Riemann Integrals and the Fundamental Theorem of Calculus

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1 Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums Riemnn Integrls Uniform Prtition Riemnn Sums Properties Fundmentl Theorem Of Clculus

2 Abstrct This lecture explins the mzing connection between the ide of n ntiderivtive or primitive of function nd the Riemnn integrl of tht sme function on some intervl. If we wnt to grph the Riemnn sums, we need to grph those rectngles we drw by hnd. To grph rectngle, we grph 4 lines. The MtLb commnd plot([x1 x2], [y1 y2]) plots line from the pir (x1, y1) to (x2, y2). So the commnd plot([x(i) x(i+1)],[f(s(i)) f(s(i))]); plots the horizontl line which is the top of our rectngle. The commnd plot([x(i) x(i)], [0 f(s(i))]); plots verticl line tht strts on the x xis t x i nd s t the function vlue f (s i).

3 The commnd plot([x(i+1) x(i+1)], [0 f(s(i))]); plots verticl line tht strts on the x xis t x i+1 nd s t the function vlue f (s i). To plot rectngle, for the first pir of prtition points, first we set the xis of our plot so we will be ble to see it. We use the xis commnd in Mtlb look it up using help! If the two x points re x1 nd x2 nd the y vlue is f (s1) where s1 is the first evlution point, we expnd the x xis to [x1 1, x2 + 1] nd expnd the y xis to [0, f (s1)]. This llows our rectngle to be seen. The commnd is xis([x1-1 x2+1 0 f((s1))+1]);. The code so fr Putting this ll together, we plot the first rectngle like this: >> h o l d on % set x i s so we cn see rectngle >> x i s ( [ P( 1 ) 1 P( 2 )+1 0 f ( E ( 1 ) ) +1]) % plot top, LHS, RHS nd bottom of rectngle >> p l o t ( [ P( 1 ) P( 2 ) ], [ f (E ( 1 ) ) f (E ( 1 ) ) ] ) ; >> p l o t ( [ P( 1 ) P( 1 ) ], [ 0 f (E ( 1 ) ) ] ) ; >> p l o t ( [ P( 2 ) P( 2 ) ], [ 0 f (E ( 1 ) ) ] ) ; >> p l o t ( [ P( 1 ) P( 2 ) ], [ 0 0 ] ) ; >> h o l d o f f We hve to force Mtlb to plot repetedly without ersing the previous plot. We use hold on nd hold off to do this. We strt with hold on nd then ll plots re kept until the hold off is used.

4 This genertes the rectnge we see below: Figure: Simple Rectngle To show the Riemnn sum pproximtion s rectngles, we use for loop in MtLb To put this ll together, h o l d on % s e t h o l d to on f o r i = 1 : 4 % grph r e c t n g l e s bottom = 0 ; top = f ( E ( i ) ) ; p l o t ( [ P( i ) P( i +1) ], [ f (E( i ) ) f (E( i ) ) ] ) ; p l o t ( [ P( i ) P( i ) ], [ bottom top ] ) ; p l o t ( [ E ( i ) E ( i ) ], [ bottom top ], r ) ; p l o t ( [ P( i +1) P( i +1) ], [ bottom top ] ) ; p l o t ( [ P( i ) P( i +1) ], [ 0 0 ] ) ; h o l d o f f % s e t h o l d o f f

5 Of course, we don t know if f cn be negtive, so we need to djust our thinking s some of the rectngles might need to point down. We do tht by setting the bottom nd top of the rectngles using n if test. bottom = 0 ; top = f ( E ( i ) ) ; i f f ( E ( i ) ) < 0 top = 0 ; bottom = f ( E ( i ) ) ; All together, we hve h o l d on % s e t h o l d to on [ s i z e P,m] = s i z e (P) ; f o r i = 1 : s i z e P 1 % grph l l t h e r e c t n g l e s bottom = 0 ; top = f ( E ( i ) ) ; i f f ( E ( i ) ) < 0 top = 0 ; bottom = f ( E ( i ) ) ; p l o t ( [ P( i ) P( i +1) ], [ f (E( i ) ) f (E( i ) ) ] ) ; p l o t ( [ P( i ) P( i ) ], [ bottom top ] ) ; p l o t ( [ E ( i ) E ( i ) ], [ bottom top ], r ) ; p l o t ( [ P( i +1) P( i +1) ], [ bottom top ] ) ; p l o t ( [ P( i ) P( i +1) ], [ 0 0 ] ) ; h o l d o f f ; We lso wnt to plce the plot of f over these rectngles.

6 h o l d on % s e t h o l d to on [ s i z e P,m] = s i z e (P) ; f o r i = 1 : s i z e P 1 % grph l l t h e r e c t n g l e s bottom = 0 ; top = f ( E ( i ) ) ; i f f ( E ( i ) ) < 0 top = 0 ; bottom = f ( E ( i ) ) ; p l o t ( [ P( i ) P( i +1) ], [ f (E( i ) ) f (E( i ) ) ] ) ; p l o t ( [ P( i ) P( i ) ], [ bottom top ] ) ; p l o t ( [ E ( i ) E ( i ) ], [ bottom top ], r ) ; p l o t ( [ P( i +1) P( i +1) ], [ bottom top ] ) ; p l o t ( [ P( i ) P( i +1) ], [ 0 0 ] ) ; y = l i n s p c e (P( 1 ),P( s i z e P ), 101) ; p l o t ( y, f ( y ) ) ; x l b e l ( x x i s ) ; y l b e l ( y x i s ) ; t i t l e ( Riemnn Sum w i t h f u n c t i o n o v e r l i d ) ; h o l d o f f ; We generte this figure: Figure: Riemnn Sum for f (x) = x 2 for Prtition {1, 1.5, 2.1, 2.8, 3.0}

7 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: 1. 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 1, P 2 nd so on with ssocited evlution sets E 1, E 2 etc., we would construct sequence of rel numbers {S(f, P 1, E 1), 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.

8 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 indepent 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 n S(f, P n, E n) = I 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 depent 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 16 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 indepent 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.

9 The symbol dt for the indepent vrible t is used s reminder tht t i+1 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 indepent vrible ws x insted of t, we would use b f (x) dx. Since for function f, the nme we give to the indepent 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 indepent 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 1. 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.

10 To sve typing, let s lern to use Mtlb function. In Mtlb s file menu, choose crete new Mtlb function which gives f u n c t i o n [ v l u e 1, v l u e 2,... ] = MyFunction ( rg1, rg2,... ) % s t u f f i n h e r e [vlue1, vlue2,...] re returned vlues the function clcultes tht we wnt to sve. (rg1, rg2,...) re things the function needs to do the clcultions. They re clled the rguments to the function. MyFunction is the nme of the function. This function must be stored in the file MyFunction.m. Our function returns the Riemnn sum, RS, nd use the rguments: our function f, the prtition P nd the Evlution set E. Since only one vlue returned [RS] cn be RS. function RS = RiemnnSum ( f, P, E) % comments l w y b e g i n w i t h % mtlb l i n e s h e r e The nme for the function RiemnnSum must be used s the file nme: i.e. we must use RiemnnSum.m s the file nme.

11 The Riemnn sum function: 1 function RS = RiemnnSum ( f, P, E) % f i n d Riemnn sum dx = d i f f (P) ; RS = sum ( f ( E ). dx ) ; [ s i z e P,m] = s i z e (P) ; %g e t s i z e o f P r t i t i o n 6 c l f ; % c l e r t h e o l d grph h o l d on % s e t h o l d to on f o r i = 1 : s i z e (P) 1 % grph r e c t n g l e s % p l o t r e c t n g l e code % p l o t f u n c t i o n code... y = l i n s p c e (P( 1 ),P( s i z e P ), 101) ; h o l d o f f ; Now to see grphiclly how the Riemnn sums converge to finite number, let s write new function: Riemnn sums using uniform prtitions nd midpoint evlution sets. 1 f u n c t i o n RS = RiemnnUniformSum ( f,, b, n ) % s e t up u n i f o r m p r t i t i o n w i t h n+1 p o i n t s d e l t x = ( b ) /n ; P = [ : d e l t x : b ] ; % mkes row v e c t o r f o r i =1:n 6 s t r t = +( i 1) d e l t x ; s to p = +i d e l t x ; E ( i ) = 0. 5 ( s t r t+s t o p ) ; % s i n t r n s p o s e o f P nd E so we use column v e c t o r s 11 % b e c u s e o r i g i n l RiemnnSum f u n c t i o n u s e s columns RS = RiemnnSum ( f, P, E ) ;

12 We cn then generte sequence of Riemnn sums for different vlues of n. We generte sequence of figures which converge to fixed vlue. >> f x ) s i n ( 3 x ) ; >> RS = RiemnnSumTwo ( f, 1,4,10) ; >> RS= RiemnnSumTwo ( f, 1,4,20) ; >> RS = RiemnnSumTwo ( f, 1,4,30) ; >> RS= RiemnnSumTwo ( f, 1,4,40) ; Figure: The Riemnn sum with uniform prtition P10 of [ 1, 4] for n = 10. The function is sin(3x) nd the Riemnn sum is

13 Figure: Riemnn sum with uniform prtition P20 of [ 1, 4] for n = 20. The function is sin(3x) nd the Riemnn sum is Figure: Riemnn sum with uniform prtition P40 of [ 1, 4] for n = 40. The function is sin(3x) nd the Riemnn sum is

14 Figure: Riemnn sum with uniform prtition P80 of [ 1, 4] for n = 80. The function is sin(3x) nd the Riemnn sum is Homework 26 For the given function f, intervl [, b] nd choice of n, you ll clculte the corresponding uniform prtition Riemnn sum using the functions RiemnnSum in file RiemnnSum.m nd RiemnnUniformSum in file RiemnnUniformSum.m. You cn downlod these functions s files from the clss web site. Sve them in your personl clss directory. Crete new word document in single spce with mtlb frgments in bold font. The document strts with your nme, MTHSC 106-Bio, Section, HW number nd the dte.

15 Homework 26 Continued Crete new word document for this homework. Do the document in single spce. Do mtlb frgments in bold font. The document strts with your nme, MTHSC 106-Bio, Section Number, Dte nd Homework number. For ech vlue of n, do sve s nd sve the figure with filenme like HW#Problem#[ ].png where [ ] is where you put the number of the grph. Something like HW17.png, HW#Problem#b.png etc. Insert this picture into the doc resizing s needed to mke it look good. Explin in the doc wht the picture shows. Homework 26 Continued Something like this: Jim Peterson MTHSC 106-B, Section Number tody s dte nd HW Number, Problem 1: Let f (t) = sin(5t) on the intervl [1, 3] with P = {1, 1.5, 2.0, 2.5, 3.0} nd E = {1.2, 1.8, 2.3, 2.8}. % dd e x p l n t i o n h e r e >> f x ) s i n ( 5 x ) ; % dd e x p l n t i o n h e r e >> RS = RiemnnUniformSum ( f, 1,4,10) % dd e x p l n t i o n h e r e >> RS = RiemnnUniformSum ( f, 1,4,20) % dd e x p l n t i o n h e r e >> RS = RiemnnUniformSum ( f, 1,4,40) % dd e x p l n t i o n h e r e >> RS = RiemnnUniformSum ( f, 1,4,80)

16 Homework 26 Continued 26.1 Let f (t) = t 2 2t + 3 on the intervl [ 2, 3] with n = 8, 16, 32 nd Let f (t) = sin(2t) on the intervl [ 1, 5] with n = 10, 40, 60 nd Let f (t) = t 2 + 8t + 5 on the intervl [ 2, 3] with n = 4, 12, 30 nd 50. 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.

17 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! To 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! To mke it esy to see, wht we re sying is this: (3 + 5x + 7x 2 ) dx = 3 1 dx + 5 x dx x 2 dx 1 Finlly, the wy we hve setup the Riemnn integrl lso mkes it esy 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. 1 f (x) dx = 0. 1

18 Now look t the wy the Riemnn sum is pictured. If we set up Riemnn sums on the intervl [1, 5], sy, it is pretty obvious tht we could brek these sums prt into Riemnn sums over [1, 3] nd [3, 5] for exmple. Then we could tke limits s usul nd see f (x) dx = f (x) dx + f (x) dx 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. One lst thing. If we mde the integrtion order go bckwrds, i.e. doing our Riemnn sums from 3 to 1 insted of 1 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 1 f (x) dx = f (x) dx 1 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!

19 There is big connection between the ide of the ntiderivtive of function f nd its Riemnn integrl. For positive function f on the finite intervl [, b], we cn construct the re under the curve function F (x) = x f (t) dt. Let s look t the difference in these res: we ssume h is positive. x+h x F (x + h) F (x) = f (t) dt f (t) dt x x+h = f (t) dt + f (t) dt x x f (t) dt where we hve used stndrd properties of the Riemnn integrl to write the first integrl s two pieces. Now subtrct to get F (x + h) F (x) = x+h f (t) dt x Now divide this difference by the chnge in x which is h. We find F (x + h) F (x) = 1 x+h f (t) dt h h x We show F (x) nd F (x + h) for smll positive h in the figure which follows.

20 F (x) is the re under this curve from to x. (, f ()) (b, f (b)) F (x) F (x + h) x x + h Figure: The Function F (x) b The difference in re, x+h x f (t) dt, is the second shded re in the figure you just looked t. We see If t is ny number between x nd x + h, the re of the rectngle with bse h nd height f (t) is f (t) h which is closely relted to the re difference. Note the difference between this re nd F (x + h) F (x) is relly smll when h is smll. We know tht f is bounded on [, b] You cn esily see tht f hs mximum vlue for the prticulr f we drw. Of course, this grph is not wht ll such bounded functions f look like, but you should be ble to get the ide tht there is number B so tht 0 < f (t) B for ll t in [, b]. Thus, we see F (x + h) F (x) x+h B dt = B h (1) x

21 From this, it follows tht We see our difference lives between 0 nd B. 0 (F (x + h) F (x)) B h And so tking the limit s h gets smll, we find 0 lim (F (x + h) F (x)) h 0 lim B h = 0. h 0 We conclude tht F is continuous t ech x in [, b] s lim (F (x + h) F (x)) = 0. h 0 It seems tht the new function F we construct by integrting the function f in this mnner lwys builds new function tht is continuous. Is F differentible t x? Let s do n estimte. We hve lower nd upper bound on the re of the middle slice in our figure. ( ) x+h ( ) min f (t) h f (t)dt mx f (t) h x t x+h x x t x+h Thus, we hve the estimte F (x + h) F (x) min f (t) mx x t x+h h f (t) x t x+h

22 If f ws continuous t x, then we must hve nd ( ) lim min f (t) h 0 x t x+h ( ) lim mx f (t) h 0 x t x+h = f (x) = f (x) Note the f we drw in our figure is continuous ll the time, but the rgument we use here only needs continuity t the point x! At ny rte, we cn infer for positive h, F (x + h) F (x) lim h 0 + h = f (x) You should be ble to believe tht similr rgument would work for negtive vlues of h: i.e., F (x + h) F (x) lim h 0 h = lim f (t) = f (x) h 0 This tells us tht F (p) exists nd equls f (x) s long s f is continuous t x s F (x + ) = F (x + h) F (x) lim = f (x) h 0 + h F (x ) = F (x + h) F (x) lim = f (x) h 0 h

23 This reltionship is clled the Fundmentl Theorem of Clculus. Our rgument works for x equls or b but we only need to look t the derivtive from one side. So the discussion is bit simpler. Our rgument used positive f but it works just fine if f hs positive nd negtive spots. Just divide f into it s postive nd negtive pieces nd pply these ides to ech piece nd then glue the result together. We cn ctully prove this using firly relxed ssumptions on f for the intervl [, b]. In generl, f need only be Riemnn Integrble on [, b] which llows for jumps in the function. But those rguments re more dvnced! Theorem Fundmentl Theorem of Clculus Let f be Riemnn Integrble on [, b]. Then the function F defined on [, b] by F (x) = x f (t) dt stisfies 1. F is continuous on ll of [, b] 2. F is differentible t ech point x in [, b] where f is continuous nd F (x) = f (x).

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