10. AREAS BETWEEN CURVES


 Douglas Rice
 6 years ago
 Views:
Transcription
1 . AREAS BETWEEN CURVES.. Ares etween curves So res ove the xxis 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 sense. But now we ll level with you nd tell you the whole truth. The oundries for n re tht s given y n integrl re the grph of function, two verticl lines nd, up to now the fourth oundry hs een the xxis. We re now going to generlise tht so tht the fourth oundry is the grph of second function. You cn still tke the second function to e y = nd tht gives the xxis so you cn still hve the xxis s oundry if you like. But you cn hve curved ottom oundry too. We re going to develop formul for the re etween two curves. There will e top y nd ottom y. Top nd ottom refer to which hs the greter y vlues. Of course curves my cross over nd this mkes the prolem just little it more complicted, ut not much. y = f(x) (top y) y = g(x) (ottom y) The shded re is the re we wnt. Now f(x) dx gives the re under the top curve, right down to the xxis nd g(x) dx gives the re under the ottom curve, right down to the xxis. Clerly the re we wnt is the difference etween the two, tht is, f(x) dx g(x) dx. But this cn e written s [f(x) g(x)] dx. If f(x) g(x) for x, the re etween y = f(x) nd y = g(x) etween x = nd x = is [f(x) g(x)] dx
2 . To rememer this, just rememer the integrl s: top limit ottom limit (top y ottom y) Now when we hve just one curve y = f(x) which lies ove the xxis from x = to x = nd the ottom oundry of the re is the xxis, then the top y is y = f(x) nd the ottom y is y =. So the re under the grph is [f(x) ] dx) = f(x) dx, s efore. But if y = f(x) lies elow the xxis this ecomes the ottom oundry nd the xxis, y = ecomes the top oundry. The re ecomes, in this cse, [ f(x)] dx = f(x) dx. Previously we explined this minus sign y sying tht res elow the xxis re negtive. Now we re in position to set the record stright. Ares re never negtive, ut integrls my e. Where the curve is elow the xxis the re is positive ut the integrl is negtive. Tht s the right wy to look t it. Whenever you hve to work out res elow, ove or etween curves lwys set it up s prolem involving the re etween two curves. If you correctly identify which is the top y nd which is the ottom y you ll never go wrong. The integrl you set up will lwys come out s positive nd will give the required re. Integrls come out negtive, when deling with res, only if you hven t set things up properly. There s lot of sloppy thinking with this topic nd students re often tught d hits. Alright we did strt you off thinking tht res elow the xis re negtive ut t the time it would hve confused you to do otherwise. At lest we re now setting the story right. Some students think tht since res re positive ut integrls cn e negtive, ll you hve to do is to put solute vlue signs round everything. Even worse is the student who, on evluting n integrl to e, writes = =. Using solute vlue signs disply ignornce s to wht is relly going on nd cn led to wrong nswers if the curves cross over. So some guidelines to void such errors re: Never fudge the sign if the integrl comes out negtive. Never use solute vlue signs in connection with res. Exmple : Find the re etween y = x, y = 4 nd x =. y = 4 y = x Solution: Top y is y= 4 nd ottom y is y = x. The re is therefore [4 x ] dx = [ 4x (/)x ] = = 5.
3 .. Wht If Curves Cross? It s solutely importnt to determine whether curves cross within the region eing considered. For if they do, wht ws the top y will ecome the ottom y nd vice vers. The intervl of integrtion will hve to e split up. Suppose we wnt the re etween the curves y = f(x) nd g(x) etween x = nd x = nd suppose the curves cross t some point c etween nd. Suppose tht g(x) f(x) for x c nd g(x) f(x) for x c. Then from x = to x = c it s g(x) which is the top y, while from x = c to x = it s f(x) tht s on top. We must rek the intervl [, ] into two pieces nd integrte seprtely over these two regions. y = g(x) y = f(x) The required re will e c c [g(x) f(x)] dx + [f(x) g(x)] dx. c Exmple : Find the re etween y = x, the xxis nd lines x = nd x =. Solution: Becuse the curve cuts the xxis t x = we must divide the intervl [, ] into two pieces. From x = to x = the top y is y = (the xxis) while from x = to x = it is y = x. The required re is [ x ] dx + [x ] dx = x4 4 + x 4 4 = =. Two things should e noted with this exmple. () If we hd simply integrted x etween nd we would hve concluded tht the re is x dx = x 4 4 = 4 4 =, which is clerly wrong. And tking solute vlues wouldn t help either. Rememer to never use solute vlues with these re questions. If you don t know which is top y nd which is ottom you re likely to come up with totlly wrong nswer. If you do, then you ll set up the integrl correctly so tht it will come out positive. () We could hve exploited symmetry to simplify the clcultion nd void ll those minus signs. The re elow the xxis is clerly the sme s the re ove so we could hve sid tht the totl x 4 re = x dx = 4 =.
4 Exmple : Find the re etween y = x nd y = x. Solution: Here we re not given ny endpoints. Tht s ecuse the grphs of y = x nd y = x cut in exctly two plces nd so they enclose region. We need to find these two points. Solving x = x we get x = or. So these re the limits of integrtion. The top y, etween nd, is y = x. The re is therefore [x x ] dx. Becuse we know wht we re doing nd hve set up the integrl correctly we cn e sure tht it will e positive. x [x x ] dx = x = = 6. This is the required re. Exmple 4: Find the re etween y = x, y = x nd x =. Solution: The curves cut when x = x, tht is when x =. So they cut in two plces, t x =. / / We hve to integrte from to nd then from to. By symmetry the integrl from to is doule tht from to. / The re is therefore / = [( x ) x ] dx + [x ( x )] dx / [ x ] dx + [x ] dx = x / x / + x x = + = (6 + ) = 6 = =. / 4
5 EXERCISES FOR CHAPTER Exercise : Find the re etween the prols y = x, y = x nd the yxis. Give your nswer s n exct vlue, in terms of, s well s n pproximtion to 4 deciml plces. Exercise : Find the re enclosed etween the curves y = x x nd x x. Exercise : Find the re enclosed etween the curves y = e x nd y = 6x x etween x = nd x =. Give your nswer s n exct vlue involving e s well s n pproximte numericl vlue to 4 deciml plces. Exercise 4: Find the re enclosed etween the curve y = 6 x nd the line y = 5 x. Give the nswer to 4 deciml plces without using your clcultor, ut insted using the fct tht, to 4 deciml plces, log.5 =.455. Exercise 5: Use integrtion to find the re etween the lines y = + x nd y = x etween x = 5 nd x =. Check your nswer using simple geometry. Exercise 6: Find the re enclosed etween the curves y = 4x + 5, y = 5x + 4 nd the xxis. Exercise 7: The Lorenz curve is grph tht s used in economics tht shows the distriution of income. If the ottom x% of households represents the ottom y% of income, the point (x, y) lies on the Lorenz curve. The line y = x corresponds to perfect equlity, ut in prctice the curve lies elow it. The most extreme cse, (perfect inequlity) would e where one household erns ll of the income nd the rest ern nothing. This corresponds to the line y = (with spike t x = ). The Gini coefficient is the rtio of the re etween the line of perfect equlity nd the oserved Lorenz curve, nd the re etween the line of perfect equlity nd the line pf perfect inequlity. It is mesure of inequlity. The higher the Gini coefficient, the more unequl is the distriution of income. Find the Gini coefficient if the Lorenz curve is y = x. line of perfect equlity oserved Lorenz curve line of perfect inequlity 5
6 SOLUTIONS FOR CHAPTER Exercise : The curves cut when x = x, tht is, when x = or x =. Over the rnge from x = to x = the top y is y = x. / So the re is / ( x ) x dx = / x dx = [ x x ] =..7. Exercise : The curves cross if x x = x x. Solving, we get x x + x =. Fctorising we get x(x )(x ) =, so the curves cross when x =, nd. The re enclosed is the region etween the curves from x = to x =. But since they cross t x = we need to split the intervl [, ] into two pieces. From x = to x = the top curve is y = x x. Then from x = to x = the top curve is y = x x. So the re is: A = [(x x ) (x x)] dx + [(x x) (x x )] dx = x 4 4 x + x + x x4 4 x = =. Exercise : The curves do not cross nd the top y is y = e x. So the re is [e x (6x x )] dx = = (e ) = e [e x 6x + x + ] dx) = e x x + x +x] Exercise 4: The curves cut when 6 x = 5 x. Solving we get 6 = 5x x nd so x 5x + 6 =. Hence (x )(x ) = nd so x =, 5. When x =.5 the vlue of y for the curve y = 6 x is.4 while for the line y = 5 x it is.5. So etween x = nd x = the top y is y = 5 x. 6
7 The re etween is therefore [(5 x) x ] dx = [ 5x x log x ] = (5 9 log ) (9 4 log ) = log + log = log(/).455 = Exercise 5: The lines cut t x =. When x =, the vlue of y for y = + x is while for y = x it is, so etween x = 5 nd x = the top y is y = x. Between x = nd x = the top y is y = x +. So the enclosed re is 5 [( x) (x + )] dx + [(x + ) ( x)] dx = [ x] dx + [x] dx = x x = ( +.5) + ( ) = 75 + = 78 = 9. We cn check this y simple geometry since the region consists of two tringles. Rememer tht the re of tringle is ½ se perpendiculr height. Here we tke the se to e the verticl sides. When x = 5, x = 6 while x + = 9. So the lrger tringle hs se of 5 nd perpendiculr height of 5. Its re is 5 5 = 75. When x =, x + = nd x = so the smller tringle hs se nd perpendiculr height of. Its re is. The totl re is therefore 75 + = 9. Exercise 6: The curves cross when 4x + 5 = 5x + 4, tht is when x =. 7
8 The curve y = 5x + 4 cuts the xxis t x = 4/5 while y = 4x + 5 cuts it t x = 5/4. We must divide the re into two portions t x = 4/5. The top y etween x = 5/4 nd x = is 4x + 5. Between x = 5/4 nd 4/5 the ottom y is y =, the xxis. Then etween x = 4/5 nd x = the ottom y is y = 5x /5 Hence the re is 4x + 5 dx + [ 4x + 5 5x + 4] dx 4/5 5/4 4/5 =. 4 (4x +5)/ + 5/4. 4 (4x +5)/. 5 (5x + 4)/ = / / / = 6 / / 9 5 / 9/ = = = = 9 =.45. 4/5 [ 9/ ] Exercise 7: The re etween the line of perfect equlity (y = x) nd the oserved Lorenz curve, is (x x dx) = x x = = 6. The re etween the line of perfect equlity (y = x) nd the line of perfect inequlity (y = ) is the re of the tringle elow the line y = x, which is clerly ½. Hence the Gini coefficient is /6 / =. 8
2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More 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 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 informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More 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 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 informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More informationEvaluating Definite Integrals. There are a few properties that you should remember in order to assist you in evaluating definite integrals.
Evluting Definite Integrls There re few properties tht you should rememer in order to ssist you in evluting definite integrls. f x dx= ; where k is ny rel constnt k f x dx= k f x dx ± = ± f x g x dx f
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 informationSection 7.1 Area of a Region Between Two Curves
Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region
More informationCalculus 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
More informationPolynomials 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
More informationReview of Gaussian Quadrature method
Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More informationFor a, b, c, d positive if a b and. ac bd. Reciprocal relations for a and b positive. If a > b then a ab > b. then
Slrs7.2ADV.7 Improper Definite Integrls 27.. D.dox Pge of Improper Definite Integrls Before we strt the min topi we present relevnt lger nd it review. See Appendix J for more lger review. Inequlities:
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 informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More information7.1 Integral as Net Change Calculus. What is the total distance traveled? What is the total displacement?
7.1 Integrl s Net Chnge Clculus 7.1 INTEGRAL AS NET CHANGE Distnce versus Displcement We hve lredy seen how the position of n oject cn e found y finding the integrl of the velocity function. The chnge
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More 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 informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
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 informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More 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 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 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 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 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 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 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 informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More 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 informationChapter 9 Definite Integrals
Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished
More information1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE
ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check
More informationChapter 6 Continuous Random Variables and Distributions
Chpter 6 Continuous Rndom Vriles nd Distriutions Mny economic nd usiness mesures such s sles investment consumption nd cost cn hve the continuous numericl vlues so tht they cn not e represented y discrete
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 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 informationBridging the gap: GCSE AS Level
Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions
More informationLab 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
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 informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationChapter 1: Logarithmic functions and indices
Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4
More informationAT100  Introductory Algebra. Section 2.7: Inequalities. x a. x a. x < a
Section 2.7: Inequlities In this section, we will Determine if given vlue is solution to n inequlity Solve given inequlity or compound inequlity; give the solution in intervl nottion nd the solution 2.7
More informationChapter 7: Applications of Integrals
Chpter 7: Applictions of Integrls 78 Chpter 7 Overview: Applictions of Integrls Clculus, like most mthemticl fields, egn with tring to solve everd prolems. The theor nd opertions were formlized lter. As
More informationAP Calculus AB Unit 5 (Ch. 6): The Definite Integral: Day 12 Chapter 6 Review
AP Clculus AB Unit 5 (Ch. 6): The Definite Integrl: Dy Nme o Are Approximtions Riemnn Sums: LRAM, MRAM, RRAM Chpter 6 Review Trpezoidl Rule: T = h ( y + y + y +!+ y + y 0 n n) **Know how to find rectngle
More information5.1 How do we Measure Distance Traveled given Velocity? Student Notes
. How do we Mesure Distnce Trveled given Velocity? Student Notes EX ) The tle contins velocities of moving cr in ft/sec for time t in seconds: time (sec) 3 velocity (ft/sec) 3 A) Lel the xxis & yxis
More informationDate Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( )
UNIT 5 TRIGONOMETRI RTIOS Dte Lesson Text TOPI Homework pr. 4 5.1 (48) Trigonometry Review WS 5.1 # 3 5, 9 11, (1, 13)doso pr. 6 5. (49) Relted ngles omplete lesson shell & WS 5. pr. 30 5.3 (50) 5.3 5.4
More informationUnit #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
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 information5: The Definite Integral
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
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 informationHow 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=
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More information( ) as a fraction. Determine location of the highest
AB Clculus Exm Review Sheet  Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if
More informationQuadratic Forms. Quadratic Forms
Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More information( ) where f ( x ) is a. AB 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 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through
More informationList all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.
Mth Anlysis CP WS 4.X Section 4.4.4 Review Complete ech question without the use of grphing clcultor.. Compre the mening of the words: roots, zeros nd fctors.. Determine whether  is root of 0. Show
More informationChapters Five Notes SN AA U1C5
Chpters Five Notes SN AA U1C5 Nme Period Section 5: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationKEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a
KEY CONCEPTS THINGS TO REMEMBER :. The re ounded y the curve y = f(), the is nd the ordintes t = & = is given y, A = f () d = y d.. If the re is elow the is then A is negtive. The convention is to consider
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 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 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 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 information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationAB Calculus Review Sheet
AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More informationpractice How would you find: e x + e x e 2x e x 1 dx 1 e today: improper integrals
prctice How would you find: dx e x + e x e 2x e x 1 dx e 2x 1 e x dx 1. Let u=e^x. Then dx=du/u. Ans = rctn ( e^x ) + C 2. Let u=e^x. Becomes u du / (u1), divide to get u/(u1)=1+1/(u1) Ans = e^x + ln
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
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
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationDiscrete Mathematics and Probability Theory Spring 2013 Anant Sahai Lecture 17
EECS 70 Discrete Mthemtics nd Proility Theory Spring 2013 Annt Shi Lecture 17 I.I.D. Rndom Vriles Estimting the is of coin Question: We wnt to estimte the proportion p of Democrts in the US popultion,
More informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
More informationA sequence is a list of numbers in a specific order. A series is a sum of the terms of a sequence.
Core Module Revision Sheet The C exm is hour 30 minutes long nd is in two sections. Section A (36 mrks) 8 0 short questions worth no more thn 5 mrks ech. Section B (36 mrks) 3 questions worth mrks ech.
More informationSection 4: Integration ECO4112F 2011
Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic
More informationf(a+h) f(a) x a h 0. This is the rate at which
M408S Concept Inventory smple nswers These questions re openended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnkoutnnswer problems! (There re plenty of those in the
More informationAPPROXIMATE INTEGRATION
APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose ntiderivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be
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 informationPART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point.
PART MULTIPLE CHOICE Circle the pproprite response to ech of the questions below. Ech question hs vlue of point.. If in sequence the second level difference is constnt, thn the sequence is:. rithmetic
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 informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More 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 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 informationHomework Assignment 3 Solution Set
Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution.
More information9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes
The Vector Product 9.4 Introduction In this section we descrie how to find the vector product of two vectors. Like the sclr product its definition my seem strnge when first met ut the definition is chosen
More informationx = a To determine the volume of the solid, we use a definite integral to sum the volumes of the slices as we let!x " 0 :
Clculus II MAT 146 Integrtion Applictions: Volumes of 3D Solids Our gol is to determine volumes of vrious shpes. Some of the shpes re the result of rotting curve out n xis nd other shpes re simply given
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationTrigonometric Functions
Exercise. Degrees nd Rdins Chpter Trigonometric Functions EXERCISE. Degrees nd Rdins 4. Since 45 corresponds to rdin mesure of π/4 rd, we hve: 90 = 45 corresponds to π/4 or π/ rd. 5 = 7 45 corresponds
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 informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More information1B40 Practical Skills
B40 Prcticl Skills Comining uncertinties from severl quntities error propgtion We usully encounter situtions where the result of n experiment is given in terms of two (or more) quntities. We then need
More informationObjectives. Materials
Techer Notes Activity 17 Fundmentl Theorem of Clculus Objectives Explore the connections between n ccumultion function, one defined by definite integrl, nd the integrnd Discover tht the derivtive of the
More information