## Sma iUj A it

Next, choose a set of w, e |.v, _ i, .v, | values to generate the largest possible value for the Riemann Sum that is. choose ft), such that /(<u,) > /(x) for all x e [xi-1, a, |. Let us refer to these values as &>, and the sum they generate as the upper sum SmM, where n

For our example, it is easy to see that Lire values «¿( are found by choosing the leftmost point of each subinterval, while the values <!>, are found by choosing the rightmost point of each subinterval (see figure 16.4). The result is that

<Smin = 0(0.25) + 0.5(0.25) + 1(0.25) 4 1.5(0.25) = 0.75

while

¿W = 0.5(0.25) + 1(0.25) + 1.5(0.25) + 2(0.25) = 1.25

Now it is clear that 5min is always an underestimate of the area under the curve, while is always an overestimate of the area under the curve. If, upon using finer and finer partitions, we discover that the values for imin and 5mu, both converge to the same number. 5*. then it makes sense intuitively to say that the area under the curve is well defined and that S* is in fact the value of this area. For our example an increase in n leads to a finer partition t A, = I/« V/') and as n —► og the values for 5„,jn and 5m:iv do in fact converge to the same value.

To see this formally, note that we choose the u>i values to generate 5m,„ by choosing the leftmost points of the subintervals [,v,- -1, .v,J, namely w, = jr,_|. For the partition jc, = i/n, where i = 0, 1, 2,3 n. this gives n n

Now. since

we get

Similarly we choose the a>, values to generate 5„,ax by choosing the rightmost points of the subintervals ( v,_|. x, J. namely d>; = x,. For the partition x, = i/n.

it follows that

Notice that, as expected. <5,niix > 5,„ln. However, as the partition becomes arbitrarily fine (i.e., as n —► oc) the two values converge to the same limit:

If this is the case, we say that the function is integrable and that this limit is the area under the curve.

Definition 16.3

A function is said to be integrable on the closed interval (a, h] if for every e > 0. there is some value \$> 0 such that

i=t for any partition on (r/,6J such that max A, < ft (i.e.. the length of the largest subinterval of the partition is less than ,5) and for any selection of points to, e i a, |. »,). We call this value the definite integral of /(a) over the interval |c/. b\ and write f j, i

Notice that for a function to be integrable on an interval («. b] it must be the case that as one lakes finer and finer partitions of the interval |a 6], using any set of points w- within the subintervals, the Riemann sum must become arbitrarily close (converge) to some value L. In our example, as n —► oo, ihe choices of ¡o, leading to either the largest possible or smallest possible values of S converge to the same value. Thus the Riemann sum converges to this same value for any arbitrary choice of tu, values.

In regards to the notation f(x)dx, the sign / resembles the letter S and implies a summation process, the value a is called the lower limit and b the upper limit of the integral, referring to the end points of the interval |a.b], and dx refers to the interval widths. A,. in the expression for¿>.

The method of finding the area under a curve using this technique is tedious, as even the si mple example above shows. However, as long as a function is continuous, we can apply the fundamental theorem of integral calculus, which is presented below.

Theorem 16.1 (Fundamental theorem of integral calculus) If the function ./"(v) is continuous on the closed interval |u. /?] and if Fix) is any antiderivative (indefinite integral) of f{x). dien f(x)dx = F(b) - F(a)

where Fib) is the antiderivative offix) evaluated at the point x = b and F(u) is the antiderivative of fix) evaluated at the point X = a. The expression [ Fib) -F(a) I is often denoted by | Fix)\ha.

Notice in this theorem that we ignore the constant of integration since, if included, it would in any case be subtracted out; that is, |Fib) + CI — |F(a) + C] -Fib) - F(a).

For our example of finding the area under the curve fix) = 2x over ihe interval |(), 1 ], the fundamental theorem of integral calculus clearly "works." That is, since Fix) = x2 is an antiderivative of fix) = 2x. we can compute Ihe area according to the expression

0 0