G THE DEFINITE INTEGRAL

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1 Theory Supplement Section G 35 G THE DEFINITE INTEGRAL Recll tht if f is continuous on [, b] the definite integrl is given by limit of left or right sums: n f(x) dx = lim f(x i ) x = lim n i= n i= f(x i ) x. This provides method for pproximting definite integrls numericlly. 6 In this section we give forml definition of the definite integrl tht mkes use of more generl sums. A Specil Cse: Monotonic Functions A function which is either incresing throughout n intervl or decresing throughout tht intervl is sid to be monotonic on the intervl. In Section 5. we sw tht if f is monotonic, the left nd right sums trp the exct vlue of the integrl between them. Let us consider Exmple on pge 249 of the textbook, which looks t the vlue of 2 t dt. The left- nd right-hnd sums for n = 2,, 5, nd 25 re listed in Tble G.. Becuse the function f(t) = /t is decresing, the left-hnd sums converge to the integrl from bove, nd the right-hnd sums converge from below. From the lst row of the tble we cn deduce tht < dt <.694, t so 2 t dt.69 to two deciml plces. The Difference Between the Upper nd Lower Estimtes To be sure tht the left- nd right-hnd sums trp unique number between them, we need to know tht the difference between them pproches zero. On pge 245 we sw tht for monotonic function f on the intervl [, b]: Difference between upper nd lower estimtes = f(b) f() t, where t = (b )/n. We cn mke this difference s smll s we wnt by choosing n lrge enough, which mkes t smll enough. Tble G. Left- nd right-hnd sums for 2 t dt n Left-hnd sum Right-hnd sum In prctice, we often pproximte integrls using more sophisticted numericl methods.

2 36 Theory Supplement Section G Tble G.2 Left- nd right-hnd sums for 2.5 sin(t 2 ) dt n Left-hnd sum Right-hnd sum When f Is Not Monotonic If f is not monotonic, the definite integrl is not lwys brcketed between the left- nd righthnd sums. For exmple, Tble G.2 gives sums for the integrl 2.5 sin(t 2 ) dt. Although sin (t 2 ) is certinly not monotonic on [, 2.5], by the time we get to n = 25, it is pretty cler tht 2.5 sin (t 2 ) dt.43 to two deciml plces. Notice, however, tht.43 does not lie between.25 nd.285, the left- nd right-hnd sums for n = 2, or even between.464 nd.453, the two sums for n =. If the integrnd is not monotonic, the left- nd right-hnd sums my both be lrger (or smller) thn the integrl. (See Problems 2 nd 3 for more exmples of this behvior.) Defining The Definite Integrl by Upper nd Lower Sums To prove properties bout integrls, it is esier if we llow the subintervls to hve different lengths. We let x i be the length of the i-th intervl, nd mke the following definition: Suppose tht f is bounded bove nd below on [, b]. A lower sum for f on the intervl [, b] is sum m i x i, i= where m i is the gretest lower bound for f on the i-th intervl. An upper sum is M i x i, i= where M i is the lest upper bound for f on the i-th intervl. See Figure G.24. Now insted of tking limit s n, we consider the lest upper nd gretest lower bounds of these sums. We mke the following definition. Definition of the Definite Integrl Suppose tht f is bounded bove nd below on [, b]. Let L be the lest upper bound for ll the lower sums for f on [, b], nd let U be the gretest lower bound for ll the upper sums. If L = U, then we sy tht f is integrble nd we define f(x) dx to be equl to the common vlue of L nd U. The lest upper bounds M i nd L nd gretest lower bounds m i nd U re gurnteed to exist by the Completeness Axiom.

3 Upper Sum (re of drk nd light rectngles) Lower Sum (re of drk rectngles) x m i M i Theory Supplement Section G 37 f(x) x i x i+ b x Figure G.24: Lower nd upper sums pproximting f(x) dx Using the Definition in Proof As n exmple, we will prove the theorem stted on pge 27 of the textbook: Theorem: The Men Vlue Inequlity for Integrls If m f(x) M for ll x in [, b], nd if f is integrble on [, b], then m(b ) f(x) dx M(b ). Geometriclly, if f is positive, this theorem sys tht the re under the grph of f is less thn the re of the rectngle of height M, nd greter thn the re of the rectngle of height m. See Figure 5.4 on pge 269 of the textbook. Proof The simplest subdivision of [, b] is the one tht consists of one subintervl, nmely, [, b] itself. Although it does not give very good pproximtion for the definite integrl, it still counts s subdivision. The lest upper bound for f on [, b] is less thn or equl to M, nd the length of the only subintervl in the subdivision is b. So the upper sum for this subdivision is less thn or equl to M(b ). Since every upper sum is n upper estimte for f(x) dx, we hve f(x) dx Upper sum M(b ). The rgument for the other inequlity is similr, using lower sums. (See Problem 9.) Problem 2 gives nother exmple of proof using the definition of the definite integrl. Continuous Functions Are Integrble In this section we will prove the following: Theorem: Continuous Functions re Integrble If f is continuous on [, b], then f(x) dx exists.

4 38 Theory Supplement Section G The Key Question By the Extreme Vlue Theorem on pge 29, continuous function on closed intervl [, b] is bounded bove nd below. Since ny lower sum is less thn or equl to ny upper sum (see Problems 3 7), it follows tht L U. (See Problem 8.) To show tht f is integrble, we need only show tht L = U, so the question is the following: Cn we find subdivision where the lower sum is s close s we like to the upper sum? If we cn, then L < U is not possibility, since then U L would be positive number, nd we would be ble to find lower nd upper sums less thn U L prt. In tht cse, either the lower sum would be bigger thn L or the upper sum less thn U. This cn t hppen, since L is n upper bound for the lower sums, nd U is lower bound for the upper sums. We hve lredy seen tht if f is monotonic we cn find upper nd lower sums tht re rbitrrily close; just tke the left nd right sums with lrge enough n. This proves tht monotonic functions re integrble. If f is not monotonic, we proceed differently. Squeezing the Integrl Between Lower nd Upper Sums For subdivision of [, b] we hve Difference between upper nd lower sums = (M i m i ) x i, where M i is the lest upper bound for f on the i-th subintervl nd m i is the gretest lower bound. The number M i m i represents the mount by which f vries on the i-th subintervl; we cll M i m i the vrition 7 of f on this subintervl. Suppose tht we could choose the subdivision so tht the vrition on ech subintervl ws less thn some smll positive number ɛ. Then Difference between upper nd lower sums < i= i= ɛ x i = ɛ x i = ɛ(b ). By choosing ɛ smll enough we would be ble to mke the difference s smll s we liked. Thus the next question is: i= Cn we find subdivision where the mximum vrition of f on ech of the subintervls is s smll s we like? Mking the Vrition Smll on Ech Subintervl Let ɛ be positive number, s smll s we like. We wnt to prove tht there is subdivision of [, b] such tht the vrition of f on ech subintervl is less thn ɛ. We will give n indirect proof: we ssume tht there is no such subdivision, nd show tht this leds to impossible consequences. Divide the intervl [, b] into two hlves. If ech hlf hs subdivision where the mximum vrition on subintervls is less thn ɛ, we cn put the two subdivisions together to form subdivision of [, b] with the sme property. So if [, b] fils to hve such subdivision, then so does one of the hlves. Choose hlf tht does not hve such subdivision nd divide it in hlf gin. Once more, one of the hlves must fil to hve subdivision where the vrition of f on ech subintervl is less thn ɛ. Continuing in this wy we find nested sequence of intervls, ech of which fils to hve such subdivision. 7 This is not the sme s the totl vrition used in more dvnced texts.

5 Theory Supplement Section G 39 By the Nested Intervl Theorem on pge, these intervls ll contin some number c. Since f is continuous t c, we cn find n intervl round c on which the vrition is less thn ɛ. One of our nested intervls must be contined in this intervl, since they get rbitrrily smll. So the vrition of f on one of the nested intervls is less thn ɛ. This is impossible, given the wy we chose ech nested intervl. So our supposition tht [, b] fils to hve the required subdivision is flse; there must be subdivision of [, b] such tht the vrition of f on ech subintervl is less thn ɛ. Summry We hve shown by contrdiction tht for every positive number ɛ, no mtter how smll, there is subdivision of [, b] such tht the vrition of f on ech subintervl is less thn ɛ. This mens we cn mke the upper nd lower sums s close s we like; hence L = U nd f(x) dx = U = L. Tht is, the continuous function f is integrble. More Generl Riemnn Sums Left- nd right-hnd sums re specil cses of Riemnn sums. For generl Riemnn sum, s with upper nd lower sums, we llow subintervls to hve different lengths. Also, insted of evluting f only t the left or right endpoint of ech subintervl, we llow it to be evluted nywhere in the subintervl. Thus, generl Riemnn sum hs the form (Vlue of f t some point in i-th subintervl) (Length of i-th subintervl). i= (See Figure G.25.) As before, we let x, x,..., x n be the endpoints of the subintervls, so the length of the i-th subintervl is x i = x i x i. For ech i we choose point c i in the i-th subintervl t which to evlute f, leding to the following definition: A generl Riemnn sum for f on the intervl [, b] is sum of the form f(c i ) x i, i= where = x < x < < x n = b, nd, for i =,..., n, x i = x i x i, nd x i c i x i. We define the error in n pproximtion to be the mgnitude of the difference between the pproximte nd the true vlues. (Notice tht error does not men mistke here.) Since the true vlue f(x) x f(c i) x i c i x i+ b x Figure G.25: A generl Riemnn sum pproximting f(x) dx

6 4 Theory Supplement Section G of the integrl is between ny upper estimte nd ny lower estimte, the error in pproximting definite integrl by Riemnn sum must be less thn the difference between the upper nd lower sums using the sme subdivision. If f(x) dx exists, there is subdivision for which the upper nd lower sums re s close s we like. So we cn pproximte the integrl rbitrrily closely by Riemnn sums. Problems for Section G. Write few sentences in support of or in opposition to the following sttement: If left-hnd sum underestimtes definite integrl by certin mount, then the corresponding right-hnd sum will overestimte the integrl by the sme mount. 2. Using the grph of 2 + cos x, for x 4π, list the following quntities in incresing order: the vlue of the integrl 4π (2+cos x) dx, the left-hnd sum with n = 2 subintervls, nd the right-hnd sum with n = 2 subintervls. 3. Sketch the grph of function f (you do not need to give formul for f) on n intervl [, b] with the property tht with n = 2 subintervls, f(x) dx < Left-hnd sum < Right-hnd sum. For Problems 4 2, find subdivision using subintervls of equl length for which the lower nd upper sums differ by less thn.. Give these sums nd n estimte for the integrl which is within.5 of the true vlue. Explin your resoning. [Except for Problem 2, ech function is monotonic over the given intervl.] π/4 5 3 x 2 dx x dx.5 dx 7. sin x dx + x 2 dθ cos θ 9. (ln x) 2 dx. e t2 dt cos 3 y dy e t ln t dt In Problems 3 7 you will show tht every lower sum for given bounded function f on n intervl [, b] is less thn every upper sum, using the ide of refinement of subdivision. Given subdivision of the intervl [, b], we cn subdivide one or more of its subintervls to obtin new subdivision. We sy tht the new subdivision is refinement of the old one. Notice tht if one subdivision s set of endpoints contins nother s, then the first is refinement of the second. 3. Show tht the lower sum for f on [, b] using given subdivision is less thn or equl to the upper sum using the sme subdivision. 4. In this problem we will show tht refining subdivision results in lower sum which is lrger thn the originl. Let f be function defined nd bounded from below on [, b], nd choose subdivision of [, b], with endpoints = x < x < < x n < x n = b. () Suppose x i y x i. Let m i be the gretest lower bound for f on [x i, x i]. Show tht m i is less thn or equl to the gretest lower bound for f on [x i, y] nd the gretest lower bound for f on [y, x i]. (b) Show tht the lower sum for f using the subdivision = x < x < < x n < x n = b is less thn or equl to the lower sum using the sme subdivision with y included. (c) Show tht the lower sum for f using the subdivision = x < x < < x n < x n = b is less thn or equl to the lower sum using ny refinement of the subdivision. 5. In this problem we will show tht refining subdivision results in n upper sum which is smller thn the originl. Let f be function defined nd bounded from bove on [, b], nd choose subdivision of [, b], with endpoints = x < x < < x n < x n = b. () Suppose x i y x i. Let M i be the lest upper bound for f on [x i, x i]. Show tht M i is greter thn or equl to the lest upper bound for f on [x i, y] nd the lest upper bound for f on [y, x i]. (b) Show tht the upper sum for f using the subdivision = x < x < < x n < x n = b is greter thn or equl to the upper sum using the sme subdivision with y included. (c) Show tht the upper sum for f using the subdivision = x < x < < x n < x n = b is greter thn or equl to the upper sum using ny refinement of the subdivision. 6. Given two subdivisions of [, b], show tht there is third one which is refinement of both. 7. Show tht ny lower sum for f on [, b] is less thn or equl to ny upper sum. [Hint: The lower sum uses one subdivision of [, b]; the upper sum uses nother. Use Problem 6 to choose common refinement of the two subdivisions nd then use Problems 3 5.]

7 Theory Supplement Section G 4 8. Let f be function defined nd bounded on [, b], let L be the lest upper bound for ll the lower sums for f on [, b], nd let U be the gretest lower bound for ll the upper sums. () Show tht if L were strictly greter thn U, then there would be lower sum tht ws strictly greter thn n upper sum. [Hint: Let ɛ = L U, nd find lower sum within ɛ/3 of L nd n upper sum within ɛ/3 of U.] (b) Deduce tht L U. 9. On pge 37 we proved one hlf of the Men Vlue Inequlity for Integrls. Prove the other hlf: tht is, if f is continuous on [, b] nd f(x) m for x in [, b], then m(b ) f(x) dx. 2. In this problem we will prove tht if f is continuous on [, b] nd if c is in [, b], then f(x) dx = c f(x) dx + c f(x) dx. () Show tht if l is lower sum for f on [, c], nd if l 2 is lower sum for f on [c, b], then l + l 2 is lower sum for f on [, b]. (b) Show tht if l is lower sum for f on [, b], then there is lower sum l for f on [, c] nd lower sum l 2 for f on [c, b] such tht l l + l 2. (c) Let L be the lest upper bound of ll the lower sums on [, b], let L be the lest upper bound of ll the lower sums on [, c], nd let L 2 be the lest upper bound of ll the lower sums on [c, ]. Use prts () nd (b) to show tht L = L + L 2. Since f is continuous on [, b], L = f(x) dx, L = c f(x) dx, nd L2 = f(x) dx. Thus you hve c proved the required sttement. 2. In this problem we use the Men Vlue Theorem to give proof of the Fundmentl Theorem of Clculus. Let f be continuous, with ntiderivtive F. We wish to show tht f(x)dx = F (b) F (). () Let [, b] be n intervl contined in the domin of f, nd let = x < x < < x n < x n = b be subdivision of [, b]. Show tht there is Riemnn sum for f using this subdivision which is equl to F (b) F (). [Hint: Apply the Men Vlue Theorem to F (b) F () = ((F (b) F (x n )) +(F (x n ) F (x n 2)) + +(F (x ) F ()). (b) Deduce tht F (b) F () = f(x) dx.