1 5 π 2. 5 π 3. 5 π π x. 5 π 4. Figure 1: We need calculus to find the area of the shaded region.

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1 . Area In order to quantify the size of a 2-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 -ais? (See figure.).75 sin().5.25 Figure : We need calculus to find the area of the shaded region. 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 2-dimensional objects, such as the sine wave in question. Instead, we can think about approimating the complicated shape with shapes that we can find the area of. The simplest and most practical 2-dimensional shape to use here is a rectangle, because it is much more fleible 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 approimation. However, if we use more rectangles, then we can get a better approimation (see figure 2). Just like with Euler s method, we can think of taking a limit as the width of these rectangles approaches, in order to find the eact 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 us continue by trying to approimate the area between the above sine wave and the - ais, over the interval [, π] (so we don t need to worry about what happens when we cross the -ais). For simplicity (and scalability), let s divide the interval into equal subintervals, of length. 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 approimated area function as A(n), where n is the number of subintervals, we will find For 3 subintervals we will find A(2) = sin() π 2 + sin(π/2)π 2 = π A(3) = sin() π 3 + sin(π/3)π 3 + sin(2π/3)π 3 = π(sin(π/3) + sin(2π/3)) 3 = π

2 sin().5 sin() (a) 2 rectangles. (b) 3 rectangles sin().5 sin() (c) rectangles. (d) 2 rectangles. Figure 2: Approimating the area underneath a sine curve. 2

3 We could continue this way to find more accurate approimations, 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() over the interval [a, b]. If we want n subintervals of equal length, then each will be of length = b a n. The approimated area underneath the curve will be A(n) = f(a) + f(a + ) f(a + (n ) ) + f(a + n ). This is a very interesting result. When written in this way, the problem of approimating the area underneath a curve looks very familiar to something we ve already done - approimating the solution to a differential equation. If we consider the differential equation where the value of F (a) is known, we find that df d = f(), F (b) = F (a) + f(a) () + f(a + ) f(a + (n ) ) + f(a + n ), and rearranging the terms we have the result F (b) F (a) = f(a) () + f(a + ) f(a + (n ) ) + f(a + n ). Above is only an approimation to F (b), but we can it as accurate as we like, simply by using small enough intervals. We can see from the above epressions that the process of approimating a solution to a differential equation is the same as approimating the area underneath a curve. Since we can make these approimations 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 (the curve we are finding the area underneath is the function representing the derivative in the differential equation). It follows that we can find the area underneath a curve through antidifferentiation, and conversely, solve differential equations by computing areas. 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()d = cos() + c, we find that the area underneath the sine curve from [, π] is cos(π) ( cos()) = + = 2, which was the value we began to approimate with rectangles. We should note that when we find the area underneath a curve in this way, we are really finding a signed area. In places where the function is above the -ais we have positive area, and in places where the function is below the -ais, we have negative area. Using this reasoning, if we calculate the signed area under a sine wave over [, 2π], we get, because on [, π] the sine function is positive, and on [π, 2π] the sine function is negative, and the negative portion is the mirror image of the positive portion of the function. To find the conventional area between a curve and the -ais we would need to compute the signed area underneath the absolute value of the function of interest, so that all negative area becomes positive. 3

4 In general it is difficult if not impossible to find a closed-form epression for an antiderivative, so it is still worthwhile to investigate the problem of approimating areas in more detail. Before we proceed, it is worth pointing out 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 approimate the area underneath a function, and look in the limit as the length of each rectangle approaches, so that they in a sense become infinitesimal in length. We use the Greek letter Σ to represent summation in a succinct form. We can write a sample sum in the form 3 i. We call i the inde of the this sum, and each 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 inde of the sum. For every different value of the inde, we have a corresponding term. We evaluate this sum by adding each of the terms together. Thus, this sum would be evaluated 3 i = , where each i is some value. If we let =, = 4, 2 = 3, and 3 = 3 then we would find 3 i = = =. 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 inde to appear in the term of a sum as follows 2 + i 2 = ( + ) + ( + ) + ( + 4) = 8. In general we can have some arbitrary mi of both the inde 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 2 = 2 2 2( + i 2 ) = 2 + i 2 = 2 8 = 6. Using this notation we can better define the problem of finding the area underneath a curve. To find the area underneath a function f over an interval [a, b], we partition the interval [a, b] into a number of subintervals. The endpoints of these subintervals are given by a = < < 2 <... < n = b, where the n+ points i define n subintervals. There is no reason in general that these subintervals need be of equal length, and in general we may not want them to be. Nevertheless, for simplicity let us begin with intervals of equal length. It then follows each subinterval will have a length of = b a n. 4

5 Finally, we need to define a height for the rectangle used to approimate the area of f over each subinterval. There is no reason we cannot choose any point within a given subinterval, but it is common, for the sake of simplicity, to choose such points uniformly. Most commonly one chooses to evaluate the function either at the left endpoint, right endpoint, or midpoint of an interval. For this illustration let us choose the left endpoint. In summary, to approimate the area beneath a general function f over an interval [a, b] using n subintervals of equal length, evaluating the function at left endpoints we have n I l = f( i ). Each term of this sum simply consists of the product of the height of the function at the left endpoint f( i ) and the width of the subinterval. Looking at the way is defined, we see that as the number of intervals n increases, the width of the intervals correspondingly decreases. For I l we are summing rectangles with a height equal to the value of the function at the left endpoint of each subinterval, so it is called a left-hand sum. If we look in the limit as n, we find the area underneath the curve, assuming the limit eists. Let s try and apply this machinery to finding the area underneath the curve f() =, over [, b]. We know that it should be b 2 2, because the resulting figure is simply a triangle. We also know this because the antiderivative of is 2 /2, and evaluating the antiderivative at the endpoints of the interval and subtracting yields b 2 /2 = b 2 /2. We should also be able to find the same result by approimating the area using rectangles, and looking in the limit as the length of the rectangles approaches. For n subintervals we find = b/n, and the area is n I l = f( + i ) = n ib n b n n = ib 2 n 2 = b2 n n 2 where we use the fact that for the sum of the first n integers, n n(n + ) i =. 2 i = b2 n(n ) n 2 = b2 2 2 ( n ), In order to find the eact area underneath the curve, we look in the limit as n of the left-hand sum. Doing so lim I b 2 l = lim n n 2 ( n ) = b2 2, which is the familiar result. Our left-hand sum is a special case of a Riemann sum, in which we made a number of simplifying assumptions. For a general Riemann sum the length of the subintervals need not all be the same. Rather than considering the limit as n, we look in the limit as the widths of the subintervals approach. Furthermore, the point within each subinterval at which the function is evaluated can be anywhere. If all possible Riemann sums of this form converge to the same value for a given function (in the limit as the width of subintervals approaches ), the function is said to be Riemann integrable. The Riemann integral is one type of definite integral, written b a f()d. 5

6 Here the definite integral represents the area underneath the function f on the interval [a, b] (see figure 3). We call a the lower limit of integration and b the upper limit of integration. f() b a f()d a b Figure 3: The definite integral is defined as the shaded area. It is often easier to think about an equivalent notion of the integral - the Darbou integral. This notion is equivalent to the Riemann integral in that a function is Riemann integrable if and only if it is Darbou integrable, so both methods define the same definite integral for a function and a given interval. For the Darbou integral we once again consider partitions of arbitrary size, and look in the limit as the width of the partitions approaches. However, rather than considering any arbitrary point within each interval to evaluate the function, we are only interested in two sums - upper and lower sums. In each subinterval, upper sums overestimate the area with each rectangle, and the lower sums underestimate the area of with each rectangle. For an upper sum, the height of a rectangle on a given interval is given by the least upper bound of the function over than interval (the smallest number that is greater than or equal to the function on the interval); lower sums are similarly given by the greatest lower bound of the function over an interval. For a function to be Darbou integrable, we simply require that the upper and lower sums converge to the same limit (if these sums converge to the same limit, all other Riemann sums must as well, because they are sandwiched between these two etreme cases). If the upper and lower Riemann sums don t converge, then we say a function is not Riemann (or Darbou) 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 following result. Theorem.. (Integrability). Let f : [a, b] R be bounded, with a finite number of discontinuities. It follows that f is Riemann integrable. This theorem tells us immediately that an incredible number of functions are Riemann integrable, and it turns out that there are much more powerful theorems than above, but which are difficult to state without developing more mathematical theory. It s worth noting that here we require boundedness of the function - when we are faced with unbounded functions Riemann integrability breaks down. The reason for this is that for an unbounded function the notions of least upper bound and greatest lower bound are meaningless, so our definition of the integral becomes meaningless (later on we will be able to remedy this problem in some cases, using improper integrals). Given that so many functions are Riemann integrable, one might ask the question of what is not Riemann integrable? 6

7 One of the most simple eamples is given as follows. Imagine a function that is for rational numbers, and for irrational numbers. Let us integrate over the interval [, ]. 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, and a lower sum of, so the Riemann integral will not eist. In truth though, there are many many more irrationals than rationals in this interval, so we would epect (using intuition from higher-level mathematics) that the integral should be. 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). 7