Math Calculus with Analytic Geometry II


 Bertram Carter
 3 years ago
 Views:
Transcription
1 orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0
2 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple x dx = π 3 4 = 9π 4
3 orem of definite Problem 9 Compute 6 x 5 dx by finding s of regions between the grph of f nd the xxis. Solution to Problem 9 xintercept 0 = x 5 x = 5 ( under f bove xxis) A = (6 5 ) ( 6 5) = 49 4
4 orem of definite Solution to Problem 9 (continued) ( bove f under xxis) 6 A = ( 5 ( )) ( ( ) + 5) = ( 9 ) (9) = 8 4 x 5 dx = A A = = 8 Exmple 0 (Are formuls seldom work) Evlute x dx. No formul for this portion of circle
5 orem of definite orem ( orem of I) If f is continuous on the intervl [, b] nd F (t) = f (t) then b f (t) dt = F (b) F (). Exmple ( ) sin 8x + x 4 = 8 cos 8x + 4x 3 so by FTOC d dx 8 cos 8x + 4x 3 dx = (sin(8 ) + 4 ) (sin(8 ) + 4 ) = sin 6 sin 8 + 5
6 orem of definite Nottion g(x) b = g(b) g() x= or if x is clerly the vrible to plug nd b into cn write g(x) b = g(b) g() With this nottion FTOC I ( orem of I) sys If f is continuous on the intervl [, b] nd F (t) = f (t) then b f (t) dt = F (t) b t=
7 orem of definite FTOC sys roughly Computing under f Suprising! Why should nd slopes be relted? Finding with derivtive f Could prove FTOC from definition of integrl (Definition 9) nd definition of the derivtive, but we won t. Mkes some sense visully
8 orem of definite Note on units for If x mesured in units u nd f (x) mesured in units u then hs units u u Exmple 3 b f (x) dx If t is the time mesured in hours P(t) number of people working in fctory t time t. b P(t) dt is people hours worked between time nd b
9 orem of definite Integrl of rte of chnge If F (t) is the rte of chnge of some quntity F (t) then FTOC I sy tht b F (t) dt = F (b) F () This is net chnge in F from time to b Exmple 4 As in Exmple 4 from lst week t time (in h) v(t) velocity (rte of chnge of position) t time t (in km/h) then 3 0 v(t) dt is distnce trveled (net chnge in position) in units (km/h) h = km
10 orem of definite Exmple 5 t time (in yers) g(t) growth rte (in m/yer) of person t ge t then 6 9 g(t) dt is chnge in height (in m) of person from ge 9 to 6.
11 orem of definite Definition 6 ( ) verge vlue of f on the intervl [, b] is f ve = b Where does this formul come from? Let s derive it using b f (t) dt. Philosophy of Estimte quntity Figure out how to improve estimte Tke limit
12 orem of definite Averge temperture Let f (t) be the temperture t time t. Let s estimte the verge temperture between m nd 0pm. (First estimte with intervls) totl time = 0 ech intervl is 0 f ve f ( + 0 (Better estimte with 3 intervls) totl time = 0 ech intervl is 0 3 f ve f ( ) ( + f ) ( ) + f + 0 ) ( ) + f
13 orem of definite Averge temperture (continued) (Better estimte with n intervls) totl time = 0 ech intervl is 0 n f ve f ( + 0 n = n f i= = 0 ( + i 0 n n f i= ) ( ) + + f + n 0 n n ) n ( + i 0 n ) 0 n
14 orem of definite Averge temperture (continued) (Tke limit to get exct verge) f ve = lim n 0 = 0 lim n = 0 0 n f i= n f i= f (t) dt ( + i 0 n ( + i 0 n ) 0 n ) 0 n We ve derived the formul in Definition 6.
15 orem of definite Problem 7 Find the verge vlue of the funtion on the intervl [, ] Solution to Problem 7 h ve = b = h(x) = 4 x b ( ) = 4 π h(x) dx 4 x dx = π
16 orem of definite Problem 8 t is time mesured in dys since Jn., 003 R(t) is the distnce from the erth to the sun t time t Wht does represent? Solution to Problem R(t) dt Averge distnce from erth to the sun in 003
17 orem of definite orem 9 If f is integrble then orem 30 If f is integrble then definite b f (x) dx = b f (x) dx b c c f (x) dx + f (x) dx = f (x) dx b
18 orem of definite orem 3 If f is integrble then Proof. b b cf (x) dx = c cf (x) dx = lim n = lim n c b f (x) dx n cf ( + i x) x i= = c lim n = c b n f ( + i x) x i= n f ( + i x) x i= f (x) dx
19 orem of definite orem 3 If f nd g re integrble then Proof. b b f (x) + g(x) dx = f (x) + g(x) dx = lim n = lim n b f (x) dx + b g(x) dx n (f ( + i x) + g( + i x)) x i= n (f ( + i x) x + g( + i x) x) i=
20 orem of definite Proof of orem 3 (continued). n = lim (f ( + i x) x + g( + i x) x) n i= ( n ) n = lim f ( + i x) x + g( + i x) x n i= i= ( ) ( ) n n = f ( + i x) x + g( + i x) x = lim n b i= f (x) dx + b g(x) dx lim n i=
21 orem of definite Problem 33 Compute 6 7 t(v) 7u(v) dv given 6 7 u(v) dv = 6 7 t(v) dv = Solution to Problem t(v) 7u(v) dv = = t(v) dv t(v) dv 7 = 7( ) = u(v) dv u(v) dv
22 Wrning orem of definite Exmple 34 f (x) = b f (x) g(x) dx b {, 0 x 0, < x b f (x) g(x) dx f (x) dx b b f (x) dx g(x) dx nd g(x) = b g(x) dx { 0, 0 x, < x f (x)g(x) dx = dx = 0 but f (x) dx 0 0 g(x) dx = =
23 orem of definite Definition 35 f is even if f ( x) = f (x) f is odd if f ( x) = f (x) Note: Most functions re neither. One function is both. orem 36 If f is even then If f is odd then nd 0 f (x) dx = 0 0 f (x) dx = 0 f (x) dx = f (x) dx f (x) dx = 0 f (x) dx
24 orem of definite Problem 37 Suppose 4 Q(s) ds = 3 0 Q(s) ds = 9 4 Evlute 0 Solution to Problem 37 0 Q(s) ds = 4 Q(s) ds Q(s) ds + = 3 + ( 9) = Q(s) ds
25 orem of definite Problem 38 Compute 8 y(u) du given y is n odd function 8 y(u) du = 7 Solution to Problem y(u) du = y(u) du y(u) du = 7 0 = 7
26 orem of definite Problem 39 Compute 7 h(r) dr given 3 h is n even function 7 h(r) dr = h(r) dr = 0 3 Solution to Problem h(r) dr = = = = 3 7 h(r) dr h(r) dr ( 7 h(r) dr 3 h(r) dr + h(r) dr h(r) dr h(r) dr 3 7 = ( 0) + (4) = 8 0 h(r) dr 7 0 ) h(r) dr
27 orem of definite orem 40 If m f (x) M for x [, b] then m(b ) orem 4 b If f (x) g(x) for x [, b] then f (x) dx M(b ) b f (x) dx b g(x) dx
28 orem of definite Problem 4 Show tht cos(x) sin(x 3 ) dx 99 Solution to Problem 4 For ll x [, 00] cos(x) sin(x 3 ) = cos(x) sin(x 3 ) so cos(x) sin(x 3 ) By orem 40 (00 ) cos(x) sin(x 3 ) dx (00 ) cos(x) sin(x 3 ) dx 99
29 orem of definite Problem 43 For ech pirs of decide which is the lrger π 4 0 cos(x) dx nd π 4 sin(x) dx 0 π π 4 cos(x) dx nd π π 4 Solution to Problem 43 sin(x) dx For ll x [0, π 4 ] cos(x) sin(x) so π 4 0 cos(x) dx π 4 0 sin(x) dx For ll x [ π 4, π ] cos(x) sin(x) so π π 4 cos(x) dx π π 4 sin(x) dx
30 orem of definite Definition 44 (An ntiderivtive) F (x) is n ntiderivtive of f (x) if F (x) = f (x). Exmple 45 so ( d dx x cos(4x + 3) ) = x cos(4x + 3) + 4x sin(4x + 3) x cos(4x + 3) is n ntiderivtive of x cos(4x + 3) + 4x sin(4x + 3) x cos(4x + 3) + 30 is nother ntiderivtive.
31 orem of definite Motivtion Why do we cre bout finding? FTOC I sys tht computing b f (t) dt is esy if we hve n ntiderivtive F of f. Exmple 46 From Exmple 45 bove 7 x cos(4x + 3) + 4x sin(4x + 3) dx = x cos(4x + 3) = (7 cos( ) + 30) ( cos(4 + 3) + 30) = 49 cos(3) cos(7)
32 orem of definite Notice in Exmple 45 we could hve dded ny constnt to x cos(4x + 3) nd we would hve hd nother ntiderivtive of x cos(4x + 3) + 4x sin(4x + 3) We usully dd n unspecified constnt to remind us tht there re mny. Definition 47 ( ntiderivtive) ntiderivtive of f (x) is the set of ll of f (x). orem 48 If f is continuous nd F (x) = f (x) then every ntiderivtive of f is of the form F (x) + C for some constnt C.
33 orem of definite Wht if f is not continuous? ntiderivtive of noncontinuous function Let n F (x) = { ln x + 4, x > 0 ln( x) + 8, x < 0 { F (x) = x, x > 0 { x, x < 0 = x, x > 0 x, x < 0 = x So F (x) is n ntiderivtive of x. Any choice of constnts (4 nd 8 weren t specil) gives sme result. Thus the ntiderivtive of x is F (x) = { ln x + C, x > 0 ln( x) + C, x < 0 = { ln x + C, x > 0 ln x + C, x < 0
34 orem of definite On the other hnd Min reson we cre bout is the FTOC. FTOC only pplies if f is integrble on [, b] x is not integrble on intervls contining 0 so in pplictions we only use one of the two constnts t time Exmple 49 3 x dx = ln x +C Exmple x dx 3 = (ln +C ) (ln 3 +C ) = ln ln 3 cnnot be evluted using FTOC Nottionl wrning By convention we sy tht F (x) + C is the ntiderivtive of f (x) whenever F (x) = f (x) even when this is techniclly incorrect.
35 orem of definite Nottion f (x) dx = F (x) + C mens F (x) + C is the ntiderivtive of f (x) Terminology Since FTOC links ntidifferentition nd integrtion we lso cll (indefinite). following sttements ll men the sme thing: f (x) = d dx F (x) f (x) dx = F (x) + C f (x) is the derivtive of F (x) F (x) + C is the ntiderivtive of f (x) F (x) + C is the indefinite integrl of f (x) F (x) + C is the integrl of f (x)
36 orem of definite Problem 5 Check the following (6x + 3e x ) cos(3x + 3e x ) dx = sin(3x + 3e x ) + C sec x dx = ln sec x + tn x + C Solution to Problem 5 d dx sin(3x + 3e x ) = (6x + 3e x ) cos(3x + 3e x ) d dx ln sec x + tn x = sec x tn x + sec x sec x + tn x tn x + sec x = (sec x) sec x + tn x = sec x Note: Similr to x ntiderivtive of sec x should hve different C for ech intervl [ (n )π, (n+)π ] but nobody does this.
37 orem of definite (See bckbord)
38 orem of definite Ech rule for differentition gives us rule for integrtion From ( ) c d dx F (x) = d dx cf (x) we get orem 5 (Constnt rule for integrtion) cf (x) dx = c f (x) dx
39 orem of definite Proof of orem 5. Suppose d dx F (x) = f (x). We hve the derivtive rule c d dx F (x) = ( ) d dx cf (x) Reinterpreting this rule s n ntiderivtive gives c d dx F (x) dx = cf (x) + C. Thus we my conclude cf (x) dx = c d dx F (x) dx = cf (x) + C = c(f (x) + C ) = c f (x) dx.
40 orem of definite From we get ( ) d dx F (x) + d dx G(x) = d dx F (x) + G(x) orem 53 (Sum rule for integrtion) f (x) + g(x) dx = f (x) dx + g(x) dx
41 orem of definite Proof of orem 53. Suppose d d dx F (x) = f (x) nd dx G(x) = g(x). We hve the derivtive rule ( ) d dx F (x) + d dx G(x) = d dx F (x) + G(x) Reinterpreting this rule s n ntiderivtive gives d dx F (x) + d dx G(x) dx = F (x) + G(x) + C. Thus we my conclude f (x) + g(x) dx = d dx F (x) + d dx G(x) dx = F (x) + G(x) + C = f (x) dx + g(x) dx Note: We drop constnts when we hve on both sides of n eqution.
42 orem of definite Bsic Integrls Ech bsic derivtive gives us bsic integrl differentition rule Tble : Bsic to memorize integrtion rule d dx x r+ = (r + )x r x r dx = r+ x r+ + C if r d dx ln x = x d dx cos x = sin x d dx sin x = cos x d dx ex = e x d dx rctn x = +x d dx rcsin x = x x dx = ln x + C sin x dx = cos x + C cos x dx = sin x + C e x dx = e x + C +x dx = rctn x + C x dx = rcsin x + C
43 orem of definite More bsic You lso know few more derivtive Tble : More bsic to memorize differentition rule d dx tn x = sec x d dx cot x = csc x d dx sec x = sec x tn x d dx csc x = csc x cot x d dx x = (ln ) x integrtion rule sec x dx = tn x + C csc x dx = cot x + C sec x tn x dx = sec x + C csc x cot x dx = csc x + C x dx = x ln + C
44 orem of definite Techniques of integrtion Advnced derivtive give us techniques of integrtion differentition technique of rule integrtion chin rule usubstitution ( 7.) product rule integrtion by prts ( 7.) We will return to these integrtion techniques lter.
45 orem of definite Problem 54 Find formul for 0e x + 7 sin x dx Solution to Problem 54 0e x + 7 sin x dx = Check your nswer! = 0 0e x dx + e x dx sin x dx (Sum rule) sin x dx (Constnt rule) = 0e x 7 cos x + C (Tble ) d dx (0ex 7 cos x) = 0e x + 7 sin x
46 orem of definite Problem 55 Find formul for 8 t 8t dt Solution to Problem 55 8 t 8t dt = 8 = 8 dt t dt 8 t 8 t dt (Sum rule) t dt (Const. rule) = 8 rcsin t 8 t3 3 + C (Tble ) Check your nswer! ( d dt 8 rcsin t 8 3 t3) = 8 t 8 3 3t = 8 t 8t
47 orem of definite Problem 56 Find formul for rcsin(3 π ) Solution to Problem 56 rcsin(3 π ) = + cos u du + sec u du rcsin(3 π ) du + sec u du (Sum rule) = rcsin(3 π ) u + tn u + C (Tbles nd ) Check your nswer! ( d rcsin(3 π ) dt ) u + tn u = rcsin(3 π ) + sec u
48 orem of definite Problem 57 Compute 3 y( 6 6 y) 3 y dy Solution to Problem 57 3 y( 6 6 y) dy = y 3 = = = 3y ( y y 6 ) dy y 3 3y 36y y 5 6 ) dy y 3 3y 6 36y y 3 6 dy 3y 6 36y y dy = y y = 8 7 y y 3 + 7y = 3 y 3 ( ) ( )
49 orem of definite Problem 58 Find n ntiderivtive G(x) of g(x) = sin x + 7 stisfying G(π) = 0. Solution to Problem 58 G(x) = sin x + 7 dx = cos x + 7x + C Use fct tht G(π) = 0 to solve for C. 0 = G(π) = cos π + C So C = 0 + cos π = 0 + ( ) = G(x) = cos x + 7x
50 orem of definite Problem 59 verge vlue of h(x) = x 3 3x on [, ] is 8 solve for. Solution to Problem 59 h ve = x 3 3x dx ( ) = ( 4 x 4 x 3 ) = ( ) [ 4 ( )4 ( ) 3 ] ( ) = ( ) = ( 3) = Use fct tht h ve = 8 to solve for. 8 = h ve = so = ± 8
Properties 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 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 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 informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationReversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b
Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite
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 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 informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
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 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 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 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 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 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 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 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 informationMath 190 Chapter 5 Lecture Notes. Professor Miguel Ornelas
Mth 19 Chpter 5 Lecture Notes Professor Miguel Ornels 1 M. Ornels Mth 19 Lecture Notes Section 5.1 Section 5.1 Ares nd Distnce Definition The re A of the region S tht lies under the grph of the 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 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 informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
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 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 informationTest 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher).
Test 3 Review Jiwen He Test 3 Test 3: Dec. 46 in CASA Mteril  Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec. 1417 in CASA You Might Be Interested
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 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 informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
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 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 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 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 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 information1 Techniques of Integration
November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.
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 informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
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 informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
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, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More information5 Accumulated Change: The Definite Integral
5 Accumulted Chnge: The Definite Integrl 5.1 Distnce nd Accumulted Chnge * How To Mesure Distnce Trveled nd Visulize Distnce on the Velocity Grph Distnce = Velocity Time Exmple 1 Suppose tht you trvel
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 informationMath 116 Calculus II
Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.
Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos(  1 2 ) = rcsin( 1 2 ) = rcsin(  1 2 ) = Cn you do similr problems? Review of Bsic Concepts
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 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 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 informationES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1  Thurs 28th Sept 17 Review of trigonometry and basic calculus
ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1  Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems
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 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 information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos(  1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin(  1 ) = π 2 6 2 6 Cn you do similr problems?
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
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 informationMATH1013 Tutorial 12. Indefinite Integrals
MATH Tutoril Indefinite Integrls The indefinite integrl f() d is to look for fmily of functions F () + C, where C is n rbitrry constnt, with the sme derivtive f(). Tble of Indefinite Integrls cf() d c
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 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 informationImproper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Improper Integrls MATH 2, Clculus II J. Robert Buchnn Deprtment of Mthemtics Spring 28 Definite Integrls Theorem (Fundmentl Theorem of Clculus (Prt I)) If f is continuous on [, b] then b f (x) dx = [F(x)]
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 informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationThe Fundamental Theorem of Calculus, Particle Motion, and Average Value
The Fundmentl Theorem of Clculus, Prticle Motion, nd Averge Vlue b Three Things to Alwys Keep In Mind: (1) v( dt p( b) p( ), where v( represents the velocity nd p( represents the position. b (2) v ( dt
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
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 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 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 informationf a L Most reasonable functions are continuous, as seen in the following theorem:
Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f
More informationCalculus AB. For a function f(x), the derivative would be f '(
lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:
More informationMath 3B: Lecture 9. Noah White. October 18, 2017
Mth 3B: Lecture 9 Noh White October 18, 2017 The definite integrl Defintion The definite integrl of function f (x) is defined to be where x = b n. f (x) dx = lim n x n f ( + k x) k=1 Properties of definite
More informationThe practical version
Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht
More informationF (x) dx = F (x)+c = u + C = du,
35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil
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 informationFinal Exam  Review MATH Spring 2017
Finl Exm  Review MATH 5  Spring 7 Chpter, 3, nd Sections 5.5.5, 5.7 Finl Exm: Tuesdy 5/9, :37:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.
More information1. Find the derivative of the following functions. a) f(x) = 2 + 3x b) f(x) = (5 2x) 8 c) f(x) = e2x
I. Dierentition. ) Rules. *product rule, quotient rule, chin rule MATH 34B FINAL REVIEW. Find the derivtive of the following functions. ) f(x) = 2 + 3x x 3 b) f(x) = (5 2x) 8 c) f(x) = e2x 4x 7 +x+2 d)
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 informationTime in Seconds Speed in ft/sec (a) Sketch a possible graph for this function.
4. Are under Curve A cr is trveling so tht its speed is never decresing during 1second intervl. The speed t vrious moments in time is listed in the tle elow. Time in Seconds 3 6 9 1 Speed in t/sec 3 37
More informationSection 14.3 Arc Length and Curvature
Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in
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 information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
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 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 informationThe Fundamental Theorem of Calculus Part 2, The Evaluation Part
AP Clculus AB 6.4 Funmentl Theorem of Clculus The Funmentl Theorem of Clculus hs two prts. These two prts tie together the concept of integrtion n ifferentition n is regre by some to by the most importnt
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
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 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 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 informationcos 3 (x) sin(x) dx 3y + 4 dy Math 1206 Calculus Sec. 5.6: Substitution and Area Between Curves
Mth 126 Clculus Sec. 5.6: Substitution nd Are Between Curves I. USubstitution for Definite Integrls A. Th m 6Substitution in Definite Integrls: If g (x) is continuous on [,b] nd f is continuous on the
More information2008 Mathematical Methods (CAS) GA 3: Examination 2
Mthemticl Methods (CAS) GA : Exmintion GENERAL COMMENTS There were 406 students who st the Mthemticl Methods (CAS) exmintion in. Mrks rnged from to 79 out of possible score of 80. Student responses showed
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 information1 Probability Density Functions
Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our
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 informationSample Problems for the Final of Math 121, Fall, 2005
Smple Problems for the Finl of Mth, Fll, 5 The following is collection of vrious types of smple problems covering sections.8,.,.5, nd.8 6.5 of the text which constitute only prt of the common Mth Finl.
More informationMath 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions
Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,
More informationTopics Covered AP Calculus AB
Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.
More informationMATH SS124 Sec 39 Concepts summary with examples
This note is mde for students in MTH124 Section 39 to review most(not ll) topics I think we covered in this semester, nd there s exmples fter these concepts, go over this note nd try to solve those exmples
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 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 informationSection 5.4 Fundamental Theorem of Calculus 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus 1
Section 5.4 Fundmentl Theorem of Clculus 2 Lectures College of Science MATHS : Clculus (University of Bhrin) Integrls / 24 Definite Integrl Recll: The integrl is used to find re under the curve over n
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 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationAn Overview of Integration
An Overview of Integrtion S. F. Ellermeyer July 26, 2 The Definite Integrl of Function f Over n Intervl, Suppose tht f is continuous function defined on n intervl,. The definite integrl of f from to is
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 informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More information