# 10. AREAS BETWEEN CURVES

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1 . 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 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 x-xis. 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 x-xis so you cn still hve the x-xis 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 x-xis nd g(x) dx gives the re under the ottom curve, right down to the x-xis. 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 x-xis from x = to x = nd the ottom oundry of the re is the x-xis, 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 x-xis this ecomes the ottom oundry nd the x-xis, 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 x-xis 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 x-xis 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 x-xis nd lines x = nd x =. Solution: Becuse the curve cuts the x-xis t x = we must divide the intervl [, ] into two pieces. From x = to x = the top y is y = (the x-xis) 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 x-xis 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 y-xis. 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 x-xis. 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 x-xis 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 x-xis. 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

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### Interpreting 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 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

### ( ) 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

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### MORE 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

### ( ) 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

### APPLICATIONS 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 cross-section in plne pssing through

### List 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

### Chapters 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

### Main topics for the First Midterm Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

### KEY 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

### Topics 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.

### 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

### Section 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

### MATH 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

### 2 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

### Riemann 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 >

### AB 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

### AP 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

### practice 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 / (u-1), divide to get u/(u-1)=1+1/(u-1) Ans = e^x + ln

### I1 = 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

### approaches 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.

### Discrete 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,

### The 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

### A 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.

### Section 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

### f(a+h) f(a) x a h 0. This is the rate at which M408S Concept Inventory smple nswers These questions re open-ended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnk-out-n-nswer problems! (There re plenty of those in the

### 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

### Lecture 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

### PART 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

### 5.5 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 nti-derivtive is not esily recognizble, then we re in

### Recitation 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

### 1 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

### INTRODUCTION 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

### Homework 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.

### 9.4. 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 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 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.