Math 409-502Harold P. [email protected] of the integralSuppose f is a bounded function on [a, b]. The upper sum for any partition is at least as big asthe lower sum for any other partition. (Compare both sums to the upper sum and the lowersum for a common refinement.)Therefore the infimum of the upper sums for all possible partitions is at least as big as thesupremum of the lower sums for all possible partitions.If f is integrable, then these two numbers are equal. The value is defined to be the integral of f ,usually denoted byRbaf (x) dx.This definition is not usually a convenient way to compute an integral!Math 409-502 November 22, 2004 — slide #2Riemann sumsA Riemann sum for a bounded function f and for a partition of an interval [a, b] has the formn∑k=1f (ck)(xk− xk−1), where each ckis some point in the subinterval [xk−1, xk].Some examples are left-hand endpoint sums, right-hand endpoint sums, and midpoint sums.For a fixed partition, every Riemann sum is between the upper sum and the lower sum for thatpartition.Therefore the integralRbaf (x) dx can be computed as a limit of Riemann sums as the mesh sizeof the partition goes to 0.Math 409-502 November 22, 2004 — slide #3ExampleShow thatR1−1sin(x) dx = 0.(The same argument will apply to any odd function.)Since the function sin(x) is integrable (because sin(x) is a continuous function), it suffices toconsider partitions that are uniformly spaced and symmetric about 0.No upper sum equals 0 and no lower sum equals 0. But choosing symmetrically locatedpoints ckmakes the Riemann sum equal to 0.Since the integral is a limit of Riemann sums each of which is 0, the integral has the value 0.Math 409-502 November 22, 2004 — slide #4Fundamental theorem of calculusVersion 1. If f is integrable on [a, b] and if F is a function whose derivative equals f , thenRbaf (x) dx = F(b) − F(a).Proof. Consider any partition a = x0, x1, . . . , xn= b of [a, b]. Write F(b) − F(a) as a telescopingsum: (F(xn) − F(xn−1))+ (F(xn−1) − F(xn−2))+ · · · + (F(x1) − F(x0)).By the mean-value theorem, the sum equals F0(cn)(xn− xn−1)+ F0(cn−1)(xn−1− xn−2)+ · · · +F0(c1)(x1− x0), which is the same as f (cn)(xn− xn−1) + f (cn−1)(xn−1− xn−2) + · · · + f (c1)(x1−x0).This is a Riemann sum for f , so by taking partitions with sufficiently small mesh we get arbi-trarily close toRbaf (x) dx.Math 409-502 November 22, 2004 — slide #5Homework• Read sections 19.1–19.3, pages 251–256.• Work on proving various versions of l’Hˆopital’s rule.Math 409-502 November 22, 2004 — slide #6Sketch of proof of l’Hˆopital’s ruleThis sketch applies to the case x → a, limx→af (x) = 0 = limx→ag(x), limx→af0(x)g0(x)= L.[Remark: f and g need not be defined at a.]Fix e > 0. By hypothesis, there exists δ > 0 such that¯¯¯f0(x)g0(x)− L¯¯¯< e/2 when 0 < |x − a| < δ.Fix such an x. By hypothesis, there exists x2so close to a thatf (x)g(x)differs fromf (x)− f (x2)g(x)− g(x2)by lessthan e/2.By Cauchy’s version of the mean-value theorem, that fraction equalsf0(c)g0(c)for some c. Hencef (x)g(x)differs from L by less than e when 0 < |x − a| < δ.Technical complication: are all the denominators non-zero?Math 409-502 November 22, 2004 — slide