Riemann Integrals and the Fundamental Theorem of Calculus


 Oswin Young
 1 years ago
 Views:
Transcription
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([x11 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 106Bio, 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 106Bio, 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 106B, 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).
Riemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationRiemann Sums and Riemann Integrals
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 Abstrct
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationSYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus
SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationThe Fundamental Theorem of Calculus
The Fundmentl Theorem of Clculus MATH 151 Clculus for Mngement J. Robert Buchnn Deprtment of Mthemtics Fll 2018 Objectives Define nd evlute definite integrls using the concept of re. Evlute definite integrls
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationIntegration by Parts Logarithms and More Riemann Sums!
Integration by Parts Logarithms and More Riemann Sums! James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 16, 2013 Outline 1 IbyP with
More informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255  Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationSection Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled?
Section 5.  Ares nd Distnces Exmple : Suppose cr trvels t constnt 5 miles per hour for 2 hours. Wht is the totl distnce trveled? Exmple 2: Suppose cr trvels 75 miles per hour for the first hour, 7 miles
More informationReview on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.
Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5.  5.3) Remrks on the course. Slide Review: Sec. 5.5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description
More informationROB EBY Blinn College Mathematics Department
ROB EBY Blinn College Mthemtics Deprtment Mthemtics Deprtment 5.1, 5.2 Are, Definite Integrls MATH 2413 Rob EbyFll 26 Weknowthtwhengiventhedistncefunction, wecnfindthevelocitytnypointbyfindingthederivtiveorinstntneous
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationCalculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties
Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ fuehr@mth.rwthchen.de Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009 Overview 1 Motivtion:
More information11 An introduction to Riemann Integration
11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationObjectives. Materials
Techer Notes Activity 17 Fundmentl Theorem of Clculus Objectives Explore the connections between n ccumultion function, one defined by definite integrl, nd the integrnd Discover tht the derivtive of the
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More informationCMDA 4604: Intermediate Topics in Mathematical Modeling Lecture 19: Interpolation and Quadrature
CMDA 4604: Intermedite Topics in Mthemticl Modeling Lecture 19: Interpoltion nd Qudrture In this lecture we mke brief diversion into the res of interpoltion nd qudrture. Given function f C[, b], we sy
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationNumerical Analysis: Trapezoidal and Simpson s Rule
nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =
More information5.2 Volumes: Disks and Washers
4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict
More informationMore Properties of the Riemann Integral
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
More information4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve
Dte: 3/14/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: Use your clcultor to solve 4 7x =250; 5 3x =500; HW Requests: Properties of Log Equtions
More informationIntegrals  Motivation
Integrls  Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is nonliner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationc n φ n (x), 0 < x < L, (1) n=1
SECTION : Fourier Series. MATH4. In section 4, we will study method clled Seprtion of Vribles for finding exct solutions to certin clss of prtil differentil equtions (PDEs. To do this, it will be necessry
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
More informationStuff You Need to Know From Calculus
Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Theorem Suppose f is continuous
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in
More informationBig idea in Calculus: approximation
Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through
More informationRead section 3.3, 3.4 Announcements:
Dte: 3/1/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: 1. f x = 3x 6, find the inverse, f 1 x., Using your grphing clcultor, Grph 1. f x,f
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More information(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35
7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationSTEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.
STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t
More informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More informationThe Evaluation Theorem
These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for inclss presenttion nd should not
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationThe Fundamental Theorem of Calculus
The Fundmentl Theorem of Clculus Professor Richrd Blecksmith richrd@mth.niu.edu Dept. of Mthemticl Sciences Northern Illinois University http://mth.niu.edu/ richrd/mth229. The Definite Integrl We define
More information