider subintervals of length 0.25, which will break [0, 1] into four subintervals. We will use both the left and right sums to approximate this integral. Il = 3 X √ √ √ √ 0.25 1 − x2 = 0.25(1 + 1 − 0.252 + 1 − 0.52 + 1 − 0.752 ) ≈ 0.873927 i=0 and Ir = 4 X √ √ √ √ 0.25 1 − x2 = 0.25( 1 − 0.252 + 1 − 0.52 + 1 − 0.752 + 0) ≈ 0.623927 i=1 In thise case the left sums are over the curve, so they lead us to a high estimate. We can compare these estimates to the actual area of the region of interest, if we recognize that we are considering one quarter of a circle of radius 1. We find π ≈ 0.785398 4 Thus, only using a mere four subintervals we have a somewhat reasonable estimate of the area under the curve. Using more subintervals, we could make a much better estimate. Using 8 subintervals, Il ≈ 0.83495 and Ir ≈ 0.709954. This analysis seems to indicate that a mere 50 to 100 subintervals would provide a pretty accurate estimate. There is another thing to note regarding Riemann sums which is rather important. The factor f (xi ) appears in each term of the sum. If all of the f (xi )’s (or enough of them) are negative, then we would find a negative area as the result of the sum. A negative area corresponds to regions that are below the x-axis, as opposed to above it. In light of this fact, let us consider the following example. Z Example 2 Evaluate 2π sin(x)dx 0 Solution As is often useful, let us consider the graph of the sine function. On [0, π] the sine function is positive, and on [π, 2π] the sine function is negative. In fact, the negative portion is the mirror image of the positive portion of the function. What conclusion can we draw? We conclude that Z 2π sin(x)dx = 0 0 because there is exactly as much ‘positive’ and ‘negative’ area summed by the Riemann sum. Using a finite sum we would not necessarily be able to reach this conclusion.