7.1 Integral as Net Change Calculus. What is the total distance traveled? What is the total displacement?


 Lindsay Ryan
 10 months ago
 Views:
Transcription
1 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 in position is displcement. To see the difference etween distnce nd displcement, consider the following sying: "Two steps forwrd nd one step ck" Wht is the totl distnce trveled? Wht is the totl displcement? To find displcement, we only need to find vt () dt. In order to find your new loction, we sy tht your new position = initil position + displcement. To find totl distnce we use vt () dtor find when the oject is moving in the negtive direction, rek the integrl into pieces nd sutrct the vlue of the integrl for the re under the curve. Exmple: Suppose the velocity of prticle moving long the x xis is given y t. vt () = 6t 18t+ 1 when ) When is the prticle moving to the right? When is it moving left? When is it stopped? ) Find the prticle's displcement for the time intervl. c) Find the prticle's totl distnce trveled y setting up ONE integrl nd using your clcultor. d) Find the prticle's totl distnce trveled without using solute vlue. 111
2 7.1 Integrl s Net Chnge Clculus Exmple: ) Integrting velocity gives ) Integrting the solute vlue of velocity gives. Consumption over Time Velocity is not the only rte in which you cn integrte to get totl. In fct if you were given function tht gve the numer of tickets per hour tht the police wrote ech dy, nd you wnted to find the totl numer of tickets in 4hour period, you could integrte. Exmple: [5 AP Clculus AB # Clcultor Allowed] The tide removes snd from Sndy Point Bech t rte modeled y the function R given y 4πt Rt () = + 5sin 5. A pumping sttion dds snd to the ech t rte modeled y the function S, given y 15t S() t = t Both R (t) nd S (t) hve units of cuic yrds per hour nd t is mesured in hours for < t < 6. At time t =, the ech contins 5 cuic yrds of snd. ) How much snd will the tide remove from the ech during this 6hour period? Indicte units of mesure. ) Write n expression for Y (t), the totl numer of cuic yrds of snd on the ech t time t. c) Find the rte t which the totl mount of snd on the ech is chnging t time t = 4. d) For < t < 6, t wht time t is the mount of snd on the ech minimum? Wht is the minimum vlue? Justify your nswers. 11
3 7.1 Integrl s Net Chnge Clculus Exmple: [4 AP Clculus AB (Form B) # Clcultor Allowed] For < t < 31, the rte of chnge of the t numer of mosquitoes on Tropicl Islnd t time t dys is modeled y Rt () = 5 tcos 5 mosquitoes per dy. There re 1 mosquitoes on Tropicl Islnd t time t =. ) Show tht the numer of mosquitoes is incresing t time t = 6. ) At time t = 6, is the numer of mosquitoes incresing t n incresing rte, or is the numer of mosquitoes incresing t decresing rte? Give reson for your nswer. c) According to the model, how mny mosquitoes will e on the islnd t time t = 31? Round your nswer to the nerest whole numer. d) To the nerest whole numer, wht is the mximum numer of mosquitoes for < t < 31? Show the nlysis tht leds to your conclusion. 113
4 7.1 Integrl s Net Chnge Clculus Exmple: [3 AP Clculus AB # Clcultor Allowed] A prticle moves long the x xis so tht its velocity t time t is given y t vt () = ( t+ 1sin ). At time t =, the prticle is t position x = 1. ) Find the ccelertion of the prticle t time t =. Is the speed of the prticle incresing t t =? Why or why not? ) Find ll times t in the open intervl < t < 3 when the prticle chnges direction. Justify your nswer. c) Find the totl distnce trveled y the prticle from time t = until time t = 3. d) During the time intervl < t < 3, wht is the gretest distnce etween the prticle nd the origin? Show the work tht leds to your nswer. 114
5 7.1 Integrl s Net Chnge Clculus Exmple: [ AP Clculus AB # Clcultor Allowed] The rte t which people enter n musement prk on given dy is modeled y the function E defined y 156 E() t = t 4t The rte t which people leve the sme musement prk on the sme dy is modeled y function L defined y 989 Lt () = t 38t Both E (t) nd L (t) re mesured in people per hour nd time t is mesured in hours fter midnight. These functions re vlid for 9 < t < 3, the hours during which the prk is open. At time t = 9, there re no people in the prk. ) How mny people hve entered the prk y 5: pm (t = 17)? Round your nswer to the nerest whole numer. ) The price of dmission to the prk is $15 until 5: pm (t = 17). After 5: pm, the price of dmission to the prk is $11. How mny dollrs re collected from dmissions to the prk on the given dy? Round your nswer to the nerest whole numer. t c) Let H () t = ( E( x) L( x) ) dx for 9 < t < 3. The vlue of H ( 17 ) to the nerest whole numer is 375. Find 9 the vlue of H '17 ( ) nd explin the mening of H ( 17) nd H '17 ( ) in the context of the prk. d) At wht time t, for 9 < t < 3, does the model predict tht the numer of people in the prk is mximum? 115
6 7. Ares in the Plne Clculus 7. AREAS IN THE PLANE Let s Review the concept of re s it reltes to clculus! Recll the re under curve cn e pproximted through the use of Riemnn sums: We cn rek the re into rectngles. Consider the one rectngle drwn. It s height is given y the function vlue of the curve t the right endpoint nd the width is given s Δ x. The re under the curve then is pproximtely the sum of the res of ALL the rectngles just like this one. f ( c) Are n k= 1 f ( c ) Δx k k Δx As the numer of rectngles, n, increses, the pproximted re gets closer to the ctul re, so we sy Are under the curve = lim f ( ck) Δ xk = f ( x) d n = k n x We cn pply this sme concept to the re etween curves. Consider the two functions f nd g elow. First: Drw rectngulr strip. Wht is the height nd width of your rectngle? Would the height nd width of the rectngle strip e different if you drew it in different plce? 1 g f Second: The re etween the curves is pproximtely the sum of ll of these rectngles. We cn write this s Third: How cn we get closer to the ACTUAL re etween the curves? Fourth: If we let the numer of rectngles pproch infinity, then we hve 116
7 7. Ares in the Plne Clculus Are of Region Between Two Curves If f nd g re continuous on [, ] nd g( x) f ( x) grphs of f nd g nd the verticl lines x = nd x = is for ll x in [, ], then the re of the region ounded y the ( ) ( ) A = f x g x dx Exmple: Find the re of the region ounded y the grphs of y= x +, y = x, x =, nd x = 1. Step 1: Drw picture nd shde the desired region. Step : Drw n ritrry rectngulr strip. Step 3: Using the re of the rectngulr strip s guide, set up nd solve n integrl to find the re etween the curves. Exmple: Find the re of the region ounded y the grphs of f ( x) = x nd g( x) = x. 117
8 7. Ares in the Plne Clculus 3 Exmple: Find the re of the region etween the grphs of f ( x) = 3x x 1x nd g ( x) = x + x. Exmple: Find the re of the region ounded y the grphs of x = 3 y nd x= y Exmple: Find the re of the region ounded y the grphs of g( x) = x, y = 4, nd x =. Exmple: The re of the region ounded y the grphs of 1 1 y 3 ( x ) x dx. 3 = x nd y = x cnnot e found y the single integrl Explin why this is so. Use symmetry to write single integrl tht does represent the re. 118
9 7.3 Volumes Clculus 7.3 VOLUMES Just like in the lst section where we found the re of one ritrry rectngulr strip nd used n integrl to dd up the res of n infinite numer of infinitely thin rectngles, we re going to pply the sme concept to finding volume. The key Find the volume of ONE ritrry "slice", nd use n integrl to dd up the volumes of n infinite numer of infinitely thing "slices". We will first pply this concept to the volume of solid with known cross section, then we will find the volumes of solids formed y revolving region out horizontl or verticl line. We will discuss three different methods of finding volumes of solids of revolution, ut first Dy 1: Volumes of Solids with Known Cross Sections First Question Wht is cross section? Imgine lof of red. Now imgine the shpe of slice through the lof of red. This shpe would e cross section. Techniclly cross section of three dimensionl figure is the intersection of plne nd tht figure. It would e like cutting n oject nd then looking t the fce of where you just cut. The cross sections we will e deling re lmost entirely perpendiculr to the x xis. Here's the sic ide You will e given region defined y numer of functions. We will grph tht region on n x nd y xis. Then we will ly they region flt nd uild upon tht region solid which hs the sme cross section no mtter where you slice it. To see some nimted views of this go to (We will wtch few of them in clss) Second question How do we find the volume of this solid tht hs een creted to hve similrly shped cross section, even though ech cross section my hve different size? We get to use clculus, of course! But first, we need to know how to find the Volume of prism. Even though every shpe my e different, we cn find the volume of prism y finding the re of the se times the "height". The "height" of our prisms will e the thickness of the slices. Once you know the volume of one slice, you just use n integrl to dd the volumes of ll the slices to get the volume of the solid. Exmple: Find the volume of the following squre "slice". Since most of the "slices" we will e deling with will hve thickness of dx, we will use tht sme thickness here. dx x Exmple: Find the volume of the following semicirculr "slice". x dx 119
10 7.3 Volumes Clculus You will lso need to e le to find the volume of equilterl tringle cross sections, isosceles right tringle cross sections, nd others. Rememer, wht you relly need is formul for the re of the se, which is just the cross sectionl shpe. Exmple: The se of solid is the region enclosed y the grph of y = e x, the coordinte xes, nd the line x = 3. If ll plne cross sections perpendiculr to the x xis re squres, then its volume is A) 1 e 6 B) 1 6 C) e 6 e D) 3 e E) 1 e 3 Exmple: The se of solid is the region in the first qudrnt enclosed y the prol y = 4x, the line x = 1, nd the x xis. Ech plne section of the solid perpendiculr to the x xis is squre. The volume of the solid is A) 4π 3 B) 16 π 5 C) 4 3 D) 16 5 E)
11 7.3 Volumes Clculus Exmple: The se of solid S is the region enclosed y the grph of y= ln x, the line x = e, nd the xxis. If the cross sections of S perpendiculr to the x xis re squres, then the volume of S is A) e B) 3 3 C) 1 D) E) ( 1) Exmple: The se of solid is region in the first qudrnt ounded y the xxis, the yxis, nd the line x+ y=8. If the cross sections of the solid perpendiculr to the x xis re semicircles, wht is the volume of the solid? A) B) C) D) 67.1 E) Exmple: The se of solid is the region in the first qudrnt enclosed y the grph of y = x nd the coordinte xes. If every cross section of the solid perpendiculr to the yxis is squre, the volume of the solid is given y A) π ( y) dy B) ( y) dy D) ( x ) dx E) ( ) C) π ( ) x dx x dx 11
12 7.3 Volumes Clculus Exmple: Let R e the region enclosed y the grphs of y= ln( x + 1) nd y = cos x. ) Find the re of R. (Clcultor ok) ) The se of solid is the region R. Ech cross section of the solid perpendiculr to the xxis is n equilterl tringle. Write n expression involving one or more integrls tht gives the volume of the solid. Do not evlute. 1
13 7.3 Volumes Clculus Dy : Volumes of Solids of Revolution: The Disc Method In finding the re of region, we drew n ritrry representtive rectngle. Keeping with the sme ide, if we revolve rectngle round line, it forms cylinder, s shown elow. The key to using the disc method will e twofold: 1) The rectngulr strip must e connected to the xis of revolution (no mtter where you drw it), nd ) the rectngulr strip must e perpendiculr to the xis of revolution. Exmple: Wht is the volume of the cylinder shown if the height of the rectngle is considered R nd the width of the rectngle is dx? Just like we did in finding the re, s we increse the numer of rectngles to infinity, the width of ech rectngle ecomes infinitely smll nd we denote this dx (if it is verticl strip) or dy (if it is horizontl strip). We then use n integrl to sum the volume of every one of these infinitely thin cylinders. This concept leds to the following: The Disc Method To find the volume of solid of revolution with the disc method, use one of the following; HORIZONTAL AXIS OF REVOLUTION VERTICAL AXIS OF REVOLUTION V = π R( x) dx V = π R( y) dy where R (x) nd R (y) re the "heights" of your representtive rectngulr strips. Exmple: Drw n pproprite rectngulr strip nd find the volume of the solid formed y revolving the region out the x xis. y = 4 x 13
14 7.3 Volumes Clculus Exmple: Find the volume of the solid formed y revolving the region out the y xis. (Drw representtive rectngulr strip) y = 16 x Exmple: Find the volume of the solid generted y revolving the region ounded y the grphs of the equtions xy = 6, y =, y = 6, nd x = 6 out the indicted lines. Sketch the region formed, nd drw representtive rectngulr strip for ech solid. ) the line x = 6. ) the line y = 6. 14
15 7.3 Volumes Clculus Dy 3: Volumes of Solids of Revolution: The Wsher Method For the disc method, the re we revolved hd to e connected to the re of rottion nd the representtive rectngle hd to e perpendiculr to the xis of revolution. For the wsher method, the representtive rectngle will still e perpendiculr to the xis of revolution, ut no longer ttched to the xis of revolution. Exmple: Sketch the figure formed y rotting the rectngle round the given line. Do you see why it's clled the wsher method? 5 3 Exmple: Wht is the volume of the figure formed ove? We will cll the outer rdius R, nd we will cll the inner rdius r. The height of the cylinder formed is just the width of the strip. Just like efore, if we hve nd infinitely thin strip, this distnce will e denoted dx (if it is verticl strip) nd dy (if it is horizontl strip). The volume of the solid formed y revolving region round the xis using the wsher method is given y π R r dx Exmple: Set up nd integrl, ut do not solve to find the volume of the solid generted y revolving the region ounded y the grphs of the equtions out the indicted lines. y= x ; y= ; x= ) the y xis ) the x xis c) the line y = 8 d) the line x = 15
16 7.3 Volumes Clculus Dy 4: Volumes of Solids of Revolution: The Shell Method We hve now used two different methods to find the volume of solid formed y revolving region out line. As with the disc nd wsher methods we will egin our discussion of the shell method y considering rectngle hving width w nd length h. The mjor difference etween the shell method nd the previous methods is tht the rectngle will e prllel to the xis of revolution. w h p : Let p e the distnce etween the xis of revolution nd the CENTER of the rectngulr strip. To find the volume of this figure, we cn proceed exctly like we did with the wsher method using the vlues of w, h, nd p. Exmple: If p is defined s ove to e the distnce etween the xis of revolution nd the center of the rectngulr strip, then wht is the rdius of the outer cylinder? Exmple: Wht is the volume of the outer cylinder? Exmple: Wht is the rdius of the inner cylinder? Exmple: Wht is the volume of the shell? If we were to rotte the figure elow round the line given, we could estimte the volume of the solid formed y finding the volume of the solid formed y rotting ech rectngulr strip nd dding these volumes together. If we were to consider infinitely mny strips, then ech strip would e so incredily thin tht the verge rdius, p, would e the distnce etween the xis of revolution nd the strip. The height of the shell formed y ech strip is just the length of the strip, h, nd the thickness of ech strip is given y dy (if the strip is horizontl) or dx (if the strip is verticl). dy h p : if the strip is dy, then oth p nd h must e written s functions of y 16
17 7.3 Volumes Clculus To find the volume of solid of revolution with the shell method, use one of the following: Horizontl Axis of Revolution Verticl Axis of Revolution V = π p( y) h( y) dy V = π p( x) h( x) dx dy h (y) p (y) dx h (x) Exmple: Let R e the region ounded y the grphs of y= x + 4, y = 8, nd x =, set up nd evlute the integrl tht gives the volume of the solid generted y revolving R out the y xis. ) Use the disc method p (x) ) Use the shell method 17
18 7.3 Volumes Clculus 1 Exmple: Let R e the region ounded y the grphs of y =, y =, x = 1, nd x = 4. In the lst exmple, we were x le to use oth the disc nd shell methods to rrive t the sme nswer. Explin why the volume formed y revolving R round the y xis is est found using the shell method insted of the disc nd wsher methods. Set up nd evlute the integrl tht gives the volume formed y revolving R round the y xis. 3 Exmple: Find the volume of the solid formed y revolving the region ounded y the grphs of y= x + x+ 1, y = 1, nd x = 1 out the line x =. Explin why it is necessry to use the shell method in this prolem. Exmple. Find the volume of the solid formed y revolving the region ounded y the grphs of y = x nd y= 4x x out the yxis. 18
19 7.3 Volumes Clculus π Exmple: [1973 AP Clculus AB #35] The region in the first qudrnt ounded y the grph of y = sec x, x =, 4 nd the xes is rotted out the xxis. Wht is the volume of the solid generted? A) π B) 1 4 π C) π D) π E) 8 π 3 Exmple: [1985 AP Clculus AB #45] The region enclosed y the grph of revolved out the yxis. The volume of the solid generted is y = x, the line x =, nd the xxis is A) 8 π B) 3 π C) 16 π D) 4π E) 16 π Exmple: [1985 AP Clculus BC #35] The region in the first qudrnt etween the xxis nd the grph of y = 6x x is rotted round the yxis. The volume of the resulting solid of revolution is given y 6 A) π( 6x x ) dx B) π ( 6x x ) dx C) π ( 6 ) 6 D) π( 3 9 y) dy E) π( ) y dy 6 x x x dx x Exmple: [1988 AP Clculus AB #3] A region in the first qudrnt is enclosed y the grphs of y = e, x = 1, nd the coordinte xes. If the region is rotted out the yxis, the volume of the solid tht is generted is represented y which of the following integrls? A) 1 x π xe dx B) 1 x π e dx C) 1 π e 4x dx D) e π yln y dy E) e π 4 ln ydy Exmple: [1988 AP Clculus AB #43] The volume of the solid otined y revolving the region enclosed y the ellipse x + 9y =9 out the xxis is A) π B) 4π C) 6π D) 9π E) 1 π 19
20 7.3 Volumes Clculus Exmple: [1988 AP Clculus BC #36] Let R e the region etween the grphs of y = 1 nd y = sin x from x = to π x =. The volume of the solid otined y revolving R out the xxis is given y A) π π xsinxdx B) π π C) π ( ) x cosxdx π 1 sinx dx D) π π π sin x dx E) π ( ) 1 sin x dx Exmple: [1993 AP Clculus AB #] Let R e the region in the first qudrnt enclosed y the grph of ( ) 1 3 y= x+ 1, the line x = 7, the xxis, nd the yxis. The volume of the solid generted when R is revolved out the yxis is given y 7 A) ( ) 3 π x + 1 dx B) ( ) 1 3 π x x+ 1 d C) ( ) 3 x π x dx 7 D) ( ) π x x+ 1 dx E) π ( y 1) dy Exmple: [1993 AP Clculus BC #19] The shded region R, shown in the figure elow, is rotted out the yxis to form solid who volume is 1 cuic inches. y y = kx x k x Of the following, which est pproximtes k? A) 1.51 B).9 C).49 D) 4.18 E)
7.1 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 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 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 informationCh AP Problems
Ch. 7.7. AP Prolems. Willy nd his friends decided to rce ech other one fternoon. Willy volunteered to rce first. His position is descried y the function f(t). Joe, his friend from school, rced ginst him,
More information[ ( ) ( )] Section 6.1 Area of Regions between two Curves. Goals: 1. To find the area between two curves
Gols: 1. To find the re etween two curves Section 6.1 Are of Regions etween two Curves I. Are of Region Between Two Curves A. Grphicl Represention = _ B. Integrl Represention [ ( ) ( )] f x g x dx = C.
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 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 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 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 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 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 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 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 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 information5.2 Volumes: Disks and Washers
4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict
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 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 informationUnit Six AP Calculus Unit 6 Review Definite Integrals. Name Period Date NONCALCULATOR SECTION
Unit Six AP Clculus Unit 6 Review Definite Integrls Nme Period Dte NONCALCULATOR SECTION Voculry: Directions Define ech word nd give n exmple. 1. Definite Integrl. Men Vlue Theorem (for definite integrls)
More informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two
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 information( ) Same as above but m = f x = f x  symmetric to yaxis. 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
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 information10. AREAS BETWEEN CURVES
. 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
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 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 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 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 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 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 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 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 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 informationAP CALCULUS Test #6: Unit #6 Basic Integration and Applications
AP CALCULUS Test #6: Unit #6 Bsi Integrtion nd Applitions A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS IN THIS PART OF THE EXAMINATION. () The ext numeril vlue of the orret
More informationPractice Final. Name: Problem 1. Show all of your work, label your answers clearly, and do not use a calculator.
Nme: MATH 2250 Clculus Eric Perkerson Dte: December 11, 2015 Prctice Finl Show ll of your work, lbel your nswers clerly, nd do not use clcultor. Problem 1 Compute the following limits, showing pproprite
More informationFINALTERM EXAMINATION 9 (Session  ) Clculus & Anlyticl GeometryI Question No: ( Mrs: )  Plese choose one f ( x) x According to PowerRule of differentition, if d [ x n ] n x n n x n n x + ( n ) x n+
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 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 informationAPPLICATIONS OF DEFINITE INTEGRALS
Chpter 6 APPICATIONS OF DEFINITE INTEGRAS OVERVIEW In Chpter 5 we discovered the connection etween Riemnn sums ssocited with prtition P of the finite closed intervl [, ] nd the process of integrtion. We
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 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 informationAP Calculus BC Review Applications of Integration (Chapter 6) noting that one common instance of a force is weight
AP Clculus BC Review Applictions of Integrtion (Chpter Things to Know n Be Able to Do Fin the re between two curves by integrting with respect to x or y Fin volumes by pproximtions with cross sections:
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 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 information( ) 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
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 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 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 informationShape and measurement
C H A P T E R 5 Shpe nd mesurement Wht is Pythgors theorem? How do we use Pythgors theorem? How do we find the perimeter of shpe? How do we find the re of shpe? How do we find the volume of shpe? How do
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 informationl 2 p2 n 4n 2, the total surface area of the
Week 6 Lectures Sections 7.5, 7.6 Section 7.5: Surfce re of Revolution Surfce re of Cone: Let C be circle of rdius r. Let P n be n nsided regulr polygon of perimeter p n with vertices on C. Form cone
More informationSpace Curves. Recall the parametric equations of a curve in xyplane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xyplne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)
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 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 informationDistance And Velocity
Unit #8  The Integrl Some problems nd solutions selected or dpted from HughesHllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl
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 base of each cylinder is called a crosssection.
6. Volume y Slicing Gol: To find the volume of olid uing econd emeter clculu Volume y CroSection Volume y Dik Volume y Wher Volume y Slicing Volume y Shell 6. Volume y Slicing 6. Volume y Slicing Gol:
More informationROB EBY Blinn College Mathematics Department
ROB EBY Blinn College Mthemtics Deprtment Mthemtics Deprtment 5.1, 5.2 Are, Definite Integrls MATH 2413 Rob EbyFll 26 Weknowthtwhengiventhedistncefunction, wecnfindthevelocitytnypointbyfindingthederivtiveorinstntneous
More informationFirst Semester Review Calculus BC
First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewiseliner function f, for 4, is shown below.
More informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in
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 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 informationTest , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes
Test 2 8., 8.2, 8.4 (density only), 8.5 (work only), 9., 9.2 nd 9.3 relted test mteril nd mteril from prior clsses Locl to Globl Perspectives Anlyze smll pieces to understnd the big picture. Exmples: numericl
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 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 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 information7.6 The Use of Definite Integrals in Physics and Engineering
Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 7.6 The Use of Definite Integrls in Physics nd Engineering It hs been shown how clculus cn be pplied to find solutions to geometric problems
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 information5 Applications of Definite Integrals
5 Applictions of Definite Integrls The previous chpter introduced the concepts of definite integrl s n re nd s limit of Riemnn sums, demonstrted some of the properties of integrls, introduced some methods
More informationCalculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION
lculus Section I Prt LULTOR MY NOT US ON THIS PRT OF TH XMINTION In this test: Unless otherwise specified, the domin of function f is ssumed to e the set of ll rel numers for which f () is rel numer..
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
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 informationNot for reproduction
AREA OF A SURFACE OF REVOLUTION cut h FIGURE FIGURE πr r r l h FIGURE A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundry of solid of revolution of the type
More informationSAINT IGNATIUS COLLEGE
SAINT IGNATIUS COLLEGE Directions to Students Tril Higher School Certificte 0 MATHEMATICS Reding Time : 5 minutes Totl Mrks 00 Working Time : hours Write using blue or blck pen. (sketches in pencil). This
More informationx 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx
. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
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 informationFORM FIVE ADDITIONAL MATHEMATIC NOTE. ar 3 = (1) ar 5 = = (2) (2) (1) a = T 8 = 81
FORM FIVE ADDITIONAL MATHEMATIC NOTE CHAPTER : PROGRESSION Arithmetic Progression T n = + (n ) d S n = n [ + (n )d] = n [ + Tn ] S = T = T = S S Emple : The th term of n A.P. is 86 nd the sum of the first
More informationStudent Session Topic: Particle Motion
Student Session Topic: Prticle Motion Prticle motion nd similr problems re on the AP Clculus exms lmost every yer. The prticle my be prticle, person, cr, etc. The position, velocity or ccelertion my be
More informationR(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of
Higher Mthemtics Ojective Test Prctice ook The digrm shows sketch of prt of the grph of f ( ). The digrm shows sketch of the cuic f ( ). R(, 8) f ( ) f ( ) P(, ) Q(, ) S(, ) Wht re the domin nd rnge of
More informationMath 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 xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More information5.1 Estimating with Finite Sums Calculus
5.1 ESTIMATING WITH FINITE SUMS Emple: Suppose from the nd to 4 th hour of our rod trip, ou trvel with the cruise control set to ectl 70 miles per hour for tht two hour stretch. How fr hve ou trveled during
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 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 informationExploring parametric representation with the TI84 Plus CE graphing calculator
Exploring prmetric representtion with the TI84 Plus CE grphing clcultor Richrd Prr Executive Director Rice University School Mthemtics Project rprr@rice.edu Alice Fisher Director of Director of Technology
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 informationPrep Session Topic: Particle Motion
Student Notes Prep Session Topic: Prticle Motion Number Line for AB Prticle motion nd similr problems re on the AP Clculus exms lmost every yer. The prticle my be prticle, person, cr, etc. The position,
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 information2. VECTORS AND MATRICES IN 3 DIMENSIONS
2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2dimensionl Vectors x A point in 3dimensionl spce cn e represented y column vector of the form y z zxis yxis z x y xxis Most of 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 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 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 informationReview Exercises for Chapter 4
_R.qd // : PM Pge CHAPTER Integrtion Review Eercises for Chpter In Eercises nd, use the grph of to sketch grph of f. To print n enlrged cop of the grph, go to the wesite www.mthgrphs.com... In Eercises
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 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 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 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 information2.4 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 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 informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the righthnd side limit equls to the lefthnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More information