' ' Š # # ' " # # # Section 5.3 The Definite Integral 275. œ x dx 1 dx x dx 1 dx. œ " # œ x dx 1 dx œ (3 1) œ ( 1 2) œ (see parts (a) and (b) above).

Similar documents
CHAPTER 5 INTEGRATION

Math 21B-B - Homework Set 2

2 f(x) dx = 1, 0. 2f(x 1) dx d) 1 4t t6 t. t 2 dt i)

MAT136H1F - Calculus I (B) Long Quiz 1. T0101 (M3) Time: 20 minutes. The quiz consists of four questions. Each question is worth 2 points. Good Luck!

MATH 10550, EXAM 3 SOLUTIONS

PRACTICE FINAL/STUDY GUIDE SOLUTIONS

Section Volumes by Slicing and Rotation About an Axis

Calculus I Practice Test Problems for Chapter 5 Page 1 of 9

AP Calculus BC Review Applications of Derivatives (Chapter 4) and f,

INTRODUCTORY MATHEMATICAL ANALYSIS

CHAPTER 4 Integration

MATH Exam 1 Solutions February 24, 2016

668 Chapter 11 Parametric Equatins and Polar Coordinates

Math 105: Review for Final Exam, Part II - SOLUTIONS

Areas and Distances. We can easily find areas of certain geometric figures using well-known formulas:

Math 113, Calculus II Winter 2007 Final Exam Solutions

y = f x x 1. If f x = e 2x tan -1 x, then f 1 = e 2 2 e 2 p C e 2 D e 2 p+1 4

Maximum and Minimum Values

CHAPTER 4 Integration

MATH 1A FINAL (7:00 PM VERSION) SOLUTION. (Last edited December 25, 2013 at 9:14pm.)

274 Chapter 4 Applications of Derivatives

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)

Math 1314 Lesson 16 Area and Riemann Sums and Lesson 17 Riemann Sums Using GeoGebra; Definite Integrals

TECHNIQUES OF INTEGRATION

AP CALCULUS AB 2003 SCORING GUIDELINES (Form B)

Math 176 Calculus Sec. 5.1: Areas and Distances (Using Finite Sums)

1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute

Name: Math 10550, Final Exam: December 15, 2007

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew

MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of

Section 13.3 Area and the Definite Integral

1 Approximating Integrals using Taylor Polynomials

For example suppose we divide the interval [0,2] into 5 equal subintervals of length

B U Department of Mathematics Math 101 Calculus I

CHAPTER 5 INTEGRATION

Calculus 2 Test File Spring Test #1

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is

Chapter 4. Fourier Series

Honors Calculus Homework 13 Solutions, due 12/8/5

THE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled


Riemann Sums y = f (x)

18.01 Calculus Jason Starr Fall 2005

Calculus 2 Test File Fall 2013

HOMEWORK #10 SOLUTIONS

CHAPTER 4 APPLICATIONS OF DERIVATIVES

Chapter 2 The Monte Carlo Method

Student s Printed Name:

For example suppose we divide the interval [0,2] into 5 equal subintervals of length

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

6.3 Testing Series With Positive Terms

4.1 SIGMA NOTATION AND RIEMANN SUMS

Math 142, Final Exam. 5/2/11.

2 ) 5. (a) (1)(3) + (1)(2) = 5 (b) {area of shaded region in Fig. 24b} < 5

2018 MAΘ National Convention Mu Individual Solutions ( ) ( ) + + +

CHAPTER 10 INFINITE SEQUENCES AND SERIES

n n 2 + 4i = lim 2 n lim 1 + 4x 2 dx = 1 2 tan ( 2i 2 x x dx = 1 2 tan 1 2 = 2 n, x i = a + i x = 2i

7.) Consider the region bounded by y = x 2, y = x - 1, x = -1 and x = 1. Find the volume of the solid produced by revolving the region around x = 3.

AP Calculus AB 2006 Scoring Guidelines Form B

Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations

Chapter 5: Take Home Test

4.1 Sigma Notation and Riemann Sums

The Definite Integral. Day 3 Riemann Sums

AP CALCULUS - AB LECTURE NOTES MS. RUSSELL

Math 21C Brian Osserman Practice Exam 2

6.) Find the y-coordinate of the centroid (use your calculator for any integrations) of the region bounded by y = cos x, y = 0, x = - /2 and x = /2.

Calculus 2 - D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.)

LESSON 2: SIMPLIFYING RADICALS

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t

AP Calculus BC 2011 Scoring Guidelines Form B

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim

MTH Assignment 1 : Real Numbers, Sequences

1988 AP Calculus BC: Section I

TEMASEK JUNIOR COLLEGE, SINGAPORE JC One Promotion Examination 2014 Higher 2

The Fundamental Theorem(s) of Calculus

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)

f t dt. Write the third-degree Taylor polynomial for G

INFINITE SEQUENCES AND SERIES

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1

Lesson 10: Limits and Continuity

Integrals, areas, Riemann sums

Lecture 6: Integration and the Mean Value Theorem. slope =

JEE ADVANCED 2013 PAPER 1 MATHEMATICS

The type of limit that is used to find TANGENTS and VELOCITIES gives rise to the central idea in DIFFERENTIAL CALCULUS, the DERIVATIVE.

THE INTEGRAL TEST AND ESTIMATES OF SUMS

Area As A Limit & Sigma Notation

Chapter 9: Numerical Differentiation

Archimedes - numbers for counting, otherwise lengths, areas, etc. Kepler - geometry for planetary motion

x x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0,

Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist.

INTEGRATION BY PARTS (TABLE METHOD)

We first write the integrand into partial fractions and then integrate. By EXAMPLE 27 we have the identity

VICTORIA JUNIOR COLLEGE Preliminary Examination. Paper 1 September 2015

SYDE 112, LECTURE 2: Riemann Sums

Math ~ Final Exam Review Guide* *This is only a guide, for your benefit, and it in no way replaces class notes, homework, or studying

AP Calculus BC 2007 Scoring Guidelines Form B

Chapter 5.4 Practice Problems

Estimation for Complete Data

Transcription:

Sectio. The Defiite Itegral 7 ( ) t dt t dt t dt t dt 6. av(f) Š at t dt Š ( ) ( ) Š Š. ( ) ( ) d ( ) d 6. (a) av(g) Š akk d d d d d ( ) Š ( ( )) Š ( ). () av(g) ˆ akk d ( ) d Š d d ( ). ( ) k k (c) av(g) Š akk d a d akk d ( ) (see parts (a) ad () aove). 6. (a) av(h) Š k k d ( ) d d. ( ) ( ) Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

76 Chapter Itegratio () av(h) ˆ k kd d Š. ( ) (c) av(h) Š k k d Œ k k d k k d ˆ ˆ aove). (see parts (a) ad () k Ö a aa aa kaa k a a c a a k c a a k k k a Ä_ m mä 6. Cosider the partitio P that sudivides the iterval Òa, Ó ito suitervals of width a ad let c e the right edpoit of each suiterval. So the partitio is P a, a, a,..., a ad c a. We get the Riema sum f c c c a. As ad P this epressio remais ca a. Thus, c d ca a. a k k Ö k a ˆ k ˆ k 8 k 8 a a k k k k k a a 6. Cosider the partitio P that sudivides the iterval Ò, Ó ito suitervals of width ad let c e the right edpoit of each suiterval. So the partitio is P,,,..., ad c k. We get the Riema sum f c k. As Ä_ ad mp mä the epressio has the value 6. Thus, d 6. 6. Cosider the partitio P that sudivides the iterval Òa, Ó ito suitervals of width a ad let c e the right k a aa aa kaa Ö k a k ˆ Š k a a a k a a ak a a kaa Š k k k k a a k k a a a a a a a Œ a a a a a a a a k k k edpoit of each suiterval. So the partitio is P a, a, a,..., a ad c a. We get the Riema sum f c c a a aa a a aa aa Ä_ m mä a a a a a a aaa aa a aaa aa a As ad P this epressio has value a a a a a a a a a a a a. Thus, d. a 66. Cosider the partitio P that sudivides the iterval Ò, Ó ito suitervals of width ad let c e the right edpoit of each suiterval. So the partitio is P Ö,,,..., ad Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley. k

Sectio. The Defiite Itegral 77 k k k k k k k k k k k k Š ˆ a a a k k k k a a a. As Ä_ ad mp mä this epressio has value 6. Thus, a d. 6 c k. We get the Riema sum fac Š ˆ ˆ a k k k k k k k k k a a a Š k k k k 6a 7a a 8. As Ä_ ad mp mä this epressio has value 8 6 7 9. Thus, a d 9. 67. Cosider the partitio P that sudivides the iterval Ò, Ó ito suitervals of width ad let c e the right edpoit of each suiterval. So the partitio is P Ö,,,..., ad c k. We get the Riema sum fac Š ˆ ˆ a k k a k ˆ ˆ k k k k 6k k 8k 6 8 Š k k Œ k k k k k k 6 a a a a a a a Š 68. Cosider the partitio P that sudivides the iterval Ò, Ó ito suitervals of width ad let c e the right edpoit of each suiterval. So the partitio is P Ö,,,..., ad c k. We get the Riema sum f c c k 6. As Ä _ ad mp m Ä this epressio has value 68. Thus, d. k a aa aa k a a a a k a a a k ˆ k Š k k k a a ka a ak a a k a a a a a a a a a a a a k a k k k k k k k a a a a a aa a a a a a a a Š 69. Cosider the partitio P that sudivides the iterval Òa, Ó ito suitervals of width a ad let c e the right edpoit of each suiterval. So the partitio is P Öa, a, a,..., a ad c a. We get the Riema sum f c c a k Š Œ a a a a a a a a a a a a aa a a a aaa a a a a a a aa a a a a a. As Ä_ ad mp mä this epressio has value a aa aa a. Thus, d. k k Ö k a k a k ˆ Š ˆ Œ k k k k k k k k 7. Cosider the partitio P that sudivides the iterval Ò, Ó ito suitervals of width ad let c e the right edpoit of each suiterval. So the partitio is P,,,..., ad c k. We get the Riema sum f c c c k k k k Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

78 Chapter Itegratio a a a Š. As Ä_ ad mp mä this epressio has value. Thus, a d. 7. To fid where, let Ê ( ) Ê or. If, the Ê a ad maimize the itegral. 7. To fid where Ÿ, let Ê a Ê or È. By the sig graph,, we ca see that Ÿ o Èß È Ê a È ad È È È miimize the itegral. 7. f() is decreasig o [ß] Ê maimum value of f occurs at Ê ma f f() ; miimum value of f occurs at Ê mi f f(). Therefore, ( ) mi f Ÿ d Ÿ ( ) ma f Ê Ÿ d Ÿ. That is, a upper oud ad a lower oud. (.).. (.) Ÿ Ÿ Ê Ÿ Ÿ... 9 Ÿ Ÿ. Ê Ÿ Ÿ 7. See Eercise 7 aove. O [ß.], ma f, mi f.8. Therefore (. ) mi f Ÿ f() d Ÿ (. ) ma f Ê Ÿ d Ÿ. O [.ß], ma f.8 ad mi f.. Therefore (.) mi f d (.) ma f d. The d d d. 7. Ÿ si a Ÿ for all Ê ( )( ) Ÿ si a d Ÿ ( )() or si d Ÿ Ê si d caot equal. 76. f() È 8 is icreasig o [ß] Ê ma f f() È 8 ad mi f f() È 8 È. Therefore, ( ) mi f Ÿ È 8 d Ÿ ( ) ma f Ê È Ÿ È 8 d Ÿ. 77. If f() o [aß ], the mi f ad ma f o [aß]. Now, ( a) mi f Ÿ f() d Ÿ ( a) ma f. The a Ê a Ê ( a) mi f Ê f() d. 78. If f() Ÿ o [aß], the mi f Ÿ ad ma f Ÿ. Now, ( a) mi f Ÿ f() d Ÿ ( a) ma f. The a Ê a Ê ( a) ma f Ÿ Ê f() d Ÿ. a a 79. si Ÿ for Ê si Ÿ for Ê (si ) d Ÿ (see Eercise 78) Ê si d d Ÿ a si d d si d Š si d. Thus a upper oud is. Ê Ÿ Ê Ÿ Ê Ÿ a Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Sectio. The Defiite Itegral 79 ß ß Ê Ê 7 Š 6 8. sec o ˆ ß Ê sec Š o ˆ ß Ê sec Š d (see Eercise 77) sice [ ] is cotaied i ˆ sec d Š d sec d Š Ê sec d d d Ê sec d ( ) Ê sec d. Thus a lower oud 7 is 6. a a a a av(f) d ( a)k ( a) f() d f() d. a a a a 8. Yes, for the followig reasos: av(f) f() d is a costat K. Thus av(f) d K d K( a) Ê 8. All three rules hold. The reasos: O ay iterval [aß ] o which f ad g are itegrale, we have: a a a a a a a a a (a) av(f g) [f() g()] d f() d g() d f() d g() d av(f) av(g) a a a a a a () av(kf) kf() d k f() d k f() d k av(f) a a a a a a (c) av(f) f() d Ÿ g() d sice f() Ÿ g() o [aß], ad g() d av(g). Therefore, av(f) Ÿ av(g). 8. (a) U ma? ma? á ma? where ma f( ), ma f( ), á, ma f( ) sice f is icreasig o [aß]; L mi? mi? á mi? where mi f( ), mi f( ), á, mi f( c) sice f is icreasig o [aß]. Therefore U L (ma mi )? (ma mi )? á (ma mi )? (f() f( ))? (f( ) f( ))? á (f( ) f( c))? (f( ) f( ))? (f() f(a))?. () U ma? ma? á ma? where ma f( ), ma f( ), á, ma f( ) sice f is icreasig o[aß]; L mi? mi? á mi? where mi f( ), mi f( ), á, mi f( c) sice f is icreasig o [aß]. Therefore U L (ma mi )? (ma mi )? á (ma mi )? (f() f( ))? (f( ) f( ))? á (f( ) f( c))? Ÿ (f() f( ))? ma (f( ) f( ))? ma á (f( ) f( c))? ma. The U L Ÿ (f( ) f( ))? ma (f() f(a))? ma kf() f(a) k? ma sice f() f(a). Thus lim (U L) lim (f() f(a))? ma, sice? ma lp l. lpl Ä lpl Ä 8. (a) U ma? ma? á ma? where ma f( ), ma f( ), á, ma f( c ) sice f is decreasig o [aß ]; L mi? mi? á mi? where mi f( ), mi f( ) ß á, mi f( ) sice f is decreasig o [aß ]. Therefore U L (ma mi )? (ma mi )? á (ma mi )? (f() f( ))? (f( ) f( ))? á (f( c ) f( ))? (f( ) f( ))? (f(a) f())?. Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

8 Chapter Itegratio () U ma? ma? á ma? where ma f( ), ma f( ), á, ma f( c ) sice f is decreasig o[aß]; L mi? mi? á mi? where mi f( ), mi f( ), á, mi f( ) sice f is decreasig o [aß]. Therefore U L (ma mi )? (ma mi )? á (ma mi )? (f() f( ))? (f( ) f( ))? á (f( c ) f( ))? Ÿ (f() f( ))? ma (f(a) f()? ma kf() f(a) k? ma sice f() Ÿ f(a). Thus lim (U L) lim kf() f(a) k? ma, sice? ma lp l. lpl Ä lpl Ä 8. (a) Partitio ß ito suitervals, each of legth? with poits,?,?, á,?. Sice si is icreasig o ß, the upper sum U is the sum of the areas of the circumscried rectagles of areas f( )? (si? )?, f( )? (si? )?, á, f( )? cos? cosˆ ˆ? si (si? )?. The U (si? si? á si? )??? cos cos ˆ ˆ ˆ cos cos ˆ cos cos ˆ si si si Š ˆ () The area is si d lim. Ä_ Î cos cos ˆ cos si Š 86. (a) The area of the shaded regio is m which is equal to L. i () The area of the shaded regio is M which is equal to U. i i i i i (c) The area of the shaded regio is the differece i the areas of the shaded regios show i the secod part of the figure ad the first part of the figure. Thus this area is U L. 87. By Eercise 86, U L M m where M maö fa o the ith suiterval ad i i i i i i i i a a i i i i i i i i i a a i i m miö f o the ith suiterval. Thus U L M m provided for each i ßÞÞÞ,. Sice a the result, U L a follows. 88. The car drove the first miles i hours ad the secod miles i hours, which meas it drove miles i 8 hours, for a average of mi/hr 7. mi/hr. I terms of average values of fuctios, the fuctio whose average value we seek is 8, Ÿ t Ÿ v(t), ad the average value is, Ÿ 8 ()() ()() 8 7.. Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

89-9. Eample CAS commads: Maple: with( plots ); with( Studet[Calculus] ); f := -> -; a := ; := ; N :=[,,, ]; P := [seq( RiemaSum( f(), =a.., partitio=, method=radom, output=plot ), =N )]: display( P, isequece=true ); Sectio. The Defiite Itegral 8 9-98. Eample CAS commads: Maple: with( Studet[Calculus] ); f := -> si(); a := ; := Pi; plot( f(), =a.., title=9(a) (Sectio.) ); N := [,, ]; () for i N do Xlist := [ a+.*(-a)/*i i=.. ]; Ylist := map( f, Xlist ); ed do: for i N do (c) Avg[] := evalf(add(y,y=ylist)/ops(ylist)); ed do; avg := FuctioAverage( f(), =a.., output=value ); evalf( avg ); FuctioAverage(f(),=a..,output=plot); (d) fsolve( f()=avg, =. ); fsolve( f()=avg, =. ); fsolve( f()=avg[], =. ); fsolve( f()=avg[], =. ); 89-98. Eample CAS commads: Mathematica: (assiged fuctio ad values for a,, ad may vary) Sums of rectagles evaluated at left-had edpoits ca e represeted ad evaluated y this set of commads Clear[, f, a,, ] {a, }={, }; =; d = ( a)/; f = Si[] ; vals =Tale[N[], {, a, d, d}]; yvals = f /. Ä vals; oes = MapThread[Lie[{{,},{, },{, },{, }]&,{vals, vals d, yvals}]; Plot[f, {, a, }, Epilog Ä oes]; Sum[yvals[[i]] d, {i,, Legth[yvals]}]//N Sums of rectagles evaluated at right-had edpoits ca e represeted ad evaluated y this set of commads. Clear[, f, a,, ] {a, }={, }; =; d = ( a)/; f = Si[] ; Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

8 Chapter Itegratio vals =Tale[N[], {, a d,, d}]; yvals = f /. Ä vals; oes = MapThread[Lie[{{,},{, },{, },{, }]&,{vals d,vals, yvals}]; Plot[f, {, a, }, Epilog Ä oes]; Sum[yvals[[i]] d, {i,,legth[yvals]}]//n Sums of rectagles evaluated at midpoits ca e represeted ad evaluated y this set of commads. Clear[, f, a,, ] {a, }={, }; =; d = ( a)/; f = Si[] ; vals =Tale[N[], {, a d/, d/, d}]; yvals = f /. Ä vals; oes = MapThread[Lie[{{,},{, },{, },{, }]&,{vals d/, vals d/, yvals}]; Plot[f, {, a, },Epilog Ä oes]; Sum[yvals[[i]] d, {i,, Legth[yvals]}]//N. THE FUNDAMENTAL THEOREM OF CALCULUS. ( ) d c d a () a( ) ( ) 6 c. ˆ ( ) d Š () Š ( ) c. a d a d Š Š a a a a. a d Š a a Š ( ) ( ) c a ( ). Š d Š Š 8 () () () 6 6 6 6. a d Š () Š ( ) ( ) c ( ) 7. ˆ È d ˆ Î 8. d ˆ ( ) Î& Î& Î 9. sec d [ ta ] ˆ ta ˆ ( ta ) È È Î. a cos d [ si ] asi a si Î Î. csc ) cot ) d ) [ csc )] ˆ csc ˆ ˆ csc ˆ È Š È Î Î. sec u ta u du [ sec u] sec ˆ sec () () Î Î Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Sectio. The Fudametal Theorem of Calculus 8 Î Î Î cos t. dt ˆ cos t dt t si t ˆ () si () ˆ ˆ si ˆ Î Î cos t. dt ˆ cos t dt t si t cî cî Î Î È 6 6 ˆ ˆ si ˆ ˆ ˆ si ˆ si si ˆ Î Î Î. ta d asec d [ta ] ˆ ta ˆ ata a Î6 Î6 Î6 6. asec ta d asec sec ta ta d asec sec ta d Î6 ˆ ˆ ˆ ˆ a È 6 6 6 6 [ ta sec ] ta sec ta sec 7. si d cos ˆ cos ˆ ˆ cos a Î8 Î8 È 8 Î Î cî t cî Î t Î 8. ˆ sec t dt a sec t t dt ta t Š ta ˆ Š ta ˆ (( ) ) Š Š È È ˆ ˆ r ( ) 8 9. (r ) dr ar r dr r r Š ( ) ( ) Š È È. (t ) t dt t t È a È at t t dt t t Š È È Š È Š È È È È Š È Š È Š È Š È ) Š È u u & u u 6 È 6 6 Š È ( ( ) ). È Š & du È Š u du u Š () c c c y y y ac ac c y c c ac ac. dy ay y dy y Š Š È È È. s Ès ds s ˆ Î s Î ds s È È Š È È ÈÈ 8 ˆ ˆ s ÉÈ 8 8 8 Î Î Î Î Î Î Î Î Î. d d ˆ d Î Î Î Î Î Î Î Î Î Š 8 a a8 8 a a8 Š a a a a 7 si sicos. d d cos d si asi a ˆ si ˆ Î si Î si Î Î Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

8 Chapter Itegratio Î Î Î 6. acos sec d acos sec d ˆ cos sec d Î Î ˆ cos sec d si ta ˆ si ˆ ˆ taˆ ˆ si a a taa 9È 6 8 7. k k d k k d k k d d d Š Š 6 c c c Î Î ( ) 8. acos kcos k d (cos cos ) d (cos cos ) d cos d [si ] Î si si È È È 9. (a) cos t dt [si t] si È si si È d d Î Ê Œ cos t dt ˆ si È cos È ˆ cos È È È d d () Œ cos t dt ˆ cos È ˆ ˆ È ˆ cos È ˆ Î d d d cos È È si si si d d d d. (a) t dt ct d si Ê Œ t dt asi si cos d d () Œ t dt a si ˆ (si ) si cos d si d Î t t t Î Î & dt dt t. (a) Èu du u du u d t t È d a Ê Œ u du ˆ t t t d d () Œ Èu du Èt ˆ at t at t dt & dt ta ) ta ) d d d) d) ta ). (a) sec y dy [ta y] ta (ta )) ta (ta )) Ê sec y dy (ta (ta ))) Œ asec (ta )) sec ) d d () Œ sec y dy asec (ta ) ˆ (ta ) ) ) asec (ta )) sec ) d) d) ta ). y È t dy dt Ê È dy. y dt Ê, d t d È. y si t dt dy d Î È si t dt Ê si (si ) d Š ˆ È ˆ ˆ È ˆ d dy d d d d d 6. y si t dt Ê Œ si t dt si t dt si a a si t dt 6 si si t dt t t dy ct t d 7. y dt dt Ê d si È Î Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Sectio. The Fudametal Theorem of Calculus 8 dy d d d 8. y Œ at dt Ê Œ at dt Œ at dt a Œ at dt si Èt d Èsi d Ècos kcos k cos dt dy d cos cos 9. y, k k Ê ˆ (si ) (cos ) sice kk dt dy. y Ê ˆ ˆ d (ta ) ˆ asec ta t d ta d sec. Ê ( ) Ê or ; Area a d a d a d ( ) ( ) Š Š ( ) Š ( ) ( ) Š Š Š ( ) 8 Š Š Š. Ê Ê ; ecause of symmetry aout the y-ais, Area Œ a d a d Š c d c d c aa () a () aa () a () d (6). Ê a Ê ( )( ) Ê,, or ; Area a d a d Š Š Š Š Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

86 Chapter Itegratio Î Î Î Î. Ê ˆ Ê or Î Î Ê or Ê or Ê or ; ) Area ˆ Î d ˆ Î Î d ˆ d c ) Î Î Î Î Î ( ) Š () Š ( ) Î Î Š () Š () Î 8 Î 8 Š (8) Š () ˆ. The area of the rectagle ouded y the lies y, y,, ad is. The area uder the curve y cos o [ ß] is ( cos ) d [ si ] ( si ) ( si ). Therefore the area of the shaded regio is. 6 6 6 6. The area uder the curve y si o Î6 & Î 6 6 6ß 6 is si d [ cos ] Î6 Î ˆ ˆ È È Š È È 6 6 6. The area of the rectagle ouded y the lies,, y si si, ad y is ˆ cos cos. Therefore the area of the shaded regio is. 7. O ß : The area of the rectagle ouded y the lies y È, y, ), ad ) is È ˆ È. The area etwee the curve y sec ) ta ) ad y is sec ) ta ) d ) [ sec )] cî Î ˆ È È Î ) ) Î ) ) ) ) È È ( sec ) sec ˆ È. Therefore the area of the shaded regio o ß is Š È. O ß : The area of the rectagle ouded y,, y È, ad y is È ˆ ) ). The area uder the curve y sec ta is sec ta d [sec ] sec sec. Therefore the area of the shaded regio o ß is Š È. Thus, the area of the total shaded regio is È È È Š È Š È. 8. The area of the rectagle ouded y the lies y, y, t, ad t is ˆ ˆ. The area uder the curve y sec t o ß is sec t dt [ta t] ta ta ˆ Î. The area cî uder the curve y t o [ß ] is t a t dt t Š Š. Thus, the total area uder the curves o ß is. Therefore the area of the shaded regio is ˆ. dy t d t 9. y dt Ê ad y( ) dt Ê (d) is a solutio to this prolem. dy c c d c. y sec t dt Ê sec ad y( ) sec t dt Ê (c) is a solutio to this prolem. dy d. y sec t dt Ê sec ad y() sec t dt Ê () is a solutio to this prolem. Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

dy Sectio. The Fudametal Theorem of Calculus 87. y dt Ê ad y() dt Ê (a) is a solutio to this prolem. t d t È. y sec t dt. y t dt Î Î. Area ˆ h ˆ h h d h cî h ˆ h ˆ Œhˆ Œhˆ ˆ ˆ h h h h h h 6 6 cî h Îk Îk k k 6. k Ê oe arch of y si k will occur over the iterval ß Ê the area si k d cos k cos ˆ k ˆ ˆ cos () k k k k dc 7. Î Ê c Î t dt Î t È; c() c() È È 9. d È 8. r Š d Š d ˆ Š Š ( ) ( ) ( ) ( ) ˆ. or 9. (a) t Ê T 8 È 7 F; t 6 Ê T 8 È 6 76 F; È t Ê T 8 8 F Î () average temperatuve Š 8 È t dt 8t a t Î Î Š 8aa Š 8aa 7 F 6. (a) t Ê H È a Î ft; t Ê H È a Î È È.7 ft; t 8 Ê H È8a8 Î ft 8 Î Î Î () average height t t dt t t Š È a 8 8 Î Î Î Î 9 8 8 Š a8 a8 Š a a 9.67 ft d d 6. f(t) dt Ê f() d f(t) dt d a d d 6. f(t) dt cos Ê f() f(t) dt cos si Ê f() cos () () si () 9 w 9 9 w 9 t () t 6. f() dt Ê f () Ê f () ; f() dt ; L() ( ) f() ( ) 8 Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

88 Chapter Itegratio 6. g() sec (t ) dt Ê g () asec a () sec a Ê g ( ) ( ) sec a( ) w w a ; g( ) sec (t ) dt sec (t ) dt ; L() ( ( )) g( ) ( ) 6. (a) True: sice f is cotiuous, g is differetiale y Part of the Fudametal Theorem of Calculus. () True: g is cotiuous ecause it is differetiale. w (c) True, sice g () f(). ww w (d) False, sice g () f (). w ww w (e) True, sice g () ad g () f (). ww w ww (f) False: g () f (), so g ever chages sig. w w w (g) True, sice g () f() ad g () f() is a icreasig fuctio of (ecause f () ). 66. Let a â e ay partitio of Òa, Óad let F e ay atiderivative of f. (a) Fa Fa i i i F a F a F a F a F a F a â F a F a F a F a F a F a F a F a F a âf a F a F a F a F a F a Fa a () Sice F is ay atiderivative of f o Òa, Ó Ê F is differetiale o Òa, Ó Ê F is cotiuous o Òa, Ó. Cosider ay suiterval Ò, Ói Òa, Ó, the y the Mea Value Theorem there is at least oe umer c i Ð, Ñ such that i i i i i w i i i i i i i i i? i i i i F a F a Faca fc a a fc a. Thus F a Fa a F a F a fc a?. i i i (c) Takig the limit of FaFaa faci? i we otai lim afafaa lim Œ faci? i a Ê F a Fa a f a d i mpmä mpmä i 67-7. Eample CAS commads: Maple: with( plots ); f := -> ^-*^+*; a := ; := ; F := uapply( it(f(t),t=a..), ); (a) p := plot( [f(),f()], =a.., leged=[y = f(),y = F()], title=67(a) (Sectio.) ): p; df := D(F); () q := solve( df()=, ); pts := [ seq( [,f()], =remove(has,evalf([q]),i) ) ]; p := plot( pts, style=poit, color=lue, symolsize=8, symol=diamod, leged=(,f()) where F ()= ): display( [p,p], title=8() (Sectio.) ); icr := solve( df()>, ); (c) decr := solve( df()<, ); df := D(f); (d) p := plot( [df(),f()], =a.., leged=[y = f (),y = F()], title=67(d) (Sectio.) ): p; q := solve( df()=, ); Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Sectio. The Fudametal Theorem of Calculus 89 pts := [ seq( [,F()], =remove(has,evalf([q]),i) ) ]; p := plot( pts, style=poit, color=lue, symolsize=8, symol=diamod, leged=(,f()) where f ()= ): display( [p,p], title=8(d) (Sectio.) ); 7-7. Eample CAS commads: Maple: a := ; u := -> ^; f := -> sqrt(-^); F := uapply( it( f(t), t=a..u() ), ); df := D(F); () cp := solve( df()=, ); solve( df()>, ); solve( df()<, ); df := D(dF); (c) solve( df()=, ); plot( F(), =-.., title=7(d) (Sectio.) ); 7. Eample CAS commads: Maple: f := `f`; q := Diff( It( f(t), t=a..u() ), ); d := value( q ); 76. Eample CAS commads: Maple: f := `f`; q := Diff( It( f(t), t=a..u() ),, ); value( q ); 67-76. Eample CAS commads: Mathematica: (assiged fuctio ad values for a, ad may vary) For trascedetal fuctios the FidRoot is eeded istead of the Solve commad. The Map commad eecutes FidRoot over a set of iitial guesses Iitial guesses will vary as the fuctios vary. Clear[, f, F] {a, }= {, }; f[_] = Si[] Cos[/] F[_] = Itegrate[f[t], {t, a, }] Plot[{f[], F[]},{, a, }] /.Map[FidRoot[F[]==, {, }] &,{,,, 6}] /.Map[FidRoot[f[]==, {, }] &,{,,,, 6}] Slightly alter aove commads for 7-8. Clear[, f, F, u] a=; f[_] = u[_] = F[_] = Itegrate[f[t], {t, a, u()}] /.Map[FidRoot[F[]==,{, }] &,{,,, }] /.Map[FidRoot[F[]==,{,}] &,{,,, }] Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

9 Chapter Itegratio After determiig a appropriate value for, the followig ca e etered = ; Plot[{F[], {, a, }]. INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE 6 6 6 6. Let u Ê du d Ê du d a a d u du u du u C C. Let u 7 Ê du 7 d Ê 7 du d 7È7 d Î 7 7 d Î a 7u du Î Î Î u du u C a7 C. Let u Ê du d Ê du d 7 a a d u du u du u C C a a. Let u Ê du d Ê du d d d u du u du u C C. Let u Ê du a6 d a d Ê du a d a a a d u du u du u C C 6. Let u Ê du d Ê du d È È È Î Î È È ˆ È Î Î Î Î d ˆ È d u du u du u C ˆ È C 7. Let u Ê du d Ê du d si d si u du cos u C cos C 8. Let u Ê du d Ê du d si a d si u du cos u C cos C 9. Let u t Ê du dt Ê du dt sec t ta t dt sec u ta u du sec u C sec t C t t t t t t. Let u cos Ê du si dt Ê du si dt ˆ cos ˆ si dt u du u C ˆ cos C. Let u r Ê du r dr Ê du 9r dr 9r dr Î Î Î È r u du ()u C 6 a r C. Let u y y Ê du ay 8y dy Ê du ay y dy y y y y dy a a u du u C ay y C Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Î Î. Let u Ê du d Ê du È d Sectio. Idefiite Itegrals ad the Sustitutio Rule 9 È si ˆ Î d si u du ˆ u si u C ˆ Î si ˆ Î C. Let u Ê du d 6 si ˆ C cos ˆ u d cos au du cos au du ˆ si u C si ˆ C. (a) Let u cot ) Ê du csc ) d ) Ê du csc ) d) u u csc ) cot ) d) u du Š C C cot ) C () Let u csc ) Ê du csc ) cot ) d ) Ê du csc ) cot ) d) u u csc ) cot ) d) u du Š C C csc ) C 6. (a) Let u 8 Ê du d Ê du d Î Î Î Š du u du ˆ u C u C È 8 C d È 8 Èu È Î d Ê Ê È8 () Let u 8 du ( 8) () d du du uc È8C d È 8 7. Let u s Ê du ds Ê du ds È s ds Èu ˆ du Î u du ˆ ˆ Î u Î C ( s) C 8. Let u s Ê du ds Ê du ds Î Î ds ˆ du u du ˆ ˆ u C Ès C Ès Èu 9. Let u ) Ê du ) d ) Ê du ) d) ) È ) d) È u ˆ du Î u du ˆ ˆ &Î u C a ) &Î C. Let u 7 y Ê du 6y dy Ê du y dy yè7 y dy Èu ˆ du u du ˆ ˆ u C a7 y C È È È du Ȉ È u u È. Let u Ê du d Ê du d Î Î Î d C C. Let u z Ê du dz Ê du dz cos (z ) dz (cos u) ˆ du cos u du si u C si (z ) C. Let u Ê du d Ê du d a ˆ sec ( ) d sec u du sec u du ta u C ta ( ) C. Let u ta Ê du sec d ta sec d u du u C ta C Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

9 Chapter Itegratio. Let u si ˆ Ê du cos ˆ d Ê du cos ˆ d & si ˆ cos ˆ & d u ( du) ˆ u C si ˆ C 6 6. Let u ta ˆ Ê du sec ˆ d Ê du sec ˆ d ( ta ˆ sec ˆ ( ) ) d u ( du) ˆ u C ta ˆ C 8 r r 8 6 r & & & u r 8 6 8 7. Let u Ê du dr Ê 6 du r dr r Š dr u (6 du) 6 u du 6 Š C Š C & r Š & & r u r Š Š 8. Let u 7 Ê du r dr Ê du r dr r 7 dr u ( du) u du C 7 C Î Î Î Î Î Î 9. Let u Ê du d Ê du d si ˆ d (si u) ˆ du si u du ( cos u) C cos ˆ C. Let u csc ˆ v Ê du csc ˆ v cot ˆ v dv Ê du csc ˆ v cot ˆ v dv csc ˆ v cot ˆ v dv du u C csc ˆ v C. Let u cos (t ) Ê du si (t ) dt Ê du si (t ) dt si (t ) du cos (t ) u u cos (t ) dt C C. Let u sec z Ê du sec z ta z dz sec z ta z Èsec z Î Î È Èu dz du u du u C sec z C t t t t t. Let u t Ê du t dt Ê du dt cos ˆ dt (cos u)( du) cos u du si u C si ˆ C È Î Î Èt. Let u t t Ê du t dt Ê du dt Èt cos ˆ Èt dt (cos u)( du) cos u du si u C si ˆ Èt C. Let u si Ê du ˆ cos ˆ d ) Ê du cos d) ) ) ) ) ) ) ) ) ) ) si cos d u du u C si C 6. Let u csc È Ê du csc È cot È ) Š ) ) Š d ) Ê du cot È) csc È) d) cos È È ) È ) ) d ) cot È) csc È) d) du u C csc È) C C È) si È) È) si È) a ˆ ˆ 6 a 7. Let u t Ê du t dt Ê du t dt t t dt u du u C t C Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

8. Let u Ê du d Sectio. Idefiite Itegrals ad the Sustitutio Rule 9 É d É d É d È u du u du u C ˆ C Î Î Î & 9. Let u Ê du d Î Î Î É d Èu du u du u C ˆ C. Let u Ê du d Ê du d É d É d È u du u du u C ˆ C Î Î Î 9 9. Let u Ê du d Ê du d É d É d É d È u du u du u C ˆ C Î Î Î 9 9 7 7 Î Î Î È Èu. Let u Ê du d Ê du d É d d du u du u C a C. Let u. The du d ad u. Thus a d au u du au u du u u C a a C. Let u. The du d ad adu d ad u. Thus È d a uèu adu a a uˆ Î u Î Î Î 8 Î Î 8 Î du u u du u u C a a C. Let u. The du d ad adu d ad u. Thus a a d 7 6 8 7 6 8 7 8 7 6 8a 7a a C au u adu au u u du u u u C Î Î 6. Let u. The du d ad u. Thus a a d au u du ˆ Î Î u u du 7 Î Î 7 Î Î 7 7 u u C a a C 7. Let u. The du d ad du d ad u. Thus È d au Èu du a a a Î Î &Î Î &Î Î &Î & & & u u du u u C u u C C & 8. Let u Êdu d ad u. So B È Î Î d au Èu du u u du &Î Î &Î & & u u C a a C a Î a u C a C 9. Let u Ê du d ad du d. Thus d d u du u du a a. Let u Ê du d ad u. Thus d a d u au du u u du a u u C a a C Î Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

9 Chapter Itegratio. (a) Let u ta Ê du sec d; v u Ê dv u du Ê 6 dv 8u du; w v Ê dw dv 8 ta sec 8u 6 dv 6 dw 6 ata au ( v) w v C u ta C d du 6 w dw 6w C C () Let u ta Ê du ta sec d Ê 6 du 8 ta sec d; v u Ê dv du 8 ta sec 6 du 6 dv 6 6 6 ta a ( u) v v u ta d C C C (c) Let u ta Ê du ta sec d Ê 6 du 8 ta sec d 8 ta sec 6 du 6 6 ta a u u ta d C C. (a) Let u Ê du d; v si u Ê dv cos u du; w v Ê dw v dv Ê dw v dv È si ( ) si ( ) cos ( ) d È si u si u cos u du vè v dv Î Î Î È Î w dw w C av C asi u C asi () C () Let u si ( ) Ê du cos ( ) d; v u Ê dv u du Ê dv u du È si ( ) si ( ) cos ( ) d u È u du È Î v dv v dv ˆ Î Î Î Î a a ˆ v C v C u C si () C (c) Let u si ( ) Ê du si ( ) cos ( ) d Ê du si ( ) cos ( ) d È si ( ) si ( ) cos ( ) d È Î u du u du ˆ Î u C Î a si ( ) C. Let u (r ) 6 Ê du 6(r )() dr Ê du (r ) dr; v È u Ê dv du Ê dv du (r ) cos È(r ) 6 È(r ) 6 Èu 6 Èu cos Èu Èu 6 6 6 dr Š ˆ du (cos v) ˆ dv si v C si Èu C si È(r ) 6 C 6. Let u cos È Ê du si È si ) ) Š ) Š d ) Ê du d) È È ) È ) si È) È) uî É cos È È Écos È È ) ) ) ) u C Écos È ) si du Î d ) d ) u du ˆ Î u C C. Let u t Ê du 6t dt Ê du t dt a s t at dt u ( du) ˆ u C u C at C; s whe t Ê ( ) C Ê 8 C Ê C Ê s t 6. Let u 8 Ê du d Ê du d Î Î y a 8 d u ( du) ˆ Î u Î C u C a 8 C; y whe Ê (8) C Ê C Ê y a 8 7. Let u t Ê du dt Î Î s 8 si ˆ t dt 8 si u du 8 ˆ u si u C ˆ t si ˆ t C; 6 ˆ ˆ 6 6 6 s 8 whe t Ê 8 si C Ê C 8 9 Ê s ˆ t si ˆ t 9 t si ˆ t 9 Î Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Sectio. Idefiite Itegrals ad the Sustitutio Rule 9 8. Let u ) Ê du d) r cos ˆ d cos u du ˆ u si u C ˆ si ˆ ) ) ) ) C; 8 ) ˆ 8 8 ) ˆ ) r si r cos ) ˆ ) 8 ) ) 8 r whe Ê si C Ê C Ê r si Ê Ê 9. Let u t Ê du dt Ê du dt ds dt si ˆ t dt (si u)( du) cos u C cos ˆ t C ; ds at t ad we have cos ˆ ds C Ê C Ê cos ˆ t dt dt Ê s ˆ cos ˆ t dt (cos u ) du si u u C si ˆ t ˆ t C ; ˆ ˆ at t ad s we have si C Ê C Ê s si ˆ t t ( ) Ê s si ˆ t t 6. Let u ta Ê du sec d Ê du sec d; v Ê dv d Ê dv d dy d sec ta d u( du) u C ta C ; dy d at ad we have C Ê C Ê ta asec sec dy d Ê y asec d asec v ˆ dv ta v v C ta C ; at ad y we have () C Ê C Ê y ta 6. Let u t Ê du dt Ê du 6 dt s 6 si t dt (si u)( du) cos u C cos t C; at t ad s we have cos C Ê C Ê s cos t Ê s ˆ cos ( ) 6 m 6. Let u t Ê du dt Ê du dt v cos t dt (cos u)( du) si u C si ( t) C ; at t ad v 8 we have 8 () C Ê C 8 Ê v si ( t) 8 Ê s ( si ( t) 8) dt si u du 8t C cos ( t) 8t C ; at t ad s we have C Ê C Ê s 8t cos ( t) Ê s() 8 cos m 6. All three itegratios are correct. I each case, the derivative of the fuctio o the right is the itegrad o the left, ad each formula has a aritrary costat for geeratig the remaiig atiderivatives. Moreover, cos si C cos C Ê C C ; also cos C C Ê C C C. Î6 Î 6 ma ma 6. (a) Š V si t dt 6 V ˆ cos ( t) V [cos cos ] V ma [ ] () V È V È () 9 volts ma rms Î6 Î6 Î6 a ma a ma ˆ cos t avma a t si t a V V si ( ) si () a V ma Î ma ma ˆ ˆ ˆ ˆ ˆ 6 (c) V si t dt V dt ( cos t) dt ds dt Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

96 Chapter Itegratio.6 SUBSTITUTION AND AREA BETWEEN CURVES. (a) Let u y Ê du dy; y Ê u, y Ê u È Î y dy u du Î u ˆ Î () ˆ Î () ˆ (8) ˆ () () Use the same sustitutio for u as i part (a); y Ê u, y Ê u È Î y dy u du Î u ˆ Î () c r È r dr Èu du Î u ˆ Î (). (a) Let u r Ê du r dr Ê du r dr; r Ê u, r Ê u () Use the same sustitutio for u as i part (a); r Ê u, r Ê u r È r dr Èu du c Î ta sec d u u du ta sec d u u du cî c. (a) Let u ta Ê du sec d; Ê u, Ê u () Use the same sustitutio as i part (a); Ê u, Ê u. (a) Let u cos Ê du si d Ê du si d; Ê u, Ê u cos si d u du cu d ( ) () a () Use the same sustitutio as i part (a); Ê u, Ê u cos si d u du t t dt u a u du 6 6 6 6. (a) u t Ê du t dt Ê du t dt; t Ê u, t Ê u () Use the same sustitutio as i part (a); t Ê u, t Ê u t a t dt u du c 6. (a) Let u t Ê du t dt Ê du t dt; t Ê u, t È7 Ê u 8 È 7 8 Î ) 8 8 8 Î t at dt u du ˆ ˆ Î u ˆ Î (8) ˆ Î () () Use the same sustitutio as i part (a); t È7 Ê u 8, t Ê u 8 Î cè 7 8 Î Î t at dt u du u du r dr u du a r 7. (a) Let u r Ê du r dr Ê du r dr; r Ê u, r Ê u () Use the same sustitutio as i part (a); r Ê u, r Ê u r & dr u du u () () a r ˆ ˆ 8 8 Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Sectio.6 Sustitutio ad Area Betwee Curves 97 Î Î 8. (a) Let u v Ê du v dv Ê du Èv dv; v Ê u, v Ê u Èv dv ˆ du u du a v u Î u Î () Use the same sustitutio as i part (a); v Ê u, v Ê u 9 9 Èv * 7 7 dv ˆ du v u a u ˆ Î 9 ˆ 8 7 È 9. (a) Let u Ê du d Ê du d; Ê u, Ê u È Î Î Î Î d du u du u () () È Èu () Use the same sustitutio as i part (a); È Ê u, È Ê u È d du cè È È u d Î Î Î Î È u du ()u () (9) È 9 9 *. (a) Let u 9 Ê du d Ê du d; Ê u 9, Ê u () Use the same sustitutio as i part (a); Ê u, Ê u 9 9 Î Î È d u du u du c È 9 9 6 Î6 ( cos t) si t dt u du () () Š u 6 6 6 6 Ê Ê Î ( cos t) si t dt u du () () Î6 Š u 6 6. (a) Let u cos t Ê du si t dt Ê du si t dt; t Ê u, t Ê u cos () Use the same sustitutio as i part (a); t u, t u cos t t t. (a) Let u ta Ê du sec dt Ê du sec dt; t Ê u ta ˆ, t Ê u ˆ t ta t sec dt u ( du) cu d Î ˆ t ta t sec dt u du cu d 8 c Î () Use the same sustitutio as i part (a); t Ê u, t Ê u c Î. (a) Let u si z Ê du cos z dz Ê du cos z dz; z Ê u, z Ê u cos z dz du È ˆ si z Èu () Use the same sustitutio as i part (a); z Ê u si ( ), z Ê u c cos z È si z Èu dz ˆ du si w dw & u du u cî ( cos w) ˆ c d ˆ Ê Ê Î si w dw u du u du ( cos w) ˆ. (a) Let u cos w Ê du si w dw Ê du si w dw; w Ê u, w Ê u () Use the same sustitutio as i part (a); w u, w u Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

98 Chapter Itegratio &. Let u t t Ê du at dt; t Ê u, t Ê u Èt& t at dt Î u du Î u Î Î () () È dy Èy 6. Let u Èy Ê du ; y Ê u, y Ê u dy du u du u Èyˆ Èy u c d ˆ ˆ 6 7. Let u cos ) Ê du si ) d ) Ê du si ) d ); ) Ê u, ) Ê u cos ˆ 6 6 Î6 Î Î Î cos ) si ) d) u ˆ du u du Š c ˆ () 8. Let u ta ˆ ) Ê du sec ˆ ) d Ê 6 du sec ˆ ) d ; Ê u ta ˆ ) ) ), ) Ê u ta 6 6 6 6 6 È Î c & cot ˆ ) sec ˆ ) & u d u (6 du) 6 Š 6 6 ) È u Î È () Î Î È Š È 9. Let u cos t Ê du si t dt Ê du si t dt; t Ê u cos, t Ê u cos 9 9 9 * Î Î ( cos t) si t dt u ˆ du Î u du ˆ &Î &Î &Î u 9. Let u si t Ê du cos t dt Ê du cos t dt; t Ê u, t Ê u Î Î Î ( si t) cos t dt u du ˆ &Î u ˆ &Î &Î () ˆ (). Let u y y y Ê du a y y dy; y Ê u, y Ê u () () () 8 8 Î ) Î ay y y ay y dy u du Î u Î Î (8) () Î y 6y y 9 y y dy Î a a u du ˆ Î u Î Î () (9) ( ) 9 *. Let u y 6y y 9 Ê du ay y dy Ê du ay y dy; y Ê u 9, y Ê u Î Î. Let u ) Ê du ) d ) Ê du È) d ); ) Ê u, ) È Ê u È Î È cos ˆ d cos u ˆ du ˆ u si u ) ) ) ˆ si () t cî c u t si. Let u Ê du t dt; t Ê u, t Ê u t si ˆ dt si u du ˆ si u ˆ si ( ) ˆ si c È È Î Î Î Î c Î Î Î 6 u () (). Let u Ê du d Ê du d; Ê u, Ê u, Ê u A d d u du u du u du u du 6. Let u cos Ê du si d; Ê u, Ê u ( cos ) si d u du u Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Sectio.6 Sustitutio ad Area Betwee Curves 99 7. Let u cos Ê du si d Ê du si d; Ê u cos ( ), Ê u cos Î Î Î Î Î &Î c A (si ) È cos d u ( du) u du u () () 8. Let u si Ê du cos d Ê du cos d; Ê u si ˆ, Ê u Because of symmetry aout, A (cos ) (si ( si )) d (si u) ˆ du si u du [ cos u] ( cos ) ( cos ) c Î cos 9. For the sketch give, a, ; f() g() cos si ; (cos ) A d ( cos ) d si [( ) ( )] Î Î Î Î Î ˆ (cos t). For the sketch give, a, ; f(t) g(t) sec t a si t sec t si t; Î Î Î Î È A sec t si t dt sec t dt si t dt sec t dt dt c Î c Î c Î c Î c Î sec t dt ( cos t) dt [ta t] [t ] È si t cî cî Î Î. For the sketch give, a, ; f() g() a ; & A a d ˆ ˆ c. For the sketch give, c, d ; f(y) g(y) y y ; 6 6 9 8 A ay y dy y dy y dy y y ( ) ( ). For the sketch give, c, d ; f(y) g(y) ay y ay y y y y; A ay y y dy y dy y dy y dy y y y ˆ () (). For the sketch give, a, ; f() g() a ; & A a d ˆ ˆ c. We wat the area etwee the lie y, Ÿ Ÿ, ad the curve y, 78?= the area of a triagle (formed y y ad y ) with ase ad height. Thus, A Š d ()() ˆ 8 6 6. We wat the area etwee the -ais ad the curve y, Ÿ Ÿ :6?= the area of a triagle (formed y, y, ad the -ais) with ase ad height. Thus, A d ()() 6 7. AREA A A A: For the sketch give, a ad we fid y solvig the equatios y ad y simultaeously for : Ê Ê ( )( ) Ê or so c c : f() g() a a Ê A a d Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Chapter Itegratio ˆ 6 8 6 ( 8 9 ) 9 ; A: For the sketch give, a ad : f() g() a a Ê A a d ˆ ˆ 6 8 c 6 8 9; Therefore, AREA A A 9 8 8. AREA A A A: For the sketch give, a ad : f() g() a a 8 Ê A a 8 d (8 6) 8; c 8 A: For the sketch give, a ad : f() g() a a 8 Ê A a8 d (6 8) 8; Therefore, AREA A A 6 8 9. AREA A A A A: For the sketch give, a ad : f() g() ( ) a c Ê A a d ˆ ˆ 8 7 ; c 6 6 A: For the sketch give, a ad : f() g() a ( ) a Ê A a d ˆ 8 ˆ 9 8 ; c A: For the sketch give, a ad : f() g() ( ) a Ê A a d ˆ 7 9 6 ˆ 8 9 8 9 ; 9 ˆ 9 8 9 6 6 6 Therefore, AREA A A A 9 9. AREA A A A A: For the sketch give, a ad : f() g() Š a c (8 ) ; Š a Ê A a d ( 8) ; A: For the sketch give, a ad we fid y solvig the equatios y ad y simultaeously for : Ê Ê ( )( ) Ê,, or so : f() g() Š a Ê A a d a A: For the sketch give, a ad : f() g() 9 8 6 8 Ê A a d ˆ 9 ˆ 8 ˆ ; Therefore, AREA A A A. a, ; f() g() a Ê A a d ˆ 8 8 ˆ 8 8 c 8 ˆ Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Sectio.6 Sustitutio ad Area Betwee Curves. a, ; f() g() a ( ) c 7 Ê A a d ˆ 9 9 ˆ. a, ; f() g() 8 Ê A a8 d 6 8 8 8 &. Limits of itegratio: Ê Ê ( ) Ê a ad ; f() g() a Ê A a d 7 9 7 8 9. Limits of itegratio: Ê Ê ( ) Ê a ad ; f() g() a 6 6 8 8 6 Ê A a d 6. Limits of itegratio: 7 Ê Ê ( )( ) Ê a ad ; f() g() a7 a Ê A a d c ˆ ˆ 6ˆ 7. Limits of itegratio: Ê Ê a a Ê ( )( )( )( ) Ê,,, ; f() g() a ad g() f() a c c c Ê A a d a d a d & & & 6 6 8 ˆ ˆ 8 ˆ ˆ ˆ 8 ˆ 8 Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Chapter Itegratio 8. Limits of itegratio: È a Ê or È a Ê or a Ê a,, a; A Èa d Èa d ca Î a Î aa aa a Î Î a aa aa a È, Ÿ 9. Limits of itegratio: y Èk k ad È, 6 y 6 or y ; for Ÿ : È 6 Ê È 6 Ê ( ) 6 Ê 7 6 Ê ( )( 6) Ê, 6 (ut 6 is ot a solutio); for : È 6 Ê 6 Ê 6 Ê ( )( 9) Ê, 9; there are three itersectio poits ad 9 A ˆ 6 È d ˆ 6 È d ˆ È 6 d c * ( 6) Î ( 6) Î Î ( 6) ( ) 6 Î 6 Î Î 9 ˆ ˆ ˆ. Limits of itegratio:, Ÿ or y k k, Ÿ Ÿ for Ÿ ad : Ê 8 8 Ê 6 Ê ; for Ÿ Ÿ : Ê 8 8 Ê Ê ; y symmetry of the graph, 6 8 6 8 6 6 6 6 A Š a d Š a d 8 ˆ ˆ 6. Limits of itegratio: c ad d ; f(y) g(y) y y Ê A y dy 9 8 y Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

. Limits of itegratio: y y Ê (y )(y ) Ê c ad d ; f(y) g(y) (y ) y y y c 8 8 9 6 Ê A ay y dy y ˆ ˆ Sectio.6 Sustitutio ad Area Betwee Curves. Limits of itegratio: y ad 6 y Ê y 6 y Ê y y Ê (y )(y ) Ê c ad d ; f(y) g(y) ˆ Š 6y y y y c y y & 6 6 ˆ 89 9 8 8 Ê A ay y dy y ˆ ˆ 8. Limits of itegratio: y ad y Ê y y Ê y Ê (y )(y ) Ê c ad d ; f(y) g(y) a y y y ay Ê A ay dy y y ˆ ˆ c ˆ. Limits of itegratio: y y ad y y 6 Ê y y y y 6 Ê y y 6 Ê ay ay Ê c ad d ; f(y) g(y) ay yay y 6 y y 6 y c 9 8 6 Ê A ay y 6 dy y 6y ˆ 9 8 ˆ Î 6. Limits of itegratio: y ad y Î Ê y y Ê c ad d ; Î f(y) g(y) a y y Ê A ˆ Î y y dy c & y &Î y y ˆ ˆ ˆ Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

Chapter Itegratio 7. Limits of itegratio: y ad kykè y Ê y kykèy Ê y y y ay Ê y y y y Ê y y Ê ay ay Ê y or y È Ê y or y Ê y or y. È Sustitutio shows that are ot solutios Ê y ; for Ÿ y Ÿ, f() g() yè y ay Î y y a y, ad y symmetry of the graph, A y yay dy c Î Î y a y c c Î a y dy y a y dy y ˆ () ˆ ˆ 8. AREA A A Limits of itegratio: y ad y y Ê y y y Ê y ay y y(y )(y ) Ê y,, : for Ÿ y Ÿ, f(y) g(y) y y y y y c y y a 6 8 8 8 7 A Ê A ay y y dy y ˆ ; for Ÿ y Ÿ, f(y) g(y) y y y Ê A y y y dy y Ê ˆ ; Therefore, A 9. Limits of itegratio: y ad y Ê Ê Ê a ( )( ) Ê a ad ; f() g() c & Ê A a d ˆ ˆ ˆ 6. Limits of itegratio: y ad y Ê Ê a () Ê ( )( ) Ê a ad ; f() g() a c 6 7 8 Ê A a d ˆ ˆ Copyright Pearso Educatio, Ic. Pulishig as Addiso-Wesley.

2024 © sciencedocbox.com Privacy Policy | Terms of Service | Feedback