# 5: The Definite Integral

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1 5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce tht the oject hs ( ) trveled during the first four seconds? The old d = r t only wors for constnt speeds Divide nd Conquer Let's te cue from the movies! As you now, movie is just series of still imges, tht re run relly fst. Ech imge is still (no movement), ut series of them give the illusion of movement. Perhps we could te our velocity eqution, chop it up into series of still imges, nd get the motion from tht Let's sy it this wy: for ny short intervl of time, the chnge in velocity is not very lrge thus, for short intervl of time, we could consider the oject to e moving with constnt velocity, nd use our old fmilir distnce eqution. There is grphic interprettion of this ide creting rectngles round the grph of ( ) We'll strt y chopping the time intervl into four pieces: v t. Notice tht the oject is t rest round time.5 seconds thus, it turns round nd heds c in the other direction. Thus, the totl distnce trveled will e different from the net distnce trveled (displcement). Keep this in mind for lter! The totl re of these rectngles is 5 the totl distnce trveled is 5 meters. This is only n pproximtion, though we're supposed to e chopping the time intervl into sections where the velocity is lmost constnt! Let's do etter: try ten intervls. Tht gives totl re of This should e little etter however, there will still e some error! Let's try once gin, with intervls: HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE OF

2 This gives n re of 4.8 this is etter, ut still not the nswer. Reductio d infinitum The nswer requires tht we consider n infinite numer of intervls! However, we'll leve tht for it, since tht requires limits. Let's te one step t time. Three Pths to the Summit The pictures tht I drew ove ll hve rectngles tht hve the left side on the function we cll these left-hnd rectngles. You could lso use midpoints (the height of the rectngle is t the midpoint of the top edge) or right sides (te wild guess!). You'll get slightly different nswers ut tht's oy; once we wor towrds the nswer, they'll ll come out the sme. Exmples [.] Estimte the re under the curve rectngles. y= e x in the intervl [, ] with four nd ten First, with four rectngles: The estimted re is.9. Now with ten rectngles: HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE OF

3 The estimted re is.66. [.] The following dt were collected from moving vehicle: time (sec) velocity (ft/sec) Estimte the distnce trveled y the vehicle during the thirty second intervl. Note tht the dt hve lredy divided the intervl into prtition! The width of ech suintervl is 5 (seconds), nd the height of the rectngle is the given velocity (feet per second). width height re This gives totl re of 4 which mens totl distnce of 4 feet. 5.: Definite Integrls The Generic Pth: Riemnn Sums Let's generlize the ides we explored in the previous section A prtition of the intervl [, ] is set of vlues ( x i ) so tht < x < x <... < xn <. If we =, then P= { x, x, x... x n } is prtition of the intervl [, ]. cll x = nd xn The th suintervl of P is [ x x ], e the width of the th suintervl of P. x. Let c e ny vlue in the th suintervl of P. Also, let If f is function which is continuous on the intervl [, ], then the Riemnn sum of f (on tht intervl) is f ( c ) n = x. Note tht this is the sum of unch of rectngle res The Definite Integrl Bc when we were developing the ide of derivtives, we creted generic point, nd then moved the point ritrrily close to the trget point we llowed the distnce etween the points to pproch zero. This limiting process gve us the slope of the curve. Now consider the rectngles res from the Riemnn sum. One might thin tht dding more points to the prtition might me the rectngle res get relly smll, nd dd up to the re HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE OF

4 under the curve however, just dding more points doesn't me ll of the rectngles smll. To do tht, you hve to dd points so tht the lrgest x pproches zero (the lrgest x is clled the norm of P, or P ). n If lim f ( c) x exists, then we sy tht f is integrle over the intervl [, ] P = vlue of the limit is clled the definite integrl of f over [, ] Nottion., nd the Tht limit nottion is terrile difficult to write, so we crete new nottion for the definite integrl of f over [, ] : ( ) f x dx. This is red s "the integrl of f from to." Evluting Definite Integrls Definite integrls re defined in such wy tht they pproximte the re etween curve nd the x-xis (if it lies ove the x-xis). However, since this is mth (nd not relity), the re etween the curve nd the x-xis for functions which lie elow the x-xis is negtive. Thus, the, is the re ove the xis plus the re elow the xis. definite integrl of f over [ ] If you relly wnt re (rel, positive re), then it's re ove the xis minus re elow the xis. Returning to our velocity exmple from while c: the totl distnce trveled will e the re etween the line v( t) = 9.8t+ 5 nd the x-xis. Fortuntely, this is just pir of tringles! First, note tht v (.55) =. The re of the left/upper tringle is ( 5 )(.55 ) =.888, nd the re of the right/lower tringle is ( 4.)( 4.55) =.88, for totl re of 4.76 ( totl distnce of 4.76 meters). The displcement is =.6 meters. Of course, rther thn doing ll of this y hnd, you could use your clcultor note tht the clcultor will do the definite integrl; not the re under the curve. My clcultor gives the vlue of the definite integrl t.6. Exmples 7 [.] Find dx. This is rectngle with se length of 8 nd height of ; the re (nd the vlue of the definite integrl) is 6. 5 [4.] Find ( ) 5 x dx. 5 =. This is tringle with se 5 nd height 5; the re is ( 5)( 5) HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 4 OF

5 [5.] Find 4 x dx. This is semicircle of rdius. The re is 4π. 5.: Definite Integrls nd Antiderivtives Properties of the Definite Integrl Reversing the limits negtes the integrl: ( ) = ( ) f x dx f x dx The integrl of point is zero: f ( x ) dx= Constnts cn e fctored out: ( ) = ( ) f x dx f x dx Integrls cn e "distriuted" cross sums nd differences: ( ) ± ( ) = ( ) ± ( ) f x g x dx f x dx g x dx c c Integrls over djcent intervls cn e comined: ( ) + ( ) = ( ) f x dx f x dx f x dx The Averge Vlue of Function We ll now how to find n verge dd up ll the vlues, nd divide y the numer of vlues. It turns out tht you cn do tht with n infinite numer of things, lso! Adding n infinite numer of things cretes n integrl, though. The verge vlue of f ( x ) over the intervl [, ] is ( ) f x dx. The Men Vlue Theorem (for Definite Integrls) There is some point ( x= c ) in the intervl [, ] so tht f ( c) = f ( x) dx. The Men Vlue Theorem for Derivtives sys tht for ny intervl, there is some point where the slope of the tngent equls the slope of the secnt. The Men Vlue Theorem for Integrls sys tht for ny intervl, there is some point where the re of rectngle equls the re under the curve (( ) ( ) ( ) f c = f x dx ). The Connection: A First Glimpse The derivtive is found with quotient of infinitely smll differences. The definite integrl is found through sum of infinitely smll products (res). Sums nd differences quotients nd products surely there is some connection! nd there is. Here's loo hed which you'll need for few of the prolems in this section. ( ) = ( ) ( ), where F( x ) is ny ntiderivtive of ( ) f x dx F F f x. HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 5 OF

6 Exmples [6.] If f ( x) dx= 7 nd g( x) dx=6, then find ( ) + ( ) Using the properties of integrls, ( ) ( ) ( ) ( ) ( ) ( ) f x g x dx. f x + g x dx= f x dx+ g x dx= = [7.] Find ( ) x x dx. Since I don't now hndy formul for the re under prol, I'll use ntiderivtives. One f x = x x is F( x) = x x x. Thus, ( ) ( ) ( ) f x dx= F F F = + =, so ( ) 6 6 x x dx= = = ntiderivtive of ( ) F ( ) = = nd ( ) [8.] Find the verge vlue of on the intervl [, 4 ]. x 4 The formul gives me dx 4. I'll hve to use n ntiderivtive gin how out x x? 4 4 dx= x = = = x : The Fundmentl Theorem of Clculus FTC, Prt x d If g( x) = f ( t) dt, then g ( x ) f ( x ) dx =. In other words, the derivtive of n integrl is just the plin function. It's s if the derivtive nd the integrl re inverse opertions The hirs on the c of your nec should e stnding on end The Integrl s Function x The ide of g( x) ( ) = f t dt needs some dditionl investigtion there re lots of potentil questions tht cn e sed out functions tht re defined s integrls of other functions! Here re just few exmples from single scenrio The grph of f ( t ) is shown elow. It consists of semicircle nd line segment. HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 6 OF

7 = f t dt. x Define A( x) ( ) A ( 6) is the re under ( ) circle. ( ) ( ) π A = π = = 9π. 4 4 A ( 8) is the re under ( ) tringle! ( ) ( ) ( )( ) f t etween x= nd 6 f t etween x= nd 8 A 8 = π = 8π x= ( ) f t dt which is qurter HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 7 OF 8 x= : ( ) f t dt. This is semicircle nd Note tht A( x ) is n incresing function s x gets lrger, the mount of re under f ( t ) eeps getting lrger. = 8 =π+. 8 The verge vlue of f ( t ) on the intervl [,8 ] is f ( t) dt A( ) FTC, Prt 8 8 If f ( x ) is ny ntiderivtive of g( x ), then g ( x ) dx= f ( ) f ( ). Note tht f ( x ) eing n ntiderivtive of g( x ) mens tht f ( x) g( x) =. Thus, this theorem is ting the integrl of n ntiderivtive. Also note tht the integrl of this derivtive resulted in the originl function (sort of). Finding Are If you wnt to use definite integrls to find the re etween curve nd the x-xis, rememer tht res elow the xis will e negtive! One wy to find the totl re is to first determine where the function is equl to zero this will divide the res into ove/elow the x-xis. Add the integrls from the sections ove the xis, nd sutrct the integrls from the sections elow the xis. Another wy to do this is to just find the re under the solute vlue of the function. The solute vlue will simply force ll of the res to e positive. This mes things relly esy especilly if you re using your clcultor Exmples π cos x dx. [9.] Find ( ) Well n ntiderivtive of cos( x ) is ( ) π ( x ) dx ( x π ) ( ) ( ) etween sin( x ) nd the x-xis is zero! sin x so cos = sin ] = sin π sin = =. Note tht this does not men tht the re [.] Find the totl re etween f ( x) = x x nd the x-xis on the intervl [,] Let's e creful nd loo t the grph..

8 We'll need to split this up! The totl re will e ( ) ( ) ( ) ( ) x x dx+ x x dx x x dx+ x x dx 4 An ntiderivtive of f ( x) = x x is F( x) x x =. Tht gives us 4 F ( x ) F ( x ) F ( x ) F ( x ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) ( ) ( ) + ( ) ( ) + ( ) F F F F F Let's strt plugging in! F( ) = 6 4= 4 = 4 F( ) = = 4 4 F = ( ) F ( ) = = 4 4 F ( ) = 6 4= 4 = = = So finlly, we get ( ) d t dt dx. π tn x. x [.] Find tn( ) Esy! The derivtive is ( ) 5.5: The Trpezoidl Rule Another Pth Now tht we hve firm connection etween integrls nd re, let's pproch the re ide gin. Before, we used rectngles to pproximte the re under the curve simply ecuse HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 8 OF

9 rectngles re very simple geometric figures. Als, they don't relly do such good jo of pproximting re under the curve until you use lot of very tiny rectngles. There is nother shpe which does etter jo the trpezoid! Divide the intervl in question s you would efore, ut now connect the function vlues to otin trpezoids. For our liner exmple (wy c t the eginning of this chpter), we get n exct nswer with s few s two trpezoids! Let's te the function f ( x) = 4 ( x ) on the intervl [ ] exmple. Here is the pproximte re with four rectngles:, 4 with four regions s n (tht's n re of 9.848) Here is the pproximte re with four trpezoids: (tht's n re of 8.788) The ctul re is 9. You cn see tht the trpezoids do much etter jo of the pproximtion! Simpson's Rule/Approximtion You cn do even etter with your pproximtion if, insted of trpezoids, you use rectngles cpped with little prolic rches. It turns out tht there is (reltively simple) formul for this nd (firly simple) formul tht generlizes the entire procedure! It's clled Simpson's Rule: To pproximte f ( x ) dx, divide the intervl [, ] into n even numer of suintervls y o+ y + y + y + + y y y n + n + n. n Hppily, this topic is not prt of the AP Exm! (me n even) nd then clculte ( ) HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 9 OF

10 Exmples [.] Use four trpezoids to estimte the vlue of Let's loo t the picture: x dx. The res of the trpezoids re + = = = + = = = + = = = + 8 = = 8 8 This gives totl re of = 6 = 4.5. (the ctul re is 4) HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE OF

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### Integrals - 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 non-liner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but

### Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

### Overview 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,

### Definite 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

### Improper 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

### Fundamental 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

### 4.4 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

### 1 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

### Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved. Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite

### Math 131. Numerical Integration Larson Section 4.6 Mth. Numericl Integrtion Lrson Section. This section looks t couple of methods for pproimting definite integrls numericlly. The gol is to get good pproimtion of the definite integrl in problems where n

### ( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists. AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find

### Math Calculus with Analytic Geometry II orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

### APPROXIMATE INTEGRATION APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

### 1 Part II: Numerical Integration Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble

### The 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

### 1 The fundamental theorems of calculus. The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Tody we provide the connection

### We 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:

### Chapters 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

### Math 116 Calculus II Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................

### Definition 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  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)

### Lab 11 Approximate Integration Nme Student ID # Instructor L Period Dte Due L 11 Approximte Integrtion Ojectives 1. To ecome fmilir with the right endpoint rule, the trpezoidl rule, nd Simpson's rule. 2. To compre nd contrst the properties

### Unit #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

### MATH , Calculus 2, Fall 2018 MATH 36-2, 36-3 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

### x = 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

### 10 Vector Integral Calculus Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve

### 10. AREAS BETWEEN CURVES . AREAS BETWEEN CURVES.. Ares etween curves So res ove the x-xis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in

### MA 124 January 18, Derivatives are. Integrals are. MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

### 38 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 x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

### Math 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

### 1 The fundamental theorems of calculus. The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Theorem Suppose f is continuous

### Z b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but... Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.

### 7.2 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

### Chapter 5. Numerical Integration Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with

### different 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

### Midpoint Approximation Midpoint Approximtion Sometimes, we need to pproximte n integrl of the form R b f (x)dx nd we cnnot find n ntiderivtive in order to evlute the integrl. Also we my need to evlute R b f (x)dx where we do

### ( ) where f ( x ) is a. AB/BC Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x). AB/ Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 Find the intersection of f ( x) nd g( x). A3 Show tht f ( x) is even. A4 Show tht

### 6.5 Numerical Approximations of Definite Integrals Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 6.5 Numericl Approximtions of Definite Integrls Sometimes the integrl of function cnnot be expressed with elementry functions, i.e., polynomil,

### How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique? XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=

### Section 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

### Unit #10 De+inite Integration & The Fundamental Theorem Of Calculus Unit # De+inite Integrtion & The Fundmentl Theorem Of Clculus. Find the re of the shded region ove nd explin the mening of your nswer. (squres re y units) ) The grph to the right is f(x) = -x + 8x )Use

### Calculus 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:

### Distance And Velocity Unit #8 - The Integrl Some problems nd solutions selected or dpted from Hughes-Hllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl 56 Chter 5: Integrtion 5.4 The Fundmentl Theorem of Clculus HISTORICA BIOGRAPHY Sir Isc Newton (64 77) In this section we resent the Fundmentl Theorem of Clculus, which is the centrl theorem of integrl 2 The Prllel Circuit Electric Circuits: Figure 2- elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is