# Math 0230 Calculus 2 Lectures

Size: px
Start display at page:

Transcription

1 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 pproches:. We consider region S tht lies between two curves y = f(x) nd y = g(x) nd between the verticl lines x = nd x = b, where f nd g re continuous nd f g on [, b]. Then we divide the region into smll verticl rectngles with bse x nd heights f(x i ) g(x i ). The corresponding Riemnn sum is [f(x i ) g(x i )] x nd the re A of S is A = lim n [f(x i ) g(x i )] x = b [f(x) g(x)] dx To solve problems on finding res between two curves lwys drw picture. Exmple. Find the re tht lies between two curves y = x + nd y = x + nd between the verticl lines x = nd x =. Solution: First, drw the picture. f(x) = x +, g(x) = x + re continuous nd f g on [, b]. Then nd (importnt!) f nd g A = [ ] [ ] x + dx = x + (x + )/ ln(x + ) = 9/ ln 4 ( ) 5/ ln = ln. We consider region S tht lies between two curves x = f(y) nd x = g(y) nd between the horizontl lines y = c nd y = d, where f nd g re continuous nd f(y) g(y) on [c, d]. Then

2 we divide the region into smll horizontl rectngles with height y nd bses f(y i ) g(y i ). The corresponding Riemnn sum is nd the re A of S is A = lim n [f(yi ) g(yi )] y [f(yi ) g(yi )] y = d c [f(y) g(y)] To solve problems on finding res between two curves lwys drw picture. Exmple. Find the re enclosed by the prbol 4x + y = nd the line y = x. Solution: The prbol hs the vertex t (, ). Points of intersection: 4x = y = 4y, y + 4y =, y = 6 nd y =. Points re ( 6, 6) nd (, ). Consider functions y x = f(y) = = y nd x = g(y) = y on [ 6, ], f(y) g(y). Then the re is 4 4 ] ] A = [ y 4 y = [y y y 6 6 = 6 8 ( ) 6 8 = = Explin why horizontl rectngles work better here compring to verticl rectngles (with x). Exmple. Sketch the region enclosed by the curves y x = nd x = e y nd lines y = nd y =. Solution: The prbol x = y hs the vertex t (, ). Two curves do not intersect inside y. Consider functions x = f(y) = e y nd x = g(y) = y on [, ], f(y) g(y). Then the re is [ A = e y y + ] ] = [e y y + y = e ( + e + ) = + e e

3 Section 7. Volumes Let S be d solid plced in the xy-plne. Let P x be plne perpendiculr to the x-xis nd P y be plne perpendiculr to the y-xis. First consider P x type of plnes. They slice S. The intersection of S nd P x is clled cross-section nd let A(x) be its re, x b. If slice is x = (b )/n thick then its volume is A(x i ) x, where x i lies in the intervl [x i, x i ]. Then the volume of S is pproximtely V A(x i ) x or exctly V = lim n A(x i ) x = Exmple. Find the volume of sphere of rdius r. b A(x) dx Solution: r x r. A cross-section t x is circle of rdius y = r x. Then A(x) = πy = π(r x ) nd V = r r π(r x ) dx = π r (r x ) dx = π ] r [r x x = π(r r /) = 4π r Now ssume we hve solid obtined by rotting bout the x-xis the region bounded by curves y = f(x) nd y =, when x b. Then ech cross-section is circle with rdius f(x) nd the volume of the solid is b V = π f (x) dx If solid is obtined by rotting bout the y-xis the region bounded by the curve x = f(y), when c y d, then ech cross-section is circle with rdius f(y) nd the volume of the solid is d V = π f (y) c Exmple. Find the volume of the solid obtined by rotting bout the y-xis the region bounded by the curves x = y nd x =, when y.

4 Solution: V = π (y ) = π y 4 = π y5 5 = π 5 Now ssume region is bounded by the curves y = f(x) nd y = g(x), x b, where f(x) g(x) [note, the second curve is not y = nymore]. In this cse cross-section is ctully wsher with the outer (lrger) rdius f(x) nd the inner (smller) rdius g(x). The re of the wsher is π[f (x) g (x)] nd the volume of the solid obtined by rottion is V = π b [f (x) g (x)] dx In the cse of rottion bout the y-xis the coressponding volume is V = π d c [f (y) g (y)] Exmple. Find the volume of the solid obtined by rotting bout the y-xis the region bounded by the curves x = y nd x = y. Solution: First, we hve to find c nd d s y coordintes of points of intersection of the curves: the eqution y = y gives two solutions c = nd d =. Also, y y when y. Hence [ ] 4y V = π [(y) (y ) ] = π [4y y 4 ] = π y5 = 64π 5 5 Solids tht we hve considered so fr re clled solids of revolution. Now consider n exmple of not solid of revolution Exmple 4. Find the volume of pyrmid whose bse is squre with side nd whose height is h. Solution: We plce the origin O t the vertex of the pyrmid nd the x-xis long its centrl xis. Every cross-section t x ( x h) is squre with the side b = x (it cn be found h from corresponding proportion). The cross-sectionl re is A(x) = b = h x. Then V = h A(x) dx = h h x dx = x h h = h h = h 4

5 Section 7. Volumes by Cylindricl Shells Some volumes cn be clculted by the method of cylindricl shells. Picture. Let there be cylindricl shell with inner rdius r nd outer rdius r. Then its volume V is the difference of the volume V of the outer cylinder nd the volume V of the inner cylinder: V = V V = πrh πrh = π(r r)h = π(r + r )(r r )h = π r + r (r r )h We put r = r + r (the verge rdius of the shell) nd r = r r, then V = πrh r Another wy: V = π(r + r) h πr h = πrh r where r is the inner rdius nd r + r is the outer rdius. Let s pply tht to clcultion of the volume V of solid S obtined by rotting bout y-xis the region bounded by y = f(x) (f(x) must be continuous nd nonnegtive f(x) ), y =, x =, nd x = b, where < < b. As usul, we divide the intervl [, b] into subintervls [x i, x i ] ech of the length x. Let x i be midpoint of ith subintervl. If the rectngle with bse [x i, x i ] nd height f( x i ) is rotted bout the y-xis, then the result is cylindricl shell with verge rdius x i, height f( x i ), nd thickness x. Its volume is nd the volume of the solid is pproximtely or exctly V V = lim n V i = π x i f( x i ) x V i = π x i f( x i ) x b π x i f( x i ) x = π xf(x) dx If solid S is obtined by rotting bout x-xis the region bounded by x = f(y) (f(y) is continuous nd nonnegtive f(y) ), x =, y = c, nd y = d, where < c < d, then its volume is d V = π yf(y) c If solid S is obtined by rotting bout the line x = l (which is prllel to the y-xis) the region bounded by y = f(x) (f(x) is continuous nd nonnegtive f(x) ), y =, x =, nd x = b, where < < b < l, then its volume is V = π b (l x)f(x) dx 5

6 Exmple. Find the volume of sphere of rdius R. Solution: The sphere is obtined by rotting bout the y-xis the region bounded by the curves y = R x nd y = R x when x R. Then r = x, h = R x ( R x ) = R x nd V = πx R x x, V = 4π R x R x dx Substitution: u = R x, du = xdx, u() = R, u(r) = V = π R u / du = π R [ ] R u / du = π u/ = 4 πr Exmple. Find the volume of the solid obtined by rotting bout the y-xis the region bounded by the curves x = y nd x = y. [It is the exmple in the section 7.] Solution: The region is bounded by the curves y = x nd y = x/ with x/ x. Before we found tht x 4. Here we pply the method of cylindricl shells: r = x, h = x x/, V = πrh r = πx( x x/) x = π(x / x /) x. Then 4 V = π ) [ ] 4 [ ] [ (x / x dx = π 5 x5/ x = π = 64π ] = 64π 5 Exmple. Find the volume of the solid obtined by rotting bout the y-xis the region bounded by the curves y = x nd y =, nd x = 4. Solution: Drw picture. =, b = 4, f(x) = x. Then 4 V = π x 4 x dx = π x / dx = π 5 x5/ 4 = 8π 5 Exmple 4. Find the volume of the solid obtined by rotting bout the x-xis the region bounded by the curves y = x, y =, nd x = 4. Solution: x = y, y. V = π y y = π 6 y = π y4 4 = 8π

7 Exmple 5. Find the volume of the solid obtined by rotting bout the line x = the region bounded by the curves y = 4x x, y = 8x x. Solution: Drw picture(!). Vertices of both prbols re t x = : (, 4) nd (, 8). Intesections of prbols: 4x x = 8x x, x 4x =, x =, x = 4. We divide [, 4] into subintervls ov width x. Let x be in one of the subinervls. Averge rdius of cylindricl shell is the distnce between x nd which is x ( ) = x +. The height is the difference between prbols: h(x) = 8x x (4x x ) = 4x x. Then V = π 4 (x + )(4x x ) dx = π 4 (8x + x x ) dx = π = π [ ] = 56 π [4x + ] 4 x x4 4 7

8 Section 7.4 Arc Length Picture. We wnt to find the length L of curve given by y = f(x), where f (x) is continuous nd x runs through the intervl [, b]. A curve is mde of smll rcs. Consider the ith rc. Its length by the Pythgoren theorem is Then or exctly L = lim n s i L = ( x i ) + ( y i ) = s i = ( ) yi + x i = x i ( ) yi + x i x i ( ) yi + x i x i b + becuse x i dx nd y i s n nd dx = f (x). ( ) b dx = + (f dx (x)) dx Exmple. Find the length L of the curve given by y = (x ) /, when x. ( ) Solution: dx = (x )/, + = + 9(x ) = 9x 8. Hence, dx L = 9x 8 dx [u = 9x 8, du = 9xdx, u() =, u() = ] = 9 u / du = 9 u/ = 7 ( ) Exmple. Find the length L of the curve given by y = x, when x. Solution: dx = x, + x ( ) = + x dx x =. Hence, x L = dx x Trig substitution: x = sin θ, θ π/, dx = cos θ dθ, x = cos θ = cos θ. Then L = π/ cos θ dθ cos θ = π/ dθ = π/ 8

9 Notice, tht the curve is qurter of the unit circle whose length is π. Exmple. Solution: Find the length L of the curve given by y = ln(cos x), when x π/. ( ) = tn x, + = + tn x = sec dx dx x = sec x. Hence, L = π/ sec x dx = ln sec x + tn x π/ = ln( + ) = ln( + ) 9

10 Section 7.6 Applictions to Physics nd Engineering Work W = F d F = m d s dt work = force x distnce (here the force is constnt) force = mss x ccelertion If the force is vrible F = f(x), then the work done in moving n object from to b is W = b f(x) dx Exmple. When prticle is locted x meters from the origin, force of x newtons cts on it. How much work is done in moving the prticle from x = to x =? Solution: W = x dx = x / dx = x x x = x = 4 Hook s Lw The force is proportionl to x: f(x) = kx, k > is the spring constnt. Exmple. A spring hs nturl length of 5 cm. If 6-N force is required to keep it stretched to length of 8 cm, how much work is required to stretch it from 5 cm to 6 cm? Solution: 8 5 = cm =. m. Then f(.) = 6, k. = 6, k =. Hence f(x) = x nd.. W = x dx = x = J Hydrosttic Pressure nd Force Suppose tht thin horizontl plte with re A sq. meters is submerged in fluid of density ρ kg per m t depth d meters below the surfce. The fluid directly bove the plte hs volume V = Ad nd the mss m = ρv = ρad. The force exerted by the fluid on the plte is F = mg = ρgad. The pressure on the plte is P = F/A = ρgd. Exmple. A plte in n irrigtion cnl is in the form of trpezoid 4 feet wide t the bottom, 8 feet wide t the top, with the height equl to feet. It is plced verticlly in the cnl nd is submerged in wter feet deep. Find the hydrosttic force in pounds on the plte. The weight density of wter is 6.5 = 5/ lb/ft. Solution : Let s plce the coordinte xes s it is shown on the picture. We use horizontl strips of equl width y nd consider strip on the depth d = y with 5 y ccording

11 to the picture. By similr tringles: = 5 d = 5 + y, = (5 + y). The length of the strip is l = 4+ = 4+ 4 (5+y) = 4 (8+y) nd its re is A = 4 (8+y) y. The pressure on the strip is P = 6.5d = 6.5y (note tht d = y since y is negtive) nd the force is F = P A = 5 y 4 (8 + y) y = 5 (8y + y ). Then the totl force is F = 5 5 (8y + y ) = 5 = 5 ] [4y + y = 5 [ 6 8 ] [ 5 ] = 5 5 = 75 lb Solution : Now we plce the coordinte xes differently. See the picture below. As before we use horizontl strips of equl width δy nd consider strip on the depth d = y with y ccording to the picture. By similr tringles: = 5 d = 5 + y, = ( + y). The length of the strip is l = 4+ = 4+ 4 (+y) = 4 (6+y) nd its re is A = 4 (6+y) y. The pressure on the strip is P = 6.5d = 6.5( y) nd the force is F = P A = 5 Then the totl force is 4 5 ( y) (6 + y) y = ( 4y y ). F = 5 ( 4y y ) = 5 ] [y y y = 5 [ ( )] = 5 45 = 5 5 = 75 lb

12 Section 7.7 Differentil Equtions Exmples. Order of DE = order of the highest derivtive. A solution. Seprble Equtions A seprble eqution is st order DE tht cn be written in the form dx = g(x) f(y) Assuming tht f(y) neq we denote h(y) =. Then the eqution becomes f(y) h(y) = g(x) dx nd we integrte both sides to get h(y) = g(x) dx nd solve the lst for y. Initil-Vlue Problem IVP = DE together with the initil condition y(x ) = y Exmple. Solve the IVP: y = x, y() =. y Solution: y = x dx, dx = x y. It is seprble eqution. y = x dx, y where C = (C C ). Then y = ± x + C. + C = x + C, y = x + C, Initil condition: y() = >. We use the solution with + sign. y() = + C = C =, C = 4. Answer: y = x + 4. Checking: y = x x + 4 = x x + 4 = x y, y() = 4 =. Exmple. Solve the IVP: y = e y (x 5), y() =. Solution: dx = x 5. It is seprble eqution. e y e y = (x 5) dx, e y = (x 5) dx, e y = x 5x + C, y = ln(x 5x + C), Initil condition: y() = ln( 6 + C) =, 6 + C =, C = 7.

13 Answer: y = ln(x 5x + 7). Logistic Growth Exmple. Solve the IVP: dt = ky (M y) (logistic DE), < y < M, y() = y. The logistic eqution models world popultion growth. Solution: y (M y) = M [ M y + It is seprble eqution. [ y + ]. So, M y ] = kt + C, M y [recll tht < y < M], Initil condition gives Notice tht lim t y(t) = M. Direction Field ln y M y y M y = A. Hence, y (M y) = k dt. [ ln y ln (M y) ] = kt + C M = M(kt + C), y M y = em(kt+c) = Ae kmt, A = e MC. y M y = y M y e kmt or y(t) = y M y + (M y )e kmt. y = F (x, y) Exmple 4. Sketch the direction field for the DE: y = xy. Solution: Notice tht on the lines x = nd y = we hve y =. Mixing Problems y = F (x, y) Exmple 4. Consider tnk with volume liters contining slt solution. Suppose solution with. kg/liter of slt flows into the tnk t rte of 5 liters/min. The solution in the tnk is well-mixed. Solution flows out of the tnk t rte of 5 liters/min. If initilly there is kg of slt in the tnk, how much slt will be in the tnk s function of time?

14 Solution: Let y(t) denote the mount of slt in kg in the tnk fter t minutes. We use fundmentl property of rtes: Totl Rte = Rte in - Rte out To find the rte in we use 5 liters kg kg (.) =.5 min liter min The rte t which slt leves the tnk is equl to the rte of flow of solution out of the tnk times the concentrtion of slt in the solution. Thus, the rte out is 5 liters min y The differentil eqution for the mount of slt is y =.5 y, y() =. kg liter = y kg min Using the method of seprtion of vrible we find dt = y, y = dt, y = dt, y = C e.5t, (C = e C ), y = Ce.5t. ln y = t + C The initil condition gives y() = C =, C =. Hence, the mount of slt in the tnk fter t minutes is given by the formul y(t) = + e.5t. Notice tht lim t y(t) = + = 4

### 7.6 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

### l 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 n-sided regulr polygon of perimeter p n with vertices on C. Form cone

### 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

### MAT187H1F Lec0101 Burbulla

Chpter 6 Lecture Notes Review nd Two New Sections Sprint 17 Net Distnce nd Totl Distnce Trvelled Suppose s is the position of prticle t time t for t [, b]. Then v dt = s (t) dt = s(b) s(). s(b) s() is

### We divide the interval [a, b] into subintervals of equal length x = b a n

Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:

### a < 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

### Math 113 Exam 1-Review

Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

### Test , 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

### Math 113 Exam 2 Practice

Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.-3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This

### Definite 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

### Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Mth 9 Course Summry/Study Guide Fll, 2005 [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)

### Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

### Math 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

### 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

### Sample 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.

### 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

### APPM 1360 Exam 2 Spring 2016

APPM 6 Em Spring 6. 8 pts, 7 pts ech For ech of the following prts, let f + nd g 4. For prts, b, nd c, set up, but do not evlute, the integrl needed to find the requested informtion. The volume of the

### A 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

### 5.2 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 cross-section or slice. In this section, we restrict

### Math 0230 Calculus 2 Lectures

Mth Clculus Lectures Chpter 9 Prmetric Equtions nd Polr Coordintes Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition Section 91 Prmetric Curves

### Disclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.

Disclimer: This is ment to help you strt studying. It is not necessrily complete list of everything you need to know. The MTH 33 finl exm minly consists of stndrd response questions where students must

### [ ( ) ( )] 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.

### Section 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

### Not 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

### 7.1 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

### The 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

### The area under the graph of f and above the x-axis 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 x-xis etween nd is denoted y f(x) dx nd clled the

### Math 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions

Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,

### Big 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:

### Math 116 Calculus II

Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................

### ( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.

AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find

### x 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

### Exam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B

PHY 249, Fll 216 Exm 1 Solutions nswer 1 is correct for ll problems. 1. Two uniformly chrged spheres, nd B, re plced t lrge distnce from ech other, with their centers on the x xis. The chrge on sphere

### JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 8 (First moments of a volume) A.J.Hobson

JUST THE MATHS UNIT NUMBER 3.8 INTEGRATIN APPLICATINS 8 (First moments of volume) b A.J.Hobson 3.8. Introduction 3.8. First moment of volume of revolution bout plne through the origin, perpendiculr to

### Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

### Math 116 Final Exam April 26, 2013

Mth 6 Finl Exm April 26, 23 Nme: EXAM SOLUTIONS Instructor: Section:. Do not open this exm until you re told to do so. 2. This exm hs 5 pges including this cover. There re problems. Note tht the problems

### MA 124 January 18, Derivatives are. Integrals are.

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

### How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?

Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

### 1 The Riemann Integral

The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

### Chapters 4 & 5 Integrals & Applications

Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions

### 4.4 Areas, Integrals and Antiderivatives

. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order

### Physics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016

Physics 333, Fll 16 Problem Set 7 due Oct 14, 16 Reding: Griffiths 4.1 through 4.4.1 1. Electric dipole An electric dipole with p = p ẑ is locted t the origin nd is sitting in n otherwise uniform electric

### 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

### Section 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

### SAINT 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

### Math& 152 Section Integration by Parts

Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

### AP 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:

### ( ) 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

### Indefinite Integral. Chapter Integration - reverse of differentiation

Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the

### Main topics for the Second Midterm

Min topics for the Second Midterm The Midterm will cover Sections 5.4-5.9, Sections 6.1-6.3, nd Sections 7.1-7.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the

### (6.5) Length and area in polar coordinates

86 Chpter 6 SLICING TECHNIQUES FURTHER APPLICATIONS Totl mss 6 x ρ(x)dx + x 6 x dx + 9 kg dx + 6 x dx oment bout origin 6 xρ(x)dx x x dx + x + x + ln x ( ) + ln 6 kg m x dx + 6 6 x x dx Centre of mss +

### ( ) 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

### Math 120 Answers for Homework 13

Mth 12 Answers for Homework 13 1. In this problem we will use the fct tht if m f(x M on n intervl [, b] (nd if f is integrble on [, b] then (* m(b f dx M(b. ( The function f(x = 1 + x 3 is n incresing

### 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

### ( ) 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

### The 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

### Distance And Velocity

Unit #8 - The Integrl Some problems nd solutions selected or dpted from Hughes-Hllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl

### Student Handbook for MATH 3300

Student Hndbook for MATH 3300 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 0.5 0 0.5 0.5 0 0.5 If people do not believe tht mthemtics is simple, it is only becuse they do not relize how complicted life is. John Louis

### Math 1132 Worksheet 6.4 Name: Discussion Section: 6.4 Work

Mth 1132 Worksheet 6.4 Nme: Discussion Section: 6.4 Work Force formul for springs. By Hooke s Lw, the force required to mintin spring stretched x units beyond its nturl length is f(x) = kx where k is positive

### x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick

### The 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

### Polynomials and Division Theory

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

### Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.

Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne 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)

### Goals: 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 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

### Math 100 Review Sheet

Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s

### 1. 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)

### Final 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, :3-7:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.

### The Wave Equation I. MA 436 Kurt Bryan

1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string

### MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK WRITTEN EXAMINATION SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/MC-UPDATES SECTION MULTIPLE CHOICE QUESTIONS QUESTION QUESTION

### APPLICATIONS 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

### JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 6 (First moments of an arc) A.J.Hobson

JUST THE MATHS UNIT NUMBER 13.6 INTEGRATION APPLICATIONS 6 (First moments of n rc) by A.J.Hobson 13.6.1 Introduction 13.6. First moment of n rc bout the y-xis 13.6.3 First moment of n rc bout the x-xis

### HIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 4 UNIT (ADDITIONAL) Time allowed Three hours (Plus 5 minutes reading time)

HIGHER SCHOOL CERTIFICATE EXAMINATION 999 MATHEMATICS UNIT (ADDITIONAL) Time llowed Three hours (Plus 5 minutes reding time) DIRECTIONS TO CANDIDATES Attempt ALL questions ALL questions re of equl vlue

### Test 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher).

Test 3 Review Jiwen He Test 3 Test 3: Dec. 4-6 in CASA Mteril - Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec. 14-17 in CASA You Might Be Interested

### Math 190 Chapter 5 Lecture Notes. Professor Miguel Ornelas

Mth 19 Chpter 5 Lecture Notes Professor Miguel Ornels 1 M. Ornels Mth 19 Lecture Notes Section 5.1 Section 5.1 Ares nd Distnce Definition The re A of the region S tht lies under the grph of the continuous

### Improper 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

### 0.1 Chapters 1: Limits and continuity

1 REVIEW SHEET FOR CALCULUS 140 Some of the topics hve smple problems from previous finls indicted next to the hedings. 0.1 Chpters 1: Limits nd continuity Theorem 0.1.1 Sndwich Theorem(F 96 # 20, F 97

### Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: olumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

### US01CMTH02 UNIT Curvature

Stu mteril of BSc(Semester - I) US1CMTH (Rdius of Curvture nd Rectifiction) Prepred by Nilesh Y Ptel Hed,Mthemtics Deprtment,VPnd RPTPScience College US1CMTH UNIT- 1 Curvture Let f : I R be sufficiently

### Review of Calculus, cont d

Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

### Practice 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

### Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors

Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector

### Mathematics Extension 2

S Y D N E Y B O Y S H I G H S C H O O L M O O R E P A R K, S U R R Y H I L L S 005 HIGHER SCHOOL CERTIFICATE TRIAL PAPER Mthemtics Extension Generl Instructions Totl Mrks 0 Reding Time 5 Minutes Attempt

### Math 42 Chapter 7 Practice Problems Set B

Mth 42 Chpter 7 Prctice Problems Set B 1. Which of the following functions is solution of the differentil eqution dy dx = 4xy? () y = e 4x (c) y = e 2x2 (e) y = e 2x (g) y = 4e2x2 (b) y = 4x (d) y = 4x

### Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...

Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting

### ragsdale (zdr82) HW2 ditmire (58335) 1

rgsdle (zdr82) HW2 ditmire (58335) This print-out should hve 22 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. 00 0.0 points A chrge of 8. µc

### Section 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

### A. Limits - L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.

A. Limits - L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c

### Review of basic calculus

Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

### 13.4 Work done by Constant Forces

13.4 Work done by Constnt Forces We will begin our discussion of the concept of work by nlyzing the motion of n object in one dimension cted on by constnt forces. Let s consider the following exmple: push

### Chapter 6 Notes, Larson/Hostetler 3e

Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

### spring from 1 cm to 2 cm is given by

Problem [8 pts] Tre or Flse. Give brief explntion or exmple to jstify yor nswer. ) [ pts] Given solid generted by revolving region bot the line x, if we re sing the shell method to compte its volme, then

### Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: Volumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

### Z b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...

Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.

### Math 107H Topics for the first exam. csc 2 x dx = cot x + C csc x cotx dx = csc x + C tan x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C

Integrtion Mth 07H Topics for the first exm Bsic list: x n dx = xn+ + C (provided n ) n + sin(kx) dx = cos(kx) + C k sec x dx = tnx + C sec x tnx dx = sec x + C /x dx = ln x + C cos(kx) dx = sin(kx) +

### A. Limits - L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. -1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.

A. Limits - L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by

### NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

### The Fundamental Theorem of Calculus

The Fundmentl Theorem of Clculus MATH 151 Clculus for Mngement J. Robert Buchnn Deprtment of Mthemtics Fll 2018 Objectives Define nd evlute definite integrls using the concept of re. Evlute definite integrls