# Riemann Sums and Riemann Integrals

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1 Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties

2 Abstrct This lecture introduces Riemnn sums nd Riemnn integrls. You should lso hve been exposed to the ide of the integrtion of function f. There re two intellectully seprte ides here: The ide of Primitive. f. This is ny function F which is differentible nd stisfies F (t) = f (t) t ll points in the domin of f. Normlly, the domin of f is finite intervl of the form [, b], lthough it could lso be n infinite intervl like ll of R or [, ) nd so on. Note tht n ntiderivtive does not require ny understnding of the process of Riemnn integrtion t ll only wht differentition is! The ide of the Riemnn integrl of function. We ll go over this now.

3 First, strt with bounded function f on finite intervl [, b]. This kind of function f need not be continuous! select finite number of distinct points from the intervl [, b], {t 0, t,,..., t n, t n}. We don t know how mny points there re, so different selection from the intervl would possibly gives us more or less points. But for convenience, we will just cll the lst point t n nd the first point t 0. These points re not rbitrry t 0 is lwys, t n is lwys b nd they re ordered like this: t 0 = < t < t 2 <... < t n < t n = b Ech prtition P defines subintervls [t 0, t ] to [t n, t n]. Ech of these subintervls hs length t i+ t i. Ech prtition P hs mximum subintervl length let s use the symbol P to denote this length. We red the symbol P s the norm of P. From ech subintervl [t i, t i+] determined by the Prtition P, select ny point you wnt nd cll it s i. This will give us the points s 0 from [t 0, t ], s from [t, t 2] nd so on up to the lst point, s n from [t n, t n]. At ech of these points, we cn evlute the function f to get the vlue f (s j). Cll these points n Evlution Set Let s denote such n evlution set by the letter E.

4 We cn sy more: If the function f ws nice enough to be positive lwys nd continuous, then the product f (s i) (t i+ t i) cn be interpreted s the re of rectngle. Then, if we dd up ll these rectngle res we get sum which is useful enough to be given specil nme: the Riemnn sum for the function f ssocited with the Prtition P nd our choice of evlution set E = {s 0,..., s n }. This sum is represented by the symbol S(f, P, E) where the things inside the prenthesis re there to remind us tht this sum depends on our choice of the function f, the prtition P nd the evlution set E. This leds to the definition of the Riemnn sum. Definition The Riemnn sum for the bounded function f, the prtition P nd the evlution set E = {s 0,..., s n } from P{t 0, t,,..., t n, t n} is defined by S(f, P, E) = n f (s i) (t i+ t i) i=0 It is misleding to write the Riemnn sum this wy s it mkes us think the n is lwys the sme when in fct it cn chnge vlue ech time we select different P. So mny of us write the definition this wy insted S(f, P, E) = f (s i) (t i+ t i) i P nd we just remember tht the choice of P will determine the size of n.

5 Let s look t n exmple of ll this. Here we see the grph of typicl function which is lwys positive on some finite intervl [, b]. (, f ()) A generic curve f on the intervl [, b] which is lwys positive. Note the re under this curve is the (b, f (b)) shded region. b Figure: The Are Under The Curve f Next, let s set the intervl to be [, 6] nd compute the Riemnn Sum for prticulr choice of Prtition P nd evlution set E. The prtition (gry) is P = {.0,.5, 2.6, 3.8, 4.3, 5.6, 6.0}. Hence, we hve subintervl lengths of t t 0 = 0.5, t 2 t =., t 3 t 2 =.2, t 4 t 3 = 0.5, t 5 t 4 =.3 nd t 6 t 5 = 0.4, giving P =.3. Thus, S(f, P, E) For the evlution set (red) E = {.,.8, 3.0, 4., 5.3, 5.8} the Riemnn sum is S(f, P, E) = 5 f (s i) (t i+ t i) i=0 = f (.) f (.8). + f (3.0).2 + f (4.) f (5.3).3 + f (5.8) 0.4

6 (, f ()) (6, f (6)) 6 Figure: A Simple Riemnn Sum We cn lso interpret the Riemnn sum s n pproximtion to the re under the curve. The prtition (gry) is P = {.0,.5, 2.6, 3.8, 4.3, 5.6, 6.0}. For the evlution set (red) E = {.,.8, 3.0, 4., 5.3, 5.8} (, f ()) (6, f (6)) 6 Figure: The Riemnn Sum As An Approximte Are

7 Exmple Exmple We let f (t) = t 2 + 6t 8 on the intervl [2, 4] with P = {2, 2.5, 3.0, 3.7, 4.0} nd E = {2.2, 2.8, 3.3, 3.8}. Solution The prtition determines subintervl lengths of t t 0 = 0.5, t 2 t = 0.5, t 3 t 2 = 0.7, nd t 4 t 3 = 0.3, giving P = 0.7. For E nd P, we hve the Riemnn sum S(f, P, E) = 3 i=0 f (si) (ti+ ti. Thus S(f, P, E) = f (2.2) f (2.8) f (3.3) f (3.8) 0.3 = Exmple Exmple Let f (t) = 3t 2 on the intervl [, 2] with P = {, 0.3, 0.6,.2, 2.0} nd E = { 0.7, 0.2, 0.9,.6}. Find the Riemnn sum. Solution The prtition determines subintervl lengths of t t 0 = 0.7, t 2 t = 0.9, t 3 t 2 = 0.6, nd t 4 t 3 = 0.8, giving P = 0.9. For the evlution set E the Riemnn sum is S(f, P, E) = 4 f (s i) (t i+ t i) i=0 = f ( 0.7) f (0.2) f (0.9) f (.6) 0.8 = 8.739

8 Homework For the given function f, prtition P nd evlution set E, do the following.. Find S(f, P, E) for the prtition P nd evlution set E. 2. Find P. 3. Sketch grph of this Riemnn sum s n pproximtion to the re under the curve f. Do nice grph with pproprite use of color. Homework Continued. Let f (t) = t on the intervl [, 3] with P = {,.5, 2.0, 2.5, 3.0} nd E = {.2,.8, 2.3, 2.8}..2 Let f (t) = t on the intervl [, 3] with P = {,.6, 2.3, 2.8, 3.0} nd E = {.2,.9, 2.5, 2.85}..3 Let f (t) = 3t 2 + 2t on the intervl [, 2] with P = {,.2,.5,.8, 2.0} nd E = {.,.3,.7,.9}.4 Let f (t) = 3t 2 + t on the intervl [, 4] with P = {,.2,.5, 2.8, 4.0} nd E = {.,.3, 2.3, 3.2}

9 We cn construct mny different Riemnn Sums for given function f. If we let the norm of the prtitions we use go to zero, the resulting Riemnn Sums often converge to fixed vlue. This fixed vlue is clled the Riemnn integrl nd in this section, we will mke this notion more precise. To define the Riemnn Integrl of f, we only need few more things:. Ech prtition P hs mximum subintervl length P, the norm of P. 2. Ech prtition P nd evlution set E determines the number S(f, P, E) by simple clcultion. 3. So if we took collection of prtitions P, P 2 nd so on with ssocited evlution sets E, E 2 etc., we would construct sequence of rel numbers {S(f, P, E ), S(f, P 2, E 2),...,, S(f, P n, E n),...}. Let s ssume the norm of the prtition P n gets smller ll the time; i.e. lim n P n = 0. We could then sk if this sequence of numbers converges to something.

10 If the sequence we construct bove converged to the sme number I no mtter wht sequence of prtitions whose norm goes to zero nd ssocited evlution sets we chose, the vlue of this limit is independent of the choices bove. This defines the Riemnn Integrl of f on [, b]. Definition Riemnn Integrbility Of A Bounded Function Let f be bounded function on the finite intervl [, b]. If there is number I so tht lim S(f, Pn, En) = I n no mtter wht sequence of prtitions {P n} with ssocited sequence of evlution sets {E n} we choose s long s lim n P n = 0, we will sy tht the Riemnn Integrl of f on [, b] exists nd equls the vlue I. The vlue I is dependent on the choice of f nd intervl [, b]. So we could denote this vlue by I (f, [, b]) or more simply s, I (f,, b). Historiclly, the ide of the Riemnn integrl ws developed using re pproximtion s n ppliction, so the summing nture of the Riemnn Sum ws denoted by the 6 th century letter S which resembled n elongted or stretched letter S which looked like wht we cll the integrl sign. Hence, the common nottion for the Riemnn Integrl of f on [, b], when this vlue exists, is b f. We usully wnt to remember wht the independent vrible of f is lso nd we wnt to remind ourselves tht this vlue is obtined s we let the norm of the prtitions go to zero.

11 The symbol dt for the independent vrible t is used s reminder tht t i+ t i is going to zero s the P 0 So it hs been very convenient to dd to the symbol b f this informtion nd use the ugmented symbol b f (t) dt insted. Hence, if the independent vrible ws x insted of t, we would use b f (x) dx. Since for function f, the nme we give to the independent vrible is mtter of personl choice, we see tht the choice of vrible nme we use in the symbol b f (t) dt is very rbitrry. Hence, it is common to refer to the independent vrible we use in the symbol b f (t) dt s the dummy vrible of integrtion. It cn be proved in more dvnced courses tht the following things re true bout the Riemnn Integrl of bounded function. Theorem Existence Of The Riemnn Integrl Let f be bounded function on the finite intervl [, b]. Then the Riemnn integrl of f on [, b], b f (t)dt exists if. f is continuous on [, b] 2. f is continuous except t finite number of points on [, b]. Further, if f nd g re both Riemnn integrble on [, b] nd they mtch t ll but finite number of points, then their Riemnn integrls mtch; i.e. b f (t)dt equls b g(t)dt.

12 If you think bout it bit, it is pretty esy to see tht we cn split up Riemnn integrls in obvious wys. For exmple, the integrl of sum should be the sum of the integrls; i.e. b b b (f (x) + g(x)) dx = f (x) dx + g(x) dx nd we should be ble to pull out constnts like so b b (c f (x)) dx = c f (x) dx This is becuse in the Riemnn sum, prtitions of pieces like tht cn be broken prt nd then the limits we wnt to do cn be tken seprtely. Look t typicl piece of Riemnn sum which for sum of functions would look (f (s i) + g(s i)) x i. We cn surely split this prt to f (s i) x i + g(s i) x i nd then dd up the pieces like usul. So we will get RS(f + g, P, E) = RS(f, P, E) + RS(g, P, E) for ny prtition nd evlution set. Now tke the limit nd we get the result! o see you cn pull out constnts, we do the sme rgument. Ech piece in the Riemnn sum hs the form (cf (s i) x i nd it is esy to see we cn whisk tht constnt c outside of the sum to find RS(cf, P, E) = c RS(f, P, E). Then we tke the limit nd voíl!

13 To mke it esy to see, wht we re sying is this: (3 + 5x + 7x 2 ) dx = 3 dx + 5 x dx x 2 ) dx Finlly, the wy we hve setup the Riemnn integrl lso mkes it to see tht if we do Riemnn integrl over n intervl of no length, the vlue should be 0 s ll the x i s re zero so the Riemnn sums re 0 nd hence the integrl is zero. For exmple, if the intervl is just [, ], we hve f (x) dx = 0. Now look t the wy the Riemnn sum is pictured. If we set up Riemnn sums on the intervl [, 5], sy, it is pretty obvious tht we could brek these sums prt into Riemnn sums over [, 3] nd [3, ] 5 for exmple. Then we could tke limits s usul nd see f (x) dx = f (x) dx + f (x) dx. 3 The rgument is bit more subtle thn this, but now is not the time to get bogged down in those detils. Subtle or not, the rgument works out nicely. And we cn split the intervl up in ny wy we wnt. So we cn sy b c b f (x) dx = f (x) dx + f (x) dx. c for ny choice of intermedite c we wnt.

14 One lst thing. If we mde the integrtion order go bckwrds, i.e. doing our Riemnn sums from 3 to insted of to 3, ll the x i s would be flipped. So the Riemnn sum would be the reverse of wht it should be nd the limiting vlue would be the negtive of wht we would normlly hve. We cn sy things like 4 f (x) dx = f (x) dx 4 nd similr things for other intervls, of course. Now you go nd ply round with these rules bit nd mke sure you re comfortble with them!

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