Chapter 12 Integral Calculus

Transcription

1 Ch - Integrl Clculus 8/3/5 Chpter Integrl Clculus Newton Leiniz. Prolem: Are under curve A very old prolem (Archimedes proposed solution! Fermt worked on it, too. Newton nd Leiniz solved it in lte 7 th century. Ide: Introduce rectngles under the curve, defined y f(, find the re of ll of those rectngles nd dd them ll up. Rigorous mthemticl detils hd to wit till the 9 th century.

2 Ch - Integrl Clculus 8/3/5. Estimting Are Under Points Wht if insted of function, we were given points: how could we use rectngles to estimte re under points? 3. Underestimting Are Here we underestimte the re y putting left corners t points: 4

3 Ch - Integrl Clculus 8/3/5. Overestimting Are Here we overestimte the re y putting right corners t points: Note: As the intervls ecome smller nd smller i.e., the prtitions of [,] ecome finer-, oth the left corner nd the right corner sed res converge. 5. Riemnn Sum In generl, integrtion is motivted s n re under curve. This geometric intuition is fine. But, we wnt to think of integrtion s form of summtion. In economics nd finnce, geometric interprettions, in generl, hve little use. But, the summtion intuition works very well. Riemnn thought of n integrl s the convergence of two sums, s the prtition of the intervl of integrtion ecomes smller. Bernhrd Riemnn (86-866, Germny 6 3

4 Ch - Integrl Clculus 8/3/5. Riemnn Sum In order to estimte n re, we need prtition of the intervl [,]. We define prtition P of the closed intervl [,] s finite set of points P = {,,,..., n } such tht = < < <... < n- < n =. If P = {,,,..., n } is prtition of the closed intervl [,] nd f is function defined on tht intervl, then the n-th Riemnn Sum of f with respect to the prtition P is defined s: R(f, P = Σ j= to n f(t j ( j - j- where t j is n ritrry numer in the intervl [ j-, j ]. But, we do not know t j. In the previous two emples, we used the left end points of the intervl [ j-, j ] (underestimtion of re nd the right points of the intervl [ j-, j ] (overestimtion of re. 7. Riemnn Sum There re two useful cses: use c j, the supremum of f( in the intervl [ j-, j ], producing the upper sum: U(f, P = Σ j= to n c j ( j - j- use d j, the infimum of f( in the intervl [ j-, j ], producing the lower sum: L(f, P = Σ j= to n d j ( j - j- Emple: U(f, P is displyed in drk rown nd L(f, P in ornge. 8 4

5 Ch - Integrl Clculus 8/3/5. Riemnn Sum Proposition: Size of Riemnn Sums Let P e prtition of the closed intervl [,], nd f(. e ounded function defined on tht intervl. Then, - The lower (upper sum is incresing (decresing with respect to refinements of prtitions --i.e. L(f, P' L(f, P or U(f, P' U(f, P for every refinement P' of the prtition P. -L(f, P R(f, P U(f, P for every prtition P Tht is, the lower sum is lwys less thn or equl to the upper sum. Q: Will U(f, P nd L(f, P ever e the sme? 9. Riemnn Integrl Suppose f(. is ounded function defined on closed, ounded intervl [, ]. Define the upper nd lower Riemnn integrls s: I * (f = inf {U(f,P: P prtition of [, ]} I * (f = sup {L(f,P: P prtition of [, ]} Then if I * (f = I * (f the function f(. is clled Riemnn integrle (Rintegrle nd the Riemnn integrl of f(. over the intervl [, ] is denoted y f(d Note: U(f, P nd L(f, P depend on the chosen prtition, while the upper nd lower integrls re independent of prtitions. But, this definition is not prcticl, since we need to find the sup nd inf over ny prtition. 5

6 Ch - Integrl Clculus 8/3/5. Riemnn Integrl Suppose f(. is ounded function defined on closed, ounded intervl [, ]. Define the upper nd lower Riemnn integrls s: I * (f = inf {U(f,P: P prtition of [, ]} I * (f = sup {L(f,P: P prtition of [, ]} Then if I * (f = I * (f the function f(. is clled Riemnn integrle (Rintegrle nd the Riemnn integrl of f(. over the intervl [, ] is denoted y f(d Note: U(f, P nd L(f, P depend on the chosen prtition, while the upper nd lower integrls re independent of prtitions. But, this definition is not prcticl, since we need to find the sup nd inf over ny prtition.. Riemnn Integrl Emple Emple: Is f(= R-integrle on [,]? It is complicted to prove tht this function is integrle. We do not hve simple condition to tell us whether this, or ny other function, is integrle. But, we should e le to generlize the proof for this prticulr emple to wider set of functions. First, note tht in the definition of upper nd lower integrl it is not necessry to tke the sup nd inf over ll prtitions: If P is prtition nd P' is refinement of P, then L(f, P' L(f, P nd U(f, P' U(f, P. Thus, prtitions with lrge intervls (lrge norms, P do not contriute to the sup or inf. We look t prtitions with smll norms. 6

7 Ch - Integrl Clculus 8/3/5. Riemnn Integrl Emple Second, tke ny ε> nd prtition P with P < ε/. Then, U(f, P - L(f, P Σ j= to n c j -d j ( j - j-, where c j is the sup of f over [ j-, j ] nd d j is the inf over tht intervl. Since f(. is incresing over [, ], we know tht the sup is chieved on the right side of ech suintervl, the inf on the left side. Then, U(f, P - L(f, P Σ j= to n c j -d j ( j - j- = Σ j= to n f( j - f( j- ( j - j- To estimte this sum, we use the Men Vlue Theorem for f( = : f( - f( f'(c - y for c etween nd y. Since f'(c for c [, ] => f( - f( - y 3. Riemnn Integrl Emple To estimte this sum, we use the Men Vlue Theorem for f( = : f( - f( f'(c - y for c etween nd y. Since f'(c for c in [, ] => f( - f( - y But P ws chosen with P < ε/ => f( j - f( j- j - j- ε / = ε. Then, U(f, P - L(f, P Σ j= to n f( j - f( j- ( j - j- ε Σ j= to n ( j - j- = ε ( - = ε. Since P ws ritrry ut with smll norm --sufficient for the upper nd lower integrls--, the upper nd lower integrl must eist nd e equl to one common limit L. 4 7

8 Ch - Integrl Clculus 8/3/5. Riemnn Integrl Emple Let s clculte L. It is esy, since now we know tht the function is integrle. Then, we tke suitle prtition to find the vlue of the integrl. For emple, tke the following prtition j = j/n for j =,,,..., n. Then, the upper sum: U(f, P = Σ j= to n c j ( j - j- = Σ j= to n f( j /n = Σ j= to n (j/n /n = /n 3 Σ j= to n j = /n 3 [/6 n (n+ (n+] = /6 (n+ (n+/n Since we know tht the upper integrl eists nd is equl to L, the limit s n goes to infinity of the ove epression must lso converge to L. Then, L = /3. 5. Riemnn Integrl Emple Emple: Is the Dirichlet function R-integrle? The Dirichlet Function (Q is the set of rtionl numers: f ( if if Q Q We hve tht U(f, P = nd L(f, P =, regrdless of P. Then, I * (f = nd I * (f =. Thus, the Dirichlet function is not R-integrle over the intervl [,]. Note: Unlike the function in the previous emple, we hve discontinuous function on the Irrtionl Numers. We hve infinite discontinuities. 6 8

9 Ch - Integrl Clculus 8/3/5. Riemnn Lemm The first emple shows tht it is difficult to estlish the integrility of given function. The second emple illustrtes tht not every function is Riemnn integrle. The Rienmnn lemm provides n esier condition to check the integrility of function. Suppose f(. is ounded function defined on the closed, ounded intervl [, ]. Then, f(. is R-integrle if nd only if for every ε> there eists t lest one prtition P such tht U(f,P - L(f,P < ε In emple, we check the ove inequlity holds for every prtition P with smll enough norm. Using Riemnn's Lemm, we only need to check the inequlity holds for one prtition. Esier! 7. Riemnn Integrl - Remrks Roughly speking, we define the Riemnn integrl s follows: - Sudivide the domin of the function (usully closed, ounded intervl into finitely mny suintervls (the prtition - Construct simple function tht hs constnt vlue on ech of the suintervls of the prtition (the Upper nd Lower sums - Tke the limit of these simple functions s you dd more nd more points to the prtition. If the limit eists, it is clled the Riemnn integrl nd the function is clled R-integrle. A function is R-integrle on [,] if: - It is continuous on [,] - It is monotone on [,] - It is ounded, with finite numer of discontinuities on [,]. 8 9

10 Ch - Integrl Clculus 8/3/5. Riemnn Integrl - Remrks For function to e R-integrle it must e ounded. If the function is unounded even t single point in n intervl [, ] it is not Riemnn integrle (ecuse the sup or inf over the suintervl tht includes the unounded vlue is infinite. For emple, f(= / over [,], unounded t =. The Riemnn integrl is sed on the concept of n "intervl", or rther on the length of suintervls [ j-, j ]. The concept of prtition pplies to n intervl. We cn tke Riemnn integrls over unions of intervls, ut nothing more complicted (sy, Cntor sets. Prtitions depend on the structure of the rel line. Thus, we cnnot define R-integrle for functions defined on more strct spces --sy, sequences, functions from N to R. 9. Riemnn-Stieljes Integrl The Riemnn Stieltjes integrl of rel-vlued function f of rel vrile with respect to rel function g is denoted y: f ( dg( defined to e the limit, s the mesh of the prtition P ={ = < < <... < n- < n = }, of the intervl [, ] pproches zero, of the pproimting sum S(f,g,P = Σ i= to n f(c i (g( j g( j-, where c i is in the i-th suintervl [ i-, i ]. The two functions f nd g re respectively clled the integrnd nd the integrtor. If g is everywhere differentile, then the Riemnn Stieltjes integrl my e different from the Riemnn integrl of f( g (. For emple, if the derivtive is unounded. But if the derivtive is continuous, they will e the sme.

11 Ch - Integrl Clculus 8/3/5. Leesgue s Theory - Introduction The Riemnn integrl is sed on prtitioning the domin [,] in suintervls [ j-, j ], picking point j * in the suintervl nd clculting the re under the curve y computing the Riemnn sums. Then, tke the limit s we dd more nd more points to the prtition. Roughly speking, Leesgue's theory, insted of prtitioning the domin, prtitions the rnge into suintervls. Bsed on these suintervls, clculte res nd sum over these res. The pproimtion improves with finer nd finer prtitions of the rnge. The Leesgue integrl will e the limit of these sums.. Leesgue s Theory - Introduction Suppose the function tkes vlues etween [c,d]. Divide rnge [c,d] into suintervls: [c=y,y ], [y,y ],...,[y N,y N =d] Define E i s the set of ll points in [,] whose vlue under f lies etween y i nd y i+ : E i =f - ([y i,y i+ ]={ [,] y i f( y i+ }. 3 Assign size to E i - mesure μ(e i. Then, the portion of the grph of y=f( etween the horizontl lines y=y i & y= y i+ will e A i, where, y i μ(e i A i y i+ μ(e i. 4 Approimte the re y picking numer y i [y i,y i+ ], nd compute: i=n y i μ(e i 5 The pproimtion improves with finer nd finer prtitions of [c,d]. The Leesgue integrl will e the limit (if it eists of these sums. The function is clled Leesgue integrle.

12 Ch - Integrl Clculus 8/3/5. Leesgue s Theory - Introduction Riemnn & Leesgue integrls: Verticl sums vs Horizontl sums. 3. Leesgue s Theory - Introduction Riemnn & Leesgue integrls: Anlogy: You hve pile of coins nd you wnt to know how much money you hve. For this purpose, you cn pick the coins rndomly, one y one, nd dd them. This is the Riemnn integrl. You cn lso sort the coins y denomintion first, nd get the totl y multiplying ech denomintion y how mny you hve of tht denomintion nd dd them up. This is the Leesgue integrl. The methods re different, ut you otin the sme result y either method. Similrly, when oth the Riemnn integrl nd the Leesgue integrl re defined, they give the sme vlue. But, there re functions for which the Leesgue integrl is defined ut the Riemnn integrl is not. In this sense, the Leesgue integrl is more generl thn the Riemnn. 4

13 Ch - Integrl Clculus 8/3/5. Leesgue s Theory - Mesure Step 3 ssigns size to the set E i. With Riemnn integrls, we use s size the length of suintervls [ j-, j ]. This is fine, ut we need to generlize the concept to more complicted sets. The generliztion, clled mesure, is function ssigning to ech set A in R m non-negtive numer, μ(a. This mesure should stisfy two conditions: It should e pplicle to intervls, unions of intervls, nd to more generl sets (sy, Cntor set. Idelly, defined for ll sets. It should shre mny of the properties of length of n intervl : - μ(a (Non-negtive; - μ(a =[,] = - ; - Invrint under trnsltion: if F=E+c={e+c e E}=>μ(F=μ(E; - Countly dditive, i.e., μ(a= UA n = Σ n (A n, where A n re pirwise disjoint sets. 5. Leesgue s Theory Leesgue Mesure Leesgue defined mesure, the Leesgue mesure, tht stisfies oth conditions: - First, define n outer mesure (sed on the infimum of set, which stisfies (: If A is ny suset of R, define the (Leesgue outer mesure of A s: λ * (A = inf {Σ l(a n } where the infimum is tken over ll countle collections of open intervls A n such tht A UA n nd l(a n is the stndrd length of the intervl A n. Note: λ * is defined for ll sets, ut, λ * is not dditive, it is sudditive i.e., λ * (F U E λ * (F + λ * (E; => not quite length. The outer mesure is rel-vlued, non-negtive, monotone nd countly sudditive set function. 6 3

14 Ch - Integrl Clculus 8/3/5. Leesgue s Theory Mesurle Sets - Second, define mesure y restricting the outer mesure to mesurle sets: A set E is (Leesgue mesurle if for every set A we hve tht λ * (A = λ * (A E + λ * (A E C If E is mesurle, the non-negtive numer μ(e = λ * (E is the (Leesgue mesure of the set E. Emple : The set R of ll rel numers is mesurle: λ*(a R + λ*(a R C = λ*(a + λ*(a Ø = λ*(a => R is mesurle. Emple : The complement of mesurle set is mesurle. Suppose E is mesurle => λ * (A = λ * (A E + λ * (A E C For E C we hve: λ * (A E C + λ * (A (E C C = λ * (A E C + λ * (A E = λ * (A => E is mesurle. 7. Leesgue s Theory Mesurle Sets This restriction mkes the mesure dditive, stisfying (, minly the dditive requirement. But, now the mesure is not defined for ll sets, since not ll sets re mesurle (the iom of choice in set theory plys role here. Leesgue s definition of mesurle sets is not very intuitive. But, it is elegnt, generl, rief nd it works. Remrk: Not every set is mesurle, ut it is fir to sy tht most sets re. Usully, the fmily of ll mesurle sets is denoted y ` (script M. ` is sigm-lger nd trnsltion invrint set contining ll intervls. 8 4

15 Ch - Integrl Clculus 8/3/5. Leesgue s Theory Properties of Mesure Properties of Leesgue mesure:. All intervls re mesurle. The mesure of n intervl: its length.. All open nd closed sets re mesurle. 3. The union nd intersection of finite or countle numer of mesurle sets is gin mesurle. 4. If A is mesurle nd A is the union of countle numer of mesurle sets A n, then μ(a Σ μ(a n. 5. If A is mesurle nd A is the union of countle numer of disjoint mesurle sets A n, then μ(a = Σ μ(a n. According to these properties, most common sets re mesurle: intervls; closed & open sets; unions & intersections of mesurle sets. But, the property tht mesure is (countl dditive implies tht not every set is mesurle. 9. Leesgue Integrl Mesurle functions In Leesgue's theory, integrls re defined for clss of functions clled mesurle functions, which re the ones for which the sets we get re mesurle sets. A function f : A [-, ] is mesurle (or mesurle on A if A ` nd the pre-imge of every intervl of the form (t, is in ` : { f(>t} ` for ll t R This is somewht comprle to one of the definitions of continuous functions: A function f is continuous if the inverse imge of every open intervl is open. However, not every mesurle function is continuous, while every continuous function is clerly mesurle. Note: Simple functions, step functions, continuous functions, nd monotonic functions re mesurle. 3 5

16 Ch - Integrl Clculus 8/3/5. Leesgue Integrl Proposition. Let f & g e mesurle etended rel-vlued functions on A ` A Then, the following functions re ll mesurle on A: f + c; c f ; f + g; f g, where c R. Note: Usul convention to void nonsense results, when f (nd/or g= /- : =. Now, we hve the tools to define, nd possily compute, the integrl for those functions: A f( λ(d tht represents, loosely, limit of integrl sums Σ i f( i λ(a i, where λ is mesure on some spce A, nd {A i } is prtition of A, nd i is point in A i. 3. Leesgue Integrl To define the Leesgue integrl, we usully follow these steps: - Define the Leesgue integrl for "simple functions." - Define the Leesgue integrl for ounded functions over sets of finite mesure. - Etend the Leesgue integrl to positive functions (not necessrily ounded. (The concept of mesurle function plys role. - Define the generl Leesgue integrl. Definition: Simple function φ(ω = Σ i= to n i I Ai (ω, where A,...,A k re mesurle sets on Ω, I Ai is n indictor function nd,..., k re rel numers. Let A,...,A k e prtition of Ω --i.e., A i s re disjoint nd A U... U A k = Ω. Then, φ(. with distinct i s ectly chrcterizes this prtition. 3 6

17 Ch - Integrl Clculus 8/3/5. Leesgue Integrl Simple functions Simple functions cn e thought of s dividing the rnge of f, where the resulting sets A n my or my not e intervls. Emple: A step function, φ(ω = i for j- < < j nd the { j } form prtition of [, ]. Upper, Lower, nd Riemnn sums re emples of step functions. Definition: Leesgue Integrl for Simple Functions Let φ( = Σ n n I An ( e simple function nd μ(a n e finite for ll n, then the Leesgue integrl of φ is defined s: φ( d = μ(a + μ(a...+ n μ(a n = Σ n n μ(a n If E is mesurle set, we define E φ( d = I E ( φ( d Recll: μ(a =[,] =. Thus, μ( j, j +d= d 33. Leesgue Integrl Simple functions Emple : Leesgue integrl of step function f(, defined s: f( = if - < < if < 4 3 if 4 8 otherwise f( d= μ(a + μ(a...+ n μ(a n = μ([-,] + μ([,4] + 3 μ([4,8] = =

18 Ch - Integrl Clculus 8/3/5. Leesgue Integrl Simple functions Emple : Leesgue integrl of f( = c f( = c cn e written s simple function f( = c I R ( Then, the Leesgue integrl of f over [,] is y definition: [, ] f( d = I [, ] ( f( d = = c I [, ] ( d = c μ([, ] = c ( - Note: The sme nswer s for the Riemnn integrl. Emple 3: Leesgue integrl of Dirichlet function on [,]. Let Q e the set of ll rtionl numers, then the Dirichlet function restricted to [, ] is the indictor function of A = Q [, ]. The set A is suset of Q, hence A is mesurle nd μ(a =. Thus, I A ( d = μ(a = Note: Not the sme nswer s in Riemnn s cse. 35. Leesgue Integrl Bounded functions We used step functions to define the R-integrl of ounded function f over n intervl [,]. Now, we use simple functions to define the L-integrl of f over set of finite mesure. Definition: Leesgue Integrl for Bounded Functions Suppose f is ounded function defined on mesurle set E with finite mesure. Define the upper nd lower Leesgue integrls s I * (f L = inf{φ( d: φ is simple nd φ f } (lower I * (f L = sup{φ( d: φ is simple nd φ f } (upper If I * (f L = I * (f L the function f is clled Leesgue integrle (L-integrle over E nd the Leesgue integrl of f over E is denoted y E f( d 36 8

19 Ch - Integrl Clculus 8/3/5. Leesgue Integrl Bounded functions Emple: Is f(= L-integrle over [,]? We know tht f( over the intervl [,]. Define sets: E j = { [,]: (j-/n f( < j/n} for j =,,..., n. Becuse f is continuous, the sets E j re mesurle, they re disjoint, nd their union (over the j's equls [,]. Define two simple functions S n ( = Σ j j/n I Ej ( s n ( = Σ j (j-/n I Ej ( Fi n integer n nd tke numer [,. Then, must e contined in ectly one set E j, nd on tht set we hve s n ( = (j-/n f( < j/n = S n ( Thus, on ll of [,], we know tht s n ( f( S n ( 37. Leesgue Integrl Bounded functions Emple (continution: Thus, on ll of [,], we know tht s n ( f( S n (. But then, I * (f L S n ( d = /n Σ j j μ(e j I * (f L s n ( d = /n Σ j (j- μ(e j Therefore, I * (f L -I * (f L /n Σ j (j - (j- μ(e j = /n Σ j μ(e j = /n μ([,] = /n Since n ws ritrry the upper nd lower Leesgue integrls must gree, hence the function f is L-integrle. Note: With few simple modifictions this emple cn e used to show tht every ounded function f, which hs the property tht the sets E j re mesurle, is L-integrle. 38 9

20 Ch - Integrl Clculus 8/3/5. Leesgue Integrl Bounded functions Emple: Vlue of the Leesgue integrl f( = over [,] Compute μ(e j using the fct tht f( = : for fied n we hve μ( E j = μ({ [,]:(j-/n < f( < j/n} = = μ({ [,]: (j-/n < < j/n} = = μ([(j-/n, j/n] = /n Then, f( d = lim /n Σ j j μ(e j = lim /n Σ j j /n = lim /n Σ j j = lim /n [n(n-/] = / => sme vlue s for the Riemnn integrl. 39. Leesgue Integrl Generl Cse We hve etended the concept of integrtion to (ounded functions defined on generl sets (mesurle sets with finite mesure without using prtitions (suintervls. The Leesgue integrl grees with the Riemnn integrl, when oth pply. This new concept removes some strnge results for emple, we cn integrte over Dirilecht functions over n intervl. But, we hve restricted our ttention to ounded functions only. To generlize the Leesgue integrl to functions tht re unounded, including functions tht my occsionlly e equl to infinity, we need the concept of mesurle function. Recll tht mesurle functions do not hve to e continuous. They my e unounded nd they cn e equl to ±. They re "lmost" continuous i.e., ecept on set of mesure less thn ε. 4

21 Ch - Integrl Clculus 8/3/5. Leesgue Integrl Generl Cse Definition: Leesgue Integrl of Non-Negtive Functions Let f e mesurle function defined on E nd h e ounded mesurle function such tht λ({: h( > } is finite, then we define E f( d = sup{ E h( d, h f } If E f( d is finite, then f is clled L-integrle over E. Definition: Generl Leesgue Integrl Let f e mesurle function. Define the positive nd negtive prts of f, respectively, s: f + ( = m(f(, f - ( = m(-f(, so tht f = f + -f -. Then, f is Leesgue integrle if f + nd f - re L-integrle nd E f( d = E f + ( d - E f - ( d 4. Leesgue Integrl - Remrks The Leesgue integrl is more generl thn the Riemnn integrl: If f(. is R-integrle, it is lso L-integrle. For most prcticl pplictions, we use the result tht for continuous functions or ounded functions with t most countly mny discontinuities over intervls [,] there is no need to distinguish etween the Leesgue or Riemnn integrl. Then, ll Riemnn integrtion techniques cn e used. But, for more complicted situtions, the Leesgue integrl is more useful. The Leesgue integrl mkes no distinction etween ounded nd unounded sets in integrtion, nd the stndrd theorems pply eqully to oth cses. 4

22 Ch - Integrl Clculus 8/3/5. Leesgue Integrl - Remrks For theoreticl purposes the Leesgue integrl provides n strction level tht simplifies proofs. But, then techniques such s integrtion y prts or sustitution my no longer pply. It plys pivotl role in the iomtic theory of proility. A proility mesure ehves nlogously to n re mesure, nd, in fct, proility mesure is mesure in the Leesgue sense. There re severl other generliztions of the Riemnn integrl: Perron, Denjoy, Henstock, etc. H. Leesgue (875-94, Frnce 43. Nottion f(: function (it must e continuous in [,]. : vrile of integrtion f( d: integrnd, : oundries f ( d 44

23 Ch - Integrl Clculus 8/3/5. Properties of Integrls Assuming f( nd g( re Riemnn integrle functions on [,], with c inside [,] nd k nd q re constnts, the following properties cn e derived (the lst three re esy if we think of Riemnn integrtion s summtion: f ( d f ( d f ( d [ kf ( qg( ] d k f ( d c f ( d f ( d c f ( d f ( d f ( d q g( d 45. Fundmentl Theorem of Clculus The fundmentl theorem of clculus sttes tht differentition nd integrtion re inverse opertions. It reltes the vlues of ntiderivtives to definite integrls. Becuse it is usully esier to compute n ntiderivtive thn to pply the definition of definite integrl, the Fundmentl Theorem of Clculus provides prcticl wy of computing definite integrls. It cn lso e interpreted s precise sttement of the fct tht differentition is the inverse of integrtion. 46 3

24 Ch - Integrl Clculus 8/3/5. Fundmentl Theorem of Clculus The Fundmentl Theorem of Clculus sttes: If function f is continuous on the intervl [, ] nd if F is function whose derivtive is f on the intervl (,, then Furthermore, for every in the intervl (,, df ( F ( f ( t dt, stisfying f ( d In other words, if function hs derivtive over rnge of numers, the integrl over tht sme rnge cn e clculted y evluting t the end points of the rnge nd sutrcting. 47. Fundmentl Theorem of Clculus: Notes The first prt is used to evlute integrls. The second prt defines the nti-derivtive. Finding the ntiderivtive is finding the integrl. Emple: Find the ntiderivtive of f( = 4 F( = 5 In generl, smll letters will e used for functions, cpitl letters for nti-derivtives. 48 4

25 Ch - Integrl Clculus 8/3/5. Physics Emple: Velocity & Accelertion Velocity, v(t, is defined s the derivtive of position, (t. Accelertion, (t, is the derivtive of v(t, nd the second derivtive of (t. The integrl of ccelertion is velocity. The integrl of velocity is position. The grph of v(t ginst time, t, shows tht the re etween two times is the distnce trveled. On the other hnd, the re under (t ginst time shows the velocity Rules of Integrtion Integrtion of the power function: n n For n d C n Integrtion of d ln( C e The Integrl of e The Constnt Multiple Rule The Sum Rule for Integrls Integrtion of sin function: e d C cf ( d c f ( d ( g( d f ( d f g( d cos sind C 5 5

26 Ch - Integrl Clculus 8/3/5.3 Rules of Integrtion: Applied Joke There's ig clculus prty, nd ll the functions re invited. ln( is tlking to some trig functions, when he sees his friend e sulking in corner. ln(: "Wht's wrong e?" e : "I'm so lonely!" ln(: "Well, you should go integrte yourself into the crowd!" e looks up nd cries, "It won't mke difference!" 5.3 Rules of Integrtion Integrtion of cosine function: sin cosd C Integrtion of d rctn C Note: The rules of differentition re complete, given set of opertions for constructing functions. But, the rules of integrtion re incomplete. We cnnot integrte simple functions like sqrt(

27 Ch - Integrl Clculus 8/3/5.3 Rules of Integrtion: Emple Evlute the following: Grph 3 3 ( e d Sum of 3 integrls Apply integrtion rules nd FTC Prt. Compute re under the curve e 3 d e 3 3 e 3 d d Integrtion y Sustitution Some integrls cnnot e esily solved y just pplying the previous rules of integrtion. Consider Grph: 3 cos( d We cn use chin rule like rgument to simplify the integrtion. For emple, let u= 3 -, then du=3 d => d = /3 du Sustituting ck into the originl integrl: cos( u du sin( u C sin( C 54 7

28 Ch - Integrl Clculus 8/3/5.4 Integrtion y Sustitution Suppose we wnt to find the integrl from - to.5. Then,.5 Grph: cos( 3 d sin( 3 [sin( C - - sin( ] Integrtion y Sustitution: Rule Theorem: Sustitution Rule Let f(. e continuous function defined on [, ], nd s(. continuously differentile function from [c, d] into [, ]. Then, f ( s( t s'( t dt s( s( f ( d If we cn identify composition of functions s well s the derivtive of one of the composed functions, we cn find the ntiderivtive nd evlute the corresponding integrl. 56 8

29 Ch - Integrl Clculus 8/3/5.4 Integrtion y Sustitution - Emple The key is to find the pproprite u. The vrile of integrtion chnges from to u. Note: It is importnt not to forget to sustitute lso d. Another emple: u 4 3 d du du 4d d u u du C Integrtion y Prts Recll the product rule of differentition: d(u v = u dv+ v du Solve for u dv: udv d ( uv vdu Integrting oth sides: The lst formul is used to integrte y prts. udv [ d ( uv vdu ] udv uv vdu Key: Selection of u & v functions. In generl, u involves logs, inverse, power, eponentil, nd trigonometric functions (in this order, LIPET. 58 9

30 Ch - Integrl Clculus 8/3/5.5 Integrtion y Prts: Emple I Emple: e d Select pproprite u & v functions ctully, select dv. Recll LIPET. We hve power function nd n eponentil function. Since power functions come efore eponentil functions, u equls. From u, get du. => u=, then du=d From dv, get v. => v=e, then dv=e d Then, simply plug this into the integrtion y prts formul Integrtion y Prts: Emple I (cont Replce u, v, du, nd dv. If the integrl on the right looks esy to compute, then simply integrte it. Otherwise, you cn integrte y prts gin, or use sustitution method. udv uv e e e vdu e d e e C ( C d 6 3

31 Ch - Integrl Clculus 8/3/5.5 Integrtion y Prts: Emple II Emple: e cos( d Select pproprite u & v functions. Eponentil functions come efore trigonometric functions. Then, u = e. From u, get du. =>u = e =>then du = e d From dv, get v. => dv =cos( d => v = sin(. Then, simply plug this into the integrtion y prts formul. e e udv uv cos( d sin( vdu e sin( d 6.5 Integrtion y Prts: Emple II (cont The epression looks more complicted thn the originl. Integrte y prts gin. The integrl of e cos( is equl to Κ (the originl integrl. Replce the integrl of e cos( with Κ nd solve for Κ. e u e v cos( e Note : e 5 e e sin( e sin( e sin( e sin( e sin( e 5 e du e sin( d dv sin( d cos( cos( d d cos( 4 cos( cos( C e cos( d 6 3

32 Ch - Integrl Clculus 8/3/5.5 Integrtion y Prts: Tricks In n integrl, do only integrtion y prts in the prts tht re not esy to integrte. If the integrtion y prts is getting out of hnd, you my hve selected the wrong u function. If you see your originl integrl in the integrl prt of the integrtion y prts, just comine the two like integrls nd solve for the integrl. Integrtion y prts cn e used to derive n effective wy to compute the vlue of n integrl numericlly, the trpezoid rule Integrtion y Prts: Trpezoid Rule Riemnn sums cn e used to pproimte n integrl, ut convergence is slow. There re mny rules designed to speed up the clcultions. The trpezoid rule is simple, with good convergence. To prove it, we need the Men Vlue Theorem for Integrtion Theorem: MVT for Integrtion Let f nd g e continuous functions defined on [,] so tht g(, then there eists numer c Є [,] with f ( g ( d f ( c g ( d Proof: Simple eercise. (Use the supremum nd infimum of f( on [,], pply Riemnn integrl properties nd then use the Intermedite Vlue Theorem for continuous functions. 64 3

33 Ch - Integrl Clculus 8/3/5.5 Integrtion y Prts: Trpezoid Rule Trpezoid Rule Let f (. e twice continuously differentile function defined on [,] nd set K = sup{ f''(, Є [, ]}. Define h = ( - /n, where n is positive integer. Then, f ( d f ( h f ( n j n j f ( f ( jh jh f ( f ( K where R ( n ( h Note: The Trpezoid Rule is useful ecuse the error, R(n, depends on the squre of h. If h is smll, h is lot smller! h R ( n h R ( n 65.5 Integrtion y Prts: Trpezoid Rule Derivtion of Trpezoid Rule First, we prove the trpezoid rule on [,]. Tht is, f ( d f ( f ( f ( c, where c [,]. Trick: Define function v( = / ( -, which hs the properties: - v( for ll Є [,] & v( = v( =. - v'( = / - => v ( =-/ & v( = ½ - v''( = - Then, f ( d v''( f ( d 66 33

34 Ch - Integrl Clculus 8/3/5.5 Integrtion y Prts: Trpezoid Rule f ( d v''( f ( d We integrte y prts with g'( = v''(: v''( f( d = v'( f( - v'( f( - v'( f'( d = = -/ f( - / f( - v'( f'( d Agin, we integrte y prts with g'( = v'(: v'( f'( d = v( f'( - v( f'( - v( f''( d = - v( f''( d =- f''(c v( d = - f''(c / where we used the MVT for Integrtion with some numer c Є [,]. Putting everything together, we hve the simple trpezoid rule: f( d = / f( + / f( + v'( f'( d = = / f( + / f( - / f''(c 67.5 Integrtion y Prts: Trpezoid Rule Generl trpezoid rule: Assume tht f (. is defined on [,]. Let h = ( - /n, pick n integer j, nd define the function u( = + jh + h for Є [,]. The composite function g( = f(u( is twice continuously differentile nd defined on [,]. We cn use the simple trpezoid rule: g( d = / g( + / g( - / g''(c But g(=f(u(=f( +jh; g(=f(u(=f( + (j+h, & g''(=h f''(. Also notice tht / g( + / g( - / g''(c = g( d = f(u( d = / h f ( u( u'( d / h ( j h jh f ( u du 68 34

35 Ch - Integrl Clculus 8/3/5.5 Integrtion y Prts: Trpezoid Rule Then, ( j h jh f ( d h [ h f ( jh f ( ( j h] f ''( c Summing this eqution from j= to j=n- completes the derivtion: 3 where 69.6 Improper Riemnn Integrls Improper Riemnn integrl: It is the limit of definite integrl s n endpoint of the intervl(s of integrtion pproches either specified rel numer tht cuses discontinuity or or or, in some cses, s oth endpoints pproch limits. Roughly, it is n integrl tht hs infinity s its limits or hs discontinuity within its limits. Emples: 5 4 d d Infinity s oundry (Prolem: domin of integrtion unounded. Discontinuity t =4 (Prolem: integrnd is unounded in the domin of integrtion. 7 35

36 Ch - Integrl Clculus 8/3/5.6 Improper Riemnn Integrls The Riemnn integrl cn often e etended y continuity, y defining the improper integrl insted s limit. With limits of infinity, use letter to replce the infinity, sy τ, nd tret it s limit (lim τ. For emple, d lim lim d With points of discontinuity, split integrl into prts. But, we cnnot integrte to the point of discontinuity, sy. Then, we integrte to ±δ nd tke limits s δ. An improper integrl converges if the limit defining it eists. It is lso possile for n improper integrl to diverge to infinity or to no prticulr vlue (oscilltion. 7.7 Applictions: Vlue of Bond Vlue of csh flows of ond, pying continuous dividend. - Mturity: T (sy, T=3 yers - Fce vlue: FV (sy, $, - Continuous discount rte: r (sy, 5% nnul - Continuous dividend rte: δ (sy, 7% nnul CF T FV e FV( e r rt rt dt FV e FV( e r CF( T 3; FV,, r.5,.7 T e rt e dt FV( r rt rt T.7 $,(.5 e.53 $

37 Ch - Integrl Clculus 8/3/5.7 Applictions: Elements of Proility X is rndom vrile (n outcome of rndom event. A rndom vrile (RV is function. X cn e discrete (recession, oom or continuous (price of IBM stock tomorrow. Given smple spce S (S: set of ll possile outcomes To ech discrete outcome A we ssocite rel numer P(A P is clled proility function nd P(A is clled the proility of the event A if Aiom : For every event A, P(A Aiom : For the Sure/Certin event S, P(S = Aiom 3: For ny numer of mutully eclusive events A, A, A 3, we hve P(A U A U A 3 U = P(A + P(A + P(A Applictions: Elements of Proility For continuous RV, the proility function is f(, such tht: f ( f ( d P( X f ( d f( is clled proility density function (pdf, or just density. The cumultive distriution function (CDF of discrete RV X, denoted s F(, is F( = P(X = i P( i The cumultive distriution function (CDF of continuous RV X, denoted s F(, is F( = P(X = i f( i 74 37

38 Ch - Integrl Clculus 8/3/5.7 Applictions: Elements of Proility Emple: Uniform Distriution: f(=constnt i.e., the proility of ny outcome is the sme. Suppose [,]. Wht is the proility tht is etween nd 5? P( X 5 5 f ( d 5 d 5 d ( 5 (5 Emple: Eponentil Distriution: f(= λ e -λ for. Let Suppose λ =3, wht is the proility tht is etween nd? P( X ( e 3 f ( d3 e e e d ( e Applictions: Elements of Proility Men nd Vrince Suppose X is continuous RV with proility density function f(. Then, the men or epected vlue of X, denoted μ, is E [ X ] f ( d The vrince of X, denoted s σ or V[X], is ( V [ X ] f ( d The stndrd devition, σ, is the squre root of V[X]

39 Ch - Integrl Clculus 8/3/5.7 Applictions: Elements of Proility Emple: Uniform Distriution: f(= /. Clculte the men nd vrince of f(, where [,]? E[ X ] V[ X ] f ( d ( 3 ( 6 f ( d d ( d ( 3 3 ( d ( u ( u du ( 3 Emple: Eponentil Distriution: f(= λ e -λ for. Clculte the men (integrtion y prts needed, u=, v=-e -λ. e E[ X ] e d e e d ( Applictions: Truncted Norml Suppose we re interested in regression, y i = i β + i, i ~N(, σ ut we only oserve the prt of the smple with y>. Model: y i = i β + i, for y i = i β + i > - Let s look t the density of i, f (., which must integrte to : ' ( d -The i s density, norml y ssumption: i ' - Then, f (. cn e written s: i f f ( d F f F i i f F ' i i f ( d i ' ( e i ( e 78 39

40 Ch - Integrl Clculus 8/3/ Applictions: Truncted Norml Now, we cn clculte the epected vlue of i : - Integrtion y sustitution: - u =η/σ nd dη= σ du. -F(u = ep(-u / - Then, E[ i ] = σ F i - f i = σ λ( i β (nd it depends on i β => E[y i y i >, i β] = i β + σ λ( i β => OLS in truncted prt is ised (omitted vriles prolem. ' ( ( ] ( [ ( ( ] [ ' ' ( ' ( ' ' i i i i i i i i i f f f F f F d e F d e F d f F d f E i i i i i integrtion y sustitution 8 Now, z=f(,, we hve two vriles of integrtion: nd y. Simple integrls represent res, doule integrls represent volume..8 Doule Integrls We wnt to know the volume defined y z=f(, on the rectngle R=[,] [c,d]

41 Ch - Integrl Clculus 8/3/5.8 Doule Integrls Similr to the intuition ehind simple integrls, we cn think of the doule integrl s sum of smll esy to clcultevolumes. 8.8 Doule Integrls y ij s column: z R ij ( i, y j f ( ij*, y ij* Smple point ( ij*, y ij* y Are of R ij is Δ A = Δ Δ y Volume of ij s column: Totl volume of ll columns: * * f (, y A m ij n i j ij f (, y A * ij * ij 8 4

42 Ch - Integrl Clculus 8/3/5.8 Doule Integrls V m n i j f (, y A * ij * ij Definition of Doule Integrl: V m n lim m, n i j f (, y * ij * ij A 83.8 Doule Integrls The doule integrl R f (, y da m n f (, y da lim R m, n i j Doule Riemnn sum: f ( m * ij n i j of f over the rectngle R is, y * ij f ( A, * ij, y * ij if the limit eists. A Note : If f is continuous then the limit eists nd the integrl is defined. Note : The definition of doule integrl does not depend on the choice of smple points. If the smple points re upper right-hnd corners then m n f (, da limm, n f ( i, y j R i j A 84 4

43 Ch - Integrl Clculus 8/3/5.8 Doule Integrls: Emple Let z=6- -y, where nd y. Estimte the volume of the solid ove the squre nd elow the grph Let s prtitioned the volume in smll (mn volumes. Ect volume: 48. m=n=4;v 4.5 m=n=8;v m=n=6;v Doule Integrls: Fuini s Theorem Theorem: Fuini s Theorem Suppose A nd B re complete mesure spces. Suppose f(, is A Bmesurle. If AB f (, d(, where the integrl is tken with respect to product mesure on the spce over A B, then f (, da f (, d dy f (, dy d AB A B B A where the lst two integrls re iterted integrls with respect to two mesures, respectively, nd the first eing n integrl with respect to 86 product of these two mesures. 43

44 Ch - Integrl Clculus 8/3/5.8 Doule Integrls: Fuini s Theorem Fuini s Theorem is very generl. For the Riemnn s cse, we hve: If f(, is continuous on rectngle R=[,] [c,d] then the doule integrl is equl to the iterted integrl. R d d f (, da f (, ddy f (, dyd Tht is, we cn compute first c y holding y constnt nd integrting over s if this were n single integrl. This cretes function with only, which we cn integrte s usul. Then, we integrte over y, gin, s usul. 87 f (, d c.8 Doule Integrls: Computtion Fuini s theorem simplifies clcultion y llowing iterted computtions. There re two wys of doing the itertion: d y y R d d f (, da f (, ddy f (, dyd c fied fied c c 88 44

45 Ch - Integrl Clculus 8/3/ Emple:.8 Doule Integrls: Computtion ( (, ( y y ydy dy dy y dy y dy d y ddy y da y f R 9.8 Doule Integrls: Computtion (Generl Cse Before, we looked t doule integrls over rectngulr region, R. Not relistic. Most regions re not rectngulr. We dpt our previous result to the generl cse. If f(, is continuous on A={(, [,] & h( y g(}, then the doule integrl is equl to the iterted integrl: y h( g( g h A dyd y f da y f ( (, (, ( A

46 Ch - Integrl Clculus 8/3/5.8 Doule Integrls: Computtion (Generl Cse Similrly, if f (, is continuous on A={(, y [c,d] & h( g(} then the doule integrl is equl to the iterted integrl: y d y c d g ( f (, da R h( A g( c h( f (, ddy 9 If f (, = φ ( ψ( then R.8 Doule Integrls: Fuini s Thorem Corollry d d f (, da ( ( ddy ( d ( dy c c Emples: R R ysin( da, e ( ( y y e A [/,] [ /, ] ddy, R [, ] [, ] 9 46

47 Ch - Integrl Clculus 8/3/5.8 Doule Integrls: Polr Coordintes Sometimes, it is esier to move from Crtesin coordintes to polr coordintes. For emple, we hve region tht is disk or portion of ring. Crtesin coordintes could e cumersome. Emples: D We cn descrie the re D s: - & - (4- y (4- D f (, da f (, da where D is disk of rdius. 4 4 Esier to descrie disk of rdius in polr coordintes: θ π & r To integrte, we need chnge of vriles: =r sin(, y=r cos( nd da = r dr dθ. f (, ddy 93.8 Doule Integrls: Polr Coordintes Let s generlize the emple. Now,: α θ β & h (θ r h (θ Then, D f (, da h ( h ( f ( r cos(, r sin( rdrd Note: da = r dr dθ (not da = dr dθ, s in the Crtesin world. DR Emple: e y ( ddy e r ( rdrd d d We use chnge of vriles: = r sin(, y = r cos( (recll r = + y nd da = r dr dθ. e r ( 94 47

48 Ch - Integrl Clculus 8/3/5.8 Doule Integrls: Properties Linerity A A [ f (, g(, ] da f (, da g(, da cf (, da c f (, da A A A A Comprison: If f(, g(, for ll (, in R, then A f (, da g(, da A Additivity: If A nd A re non-overlpping regions then f (, y da f (, y da A A A f (, y da 95.9 Computtionl Science vs. Clculus Clculus tells you how to compute precise integrls & derivtives when you know the eqution (nlyticl form for prolem; for emple, for the indefinite integrl: (-t + t + 4 dt = - t 3 /3 + 5 t + 4t + C It turns out tht mny integrl do not hve nlyticl solutions or re complicted to compute, especilly when we move to more thn 3 dimensions. For these prolems, we rely on numericl pproimtions. Computtionl science provides methods for estimting integrls nd derivtives from ctul dt