More Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologil Sienes nd Deprtment of Mthemtil Sienes Clemson University Februry 15, 2018 Outline More Riemnn Integrl Properties The Fundmentl Theorem Of Clulus Antiderivtives
We first wnt to estblish the fmilir summtion property of the Riemnn integrl over n intervl [, b] = [, ] [, b]. Most of the tehnil work for this result is done in the following Lemm. Lemm Let f B[, b] nd let (, b). Let f (x) dx = L(f ) nd f (x) dx = U(f ) denote the lower nd upper Drboux integrls of f on [, b], respetively. Then we hve f (x)dx = f (x)dx + f (x)dx f (x)dx = f (x)dx + f (x)dx. We prove the result for the upper integrls s the lower integrl se is similr. Let π be given by π = {x0 =, x1,..., xp = b}. (Cse (1): is prtition point): Thus, there is some index 1 k0 p 1 suh tht xk0 =. For ny intervl [α, β], let Uα(f β, π) denote the upper sum of f for the prtition π over [α, β]. Now, we n rewrite π s π = {x0, x1,..., xk0} {xk0, xk0+1,..., xp}. Let π1 = {x0,..., xk0} nd π2 = {xk0,..., xp}. Then π1 Π[, ], π2 Π[, b], nd U b (f, π) = U (f, π1) + U b (f, π2) f (x)dx + f (x)dx, by the definition of the upper sum.
(Cse is not in the prtition): Now, if is not in π, we n refine π by dding, obtining the prtition π = {x0, x1,...,,, xk0+1,..., xp}. xk0 Splitting up π t s we did before into π1 nd π2, we see tht π = π1 π2 where π1 = {x0,..., xk0, } nd π2 = {, xk0+1,..., xp}. Thus, by our properties of upper sums, we see tht U b (f, π) U b (f, π ) = U (f, π1)+u b (f, π2) f (x)dx + f (x)dx. Now ombine the ses to find for ny prtition π, we hve U b (f, π) f (x)dx + f (x)dx, whih implies tht f (x)dx f (x)dx + f (x)dx. Now we wnt to show the reverse inequlity. Let ɛ > 0 be given. By the definition of the upper integrl, there exists π1 Π[, ] nd π2 Π[, b] suh tht U (f, π1) < f (x)dx + ɛ 2, Ub (f, π2) < f (x)dx + ɛ 2. Let π = π1 π2 Π[, b]. It follows tht U b (f, π) = U (f, π1) + U b (f, π2) < f (x)dx + f (x)dx + ɛ.
But, by definition, we hve for ll π. Hene, we see tht f (x)dx U b (f, π) f (x)dx < f (x)dx + f (x)dx + ɛ. Sine ɛ ws rbitrry, this proves the reverse inequlity we wnted. We n onlude, then, tht f (x)dx = f (x)dx + f (x)dx. Theorem If f RI [, b] nd (, b), then f RI [, ] nd f RI [, b]. Let ɛ > 0 be given. Then there is prtition π 0 Π[, b] suh tht U b (f, π) L b (f, π) < ɛ for ny refinement, π, of π 0. Let π 0 be given by π 0 = {x 0 =, x 1,..., x p = b}. Define π 0 = π0 {}, so there is some index k0 suh tht x k0 x k0+1. Let π 1 = {x 0,..., x k0, } nd π 2 = {, x k0+1,..., x p}.
Then π1 Π[, ] nd π2 Π[, b]. Let π 1 be refinement of π1. Then π 1 π2 is refinement of π0, nd it follows tht U (f, π 1) L (f, π 1) = (Mj mj) xj π 1 (Mj mj) xj π 1 π2 But, sine π 1 π2 refines π0, we hve U b (f, π 1 π2) L b (f, π 1 π2). implying tht U b (f, π 1 π2) L b (f, π 1 π2) < ɛ, U (f, π 1) L (f, π 1) < ɛ for ll refinements, π 1, of π1. Thus, f stisfies Riemnn s riterion on [, ], nd f RI [, ]. The proof on [, b] is done in extly the sme wy. Theorem If f RI [, b] nd (, b), then f (x)dx = f (x)dx + f (x)dx. Sine f RI [, b], we know tht f (x)dx = f (x)dx.
Further, we lso know tht f RI [, ] nd f RI [, b] for ny (, b). Thus, f (x)dx = f (x)dx = f (x)dx f (x)dx. So, pplying the Lemm, we onlude tht, for ny (, b), f (x)dx = = f (x) dx = f (x) dx + f (x) dx f (x) dx + f (x) dx. Theorem Let f RI [, b]. Define F : [, b] R by F (x) = x f (t)dt. Then (i) F C[, b]; (ii) if f is ontinuous t [, b], then F is differentible t nd F () = f (). First, note tht f RI [, b] f R[, x] for ll x [, b], by our previous results. Hene, F is well-defined. We will prove the results in order. (i): Now, let x, y [, b] be suh tht x < y. Then inf [x,y] f (t) (y x) y x f (t)dt sup [x,y] f (t) (y x)
whih implies tht y F (y) F (x) = f (t)dt f (y x). x A similr rgument shows tht if y, x [, b] stisfy y < x, then y F (y) F (x) = f (t)dt f (x y). x Let ɛ > 0 be given. Then if we hve ɛ x y < f +1, f F (y) F (x) f y x < f +1 ɛ < ɛ. Thus, F is ontinuous t x nd, onsequently, on [, b]. (ii): Finlly, ssume f is ontinuous t [, b], nd let ɛ > 0 be given. Then there exists δ > 0 suh tht x ( δ, + δ) [, b] implies f (x) f () < ɛ/2. Pik h suh tht 0 < h < δ nd + h [, b]. Let s ssume, for onreteness, tht h > 0. Define m = inf f (t) nd M = sup f (t). [,+h] [,+h] If < x < + h, then we hve x ( δ, + δ) [, b] nd ɛ/2 < f (x) f () < ɛ/2. Tht is, f () ɛ 2 < f (x) < f () + ɛ 2 x [, + h].
Hene, m f () ɛ/2 nd M f () + ɛ/2. Now, we lso know tht Thus, we hve +h mh f (t)dt Mh. +h F ( + h) F () f (t)dt = f (t)dt +h f (t)dt =. h h h Combining inequlities, we find yielding f () ɛ 2 if x [, + h]. F ( + h) F () m M f () + ɛ h 2 F ( + h) F () f () h ɛ 2 < ɛ Thus, sine ɛ ws rbitrry, this shows (F + () = f (). The se where h < 0 is hndled in extly the sme wy whih tells us (F () = f (). Combining, we hve F () = f (). Note tht if = or = b, we need only onsider the definition of the derivtive from one side. Comment We ll F (x) the indefinite integrl of f. F is lwys better behved thn f, sine integrtion is smoothing opertion. We n see tht f need not be ontinuous, but, s long s it is integrble, F is lwys ontinuous. The next result is one of the mny men vlue theorems in the theory of integrtion. It is more generl form of the stndrd men vlue theorem given in beginning lulus lsses.
Theorem The Men Vlue Theorem For Riemnn Integrls: Let f C[, b], nd let g 0 be integrble on [, b]. Then there is point, [, b], suh tht f (x)g(x)dx = f () g(x)dx. Sine f is ontinuous, it is lso integrble. Hene, fg is integrble. Let m nd M denote the lower nd upper bounds of f on [, b], respetively. Then mg(x) f (x)g(x) Mg(x) for ll x [, b]. Sine the integrl preserves order, we hve m g(x)dx f (x)g(x)dx M g(x)dx. If the integrl of g on [, b] is 0, then the integrl of fg is 0. Hene, in this se, hoose ny [, b]. If the integrl of g is not 0, then it must be positive, sine g 0. Hene, we hve, in this se, m f (x)g(x)dx b g(x)dx M. Now, f is ontinuous, so it ttins M nd m t some points. Hene, by the Intermedite Vlue Theorem there is [, b] with This implies the desired result. f () = f (x)g(x)dx b g(x)dx.
The next result is nother stndrd men vlue theorem from bsi lulus. This result n be interpreted s stting tht integrtion is n verging proess. Theorem Averge Vlue For Riemnn Integrls If f C[, b], then there is point [, b] suh tht 1 b f (x)dx = f (). Apply g(x) = 1 in the previous theorem. We need better wys to lulte Riemnn Integrls. A gret wy to do it strts with the ide of ntiderivtives. Definition Let f : [, b] R be bounded funtion. Let G : [, b] R be suh tht G exists on [, b] nd G (x) = f (x) for ll x [, b]. Suh funtion is lled n ntiderivtive or primitive of f. The ide of n ntiderivtive is intelletully distint from the Riemnn integrl of bounded funtion f. Consider the following funtion f defined on [ 1, 1]. f (x) = { x 2 sin(1/x 2 ), x 0, x [ 1, 1] 0, x = 0
It is esy to see tht this funtion hs removble disontinuity t 0. Moreover, f is even differentible on [ 1, 1] with derivtive f (x) = { 2x sin(1/x 2 ) (2/x) os(1/x 2 ), x 0, x [ 1, 1] 0, x = 0 Note f is not bounded on [ 1, 1] nd hene it n not be Riemnn Integrble. Now to onnet this to the ide of ntiderivtives, just relbel the funtions. Let g be defined by g(x) = { 2x sin(1/x 2 ) (2/x) os(1/x 2 ), x 0, x [ 1, 1] 0, x = 0 then define G by G(x) = { x 2 sin(1/x 2 ), x 0, x [ 1, 1] 0, x = 0 We see tht G is the ntiderivtive of g even though g itself does not hve Riemnn integrl. Agin, the point is tht the ide of the ntiderivtive of funtion is intelletully distint from tht of being Riemnn integrble. Homework 13 13.1 You n guess the ntiderivtive of x is x 2 /2 + C. Use indution to proves the ntiderivtive of x n for integers n 1 is x n+1 /(n + 1) + C. 13.2 You n guess the ntiderivtive of 1/x 2 is 1/x + C. Use indution to proves the ntiderivtive of x n for integers n < 1 is x n+1 /(n + 1) + C. 13.3 Wht is the most generl ntiderivtive of 1/x? 13.4 Let f be defined by f (x) = { x sin(1/x), x (0, 1] 0, x = 0 Prove 1 0 f (x)dx 1/2.