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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Checkpoint

1.1

i=362i=23+24+25+26=120i=362i=23+24+25+26=120

1.2

15,550

1.3

440

1.4

The left-endpoint approximation is 0.7595. The right-endpoint approximation is 0.6345. See the below Exercise 7.27.

Two graphs side by side showing the left-endpoint approximation ad right-endpoint approximation of the area under the curve f(x) = 1/x from 1 to 2 with endpoints spaced evenly at .25 units. The heights of the left-endpoint approximation one are determined by the values of the function at the left endpoints, and the height of the right-endpoint approximation one are determined by the values of the function at the right endpoints.
1.5
  1. Upper sum=8.0313.Upper sum=8.0313.

  2. A graph of the function f(x) = 10 − x^2 from 0 to 2. It is set up for a right endpoint approximation over the area [1,2], which is labeled a=x0 to x4. It is an upper sum.
1.6

A1.125A1.125

1.7

6

1.8

18 square units

1.9

6

1.10

18

1.11

613x3dx413x2dx+213xdx133dx613x3dx413x2dx+213xdx133dx

1.12

−7

1.13

3

1.14

Average value=1.5;c=3Average value=1.5;c=3

1.15

c=3c=3

1.16

g(r)=r2+4g(r)=r2+4

1.17

F(x)=3x2cosx3F(x)=3x2cosx3

1.18

F(x)=2xcosx2cosxF(x)=2xcosx2cosx

1.19

724724

1.20

Kathy still wins, but by a much larger margin: James skates 24 ft in 3 sec, but Kathy skates 29.3634 ft in 3 sec.

1.21

103103

1.22

Net displacement: e2920.8055m;e2920.8055m; total distance traveled: 4ln47.5+e221.7404ln47.5+e221.740 m

1.23

17.5 mi

1.24

645645

1.25

3x2(x33)2dx=13(x33)3+C3x2(x33)2dx=13(x33)3+C

1.26

(x3+5)1030+C(x3+5)1030+C

1.27

1sint+C1sint+C

1.28

cos4t4+Ccos4t4+C

1.29

913913

1.30

23π0.212223π0.2122

1.31

x2e−2x3dx=16e−2x3+Cx2e−2x3dx=16e−2x3+C

1.32

ex(3ex2)2dx=19(3ex2)3ex(3ex2)2dx=19(3ex2)3

1.33

2x3ex4dx=12ex42x3ex4dx=12ex4

1.34

1204eudu=12(e41)1204eudu=12(e41)

1.35

Q(t)=2tln2+8.557.Q(t)=2tln2+8.557. There are 20,099 bacteria in the dish after 3 hours.

1.36

There are 116 flies.

1.37

121x3e4x−2dx=18[e4e]121x3e4x−2dx=18[e4e]

1.38

ln|x+2|+Cln|x+2|+C

1.39

xln3(lnx1)+Cxln3(lnx1)+C

1.40

14sin−1(4x)+C14sin−1(4x)+C

1.41

sin−1(x3)+Csin−1(x3)+C

1.42

110tan−1(2x5)+C110tan−1(2x5)+C

1.43

14tan−1(x4)+C14tan−1(x4)+C

1.44

π8π8

Section 1.1 Exercises

1.

a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting j=i1.j=i1. d. They are equal; the first sum factors the terms of the second.

3.

38530=35538530=355

5.

15(−12)=2715(−12)=27

7.

5(15)+4(−12)=275(15)+4(−12)=27

9.

j=150j22j=150j=(50)(51)(101)62(50)(51)2=40,375j=150j22j=150j=(50)(51)(101)62(50)(51)2=40,375

11.

4k=125k2100k=125k=4(25)(26)(51)950(25)(26)=−10,4004k=125k2100k=125k=4(25)(26)(51)950(25)(26)=−10,400

13.

R4=0.25R4=0.25

15.

R6=0.372R6=0.372

17.

L4=2.20L4=2.20

19.

L8=0.6875L8=0.6875

21.

L6=9.000=R6.L6=9.000=R6. The graph of f is a triangle with area 9.

23.

L6=13.12899=R6.L6=13.12899=R6. They are equal.

25.

L10=410i=1104(−2+4(i1)10)L10=410i=1104(−2+4(i1)10)

27.

R100=e1100i=1100ln(1+(e1)i100)R100=e1100i=1100ln(1+(e1)i100)

29.
A graph of the given function on the interval [0, 1]. It is set up for a left endpoint approximation and is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity, but this behavior persists for more rectangles.


R100=0.33835,L100=0.32835.R100=0.33835,L100=0.32835. The plot shows that the left Riemann sum is an underestimate because the function is increasing. Similarly, the right Riemann sum is an overestimate. The area lies between the left and right Riemann sums. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.

31.
A graph of the given function over [-1,1] set up for a left endpoint approximation. It is an underestimate since the function is increasing. Ten rectangles are shown for visual clarity, but this behavior persists for more rectangles.


L100=−0.02,R100=0.02.L100=−0.02,R100=0.02. The left endpoint sum is an underestimate because the function is increasing. Similarly, a right endpoint approximation is an overestimate. The area lies between the left and right endpoint estimates.

33.
A graph of the given function over the interval -1 to 1 set up for a left endpoint approximation. It is an underestimate since the function is increasing. Ten rectangles are shown for isual clarity, but this behavior persists for more rectangles.


L100=3.555,R100=3.670.L100=3.555,R100=3.670. The plot shows that the left Riemann sum is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.

35.

The sum represents the cumulative rainfall in January 2009.

37.

The total mileage is 7×i=125(1+(i1)10)=7×25+710×12×25=385mi.7×i=125(1+(i1)10)=7×25+710×12×25=385mi.

39.

Add the numbers to get 8.1-in. net increase.

41.

309,389,957

43.

L8=3+2+1+2+3+4+5+4=24L8=3+2+1+2+3+4+5+4=24

45.

L8=3+5+7+6+8+6+5+4=44L8=3+5+7+6+8+6+5+4=44

47.

L101.7604,L301.7625,L501.76265L101.7604,L301.7625,L501.76265

49.

R1=−1,L1=1,R10=−0.1,L10=0.1,L100=0.01,R1=−1,L1=1,R10=−0.1,L10=0.1,L100=0.01, and R100=−0.1.R100=−0.1. By symmetry of the graph, the exact area is zero.

51.

R1=0,L1=0,R10=2.4499,L10=2.4499,R100=2.1365,L100=2.1365R1=0,L1=0,R10=2.4499,L10=2.4499,R100=2.1365,L100=2.1365

53.

If [c,d][c,d] is a subinterval of [a,b][a,b] under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is f(c)(dc).f(c)(dc). But, f(c)f(x)f(c)f(x) for cxd,cxd, so the area under the graph of f between c and d is f(c)(dc)f(c)(dc) plus the area below the graph of f but above the horizontal line segment at height f(c),f(c), which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of f on [a,b].[a,b].

55.

LN=baNi=1Nf(a+(ba)i1N)=baNi=0N1f(a+(ba)iN)LN=baNi=1Nf(a+(ba)i1N)=baNi=0N1f(a+(ba)iN) and RN=baNi=1Nf(a+(ba)iN).RN=baNi=1Nf(a+(ba)iN). The left sum has a term corresponding to i=0i=0 and the right sum has a term corresponding to i=N.i=N. In RNLN,RNLN, any term corresponding to i=1,2,…,N1i=1,2,…,N1 occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with RNLN=baN(f(a+(ba))NN)(f(a)+(ba)0N)=baN(f(b)f(a)).RNLN=baN(f(a+(ba))NN)(f(a)+(ba)0N)=baN(f(b)f(a)).

57.

Graph 1: a. L(A)=0,B(A)=20;L(A)=0,B(A)=20; b. U(A)=20.U(A)=20. Graph 2: a. L(A)=9;L(A)=9; b. B(A)=11,U(A)=20.B(A)=11,U(A)=20. Graph 3: a. L(A)=11.0;L(A)=11.0; b. B(A)=4.5,U(A)=15.5.B(A)=4.5,U(A)=15.5.

59.

Let A be the area of the unit circle. The circle encloses n congruent triangles each of area sin(2πn)2,sin(2πn)2, so n2sin(2πn)A.n2sin(2πn)A. Similarly, the circle is contained inside n congruent triangles each of area BH2=12(cos(πn)+sin(πn)tan(πn))sin(2πn),BH2=12(cos(πn)+sin(πn)tan(πn))sin(2πn), so An2sin(2πn)(cos(πn))+sin(πn)tan(πn).An2sin(2πn)(cos(πn))+sin(πn)tan(πn). As n,n2sin(2πn)=πsin(2πn)(2πn)π,n,n2sin(2πn)=πsin(2πn)(2πn)π, so we conclude πA.πA. Also, as n,cos(πn)+sin(πn)tan(πn)1,n,cos(πn)+sin(πn)tan(πn)1, so we also have Aπ.Aπ. By the squeeze theorem for limits, we conclude that A=π.A=π.

Section 1.2 Exercises

61.

02(5x23x3)dx02(5x23x3)dx

63.

01cos2(2πx)dx01cos2(2πx)dx

65.

01xdx01xdx

67.

36xdx36xdx

69.

12xlog(x2)dx12xlog(x2)dx

71.

1+2·2+3·3=141+2·2+3·3=14

73.

14+9=614+9=6

75.

12π+9=102π12π+9=102π

77.

The integral is the area of the triangle, 1212

79.

The integral is the area of the triangle, 9.

81.

The integral is the area 12πr2=2π.12πr2=2π.

83.

The integral is the area of the “big” triangle less the “missing” triangle, 912.912.

85.

L=2+0+10+5+4=21,R=0+10+10+2+0=22,L+R2=21.5L=2+0+10+5+4=21,R=0+10+10+2+0=22,L+R2=21.5

87.

L=0+4+0+4+2=10,R=4+0+2+4+0=10,L+R2=10L=0+4+0+4+2=10,R=4+0+2+4+0=10,L+R2=10

89.

24f(x)dx+24g(x)dx=83=524f(x)dx+24g(x)dx=83=5

91.

24f(x)dx24g(x)dx=8+3=1124f(x)dx24g(x)dx=8+3=11

93.

424f(x)dx324g(x)dx=32+9=41424f(x)dx324g(x)dx=32+9=41

95.

The integrand is odd; the integral is zero.

97.

The integrand is antisymmetric with respect to x=3.x=3. The integral is zero.

99.

112+1314=712112+1314=712

101.

01(16x+12x28x3)dx=(x-3x2+4x3-2x4)=(1-3+4-2)(0-0+0-0)=001(16x+12x28x3)dx=(x-3x2+4x3-2x4)=(1-3+4-2)(0-0+0-0)=0

103.

754=234754=234

105.

The integrand is negative over [−2,3].[−2,3].

107.

xx2xx2 over [1,2],[1,2], so 1+x1+x21+x1+x2 over [1,2].[1,2].

109.

cos(t)22.cos(t)22. Multiply by the length of the interval to get the inequality.

111.

fave=0;c=0fave=0;c=0

113.

3232 when c=±32c=±32

115.

fave=0;c=π2,3π2fave=0;c=π2,3π2

117.

L100=1.294,R100=1.301;L100=1.294,R100=1.301; the exact average is between these values.

119.

L100×(12)=0.5178,R100×(12)=0.5294L100×(12)=0.5178,R100×(12)=0.5294

121.

L1=0,L10×(12)=8.743493,L100×(12)=12.861728.L1=0,L10×(12)=8.743493,L100×(12)=12.861728. The exact answer 26.799,26.799, so L100 is not accurate.

123.

L1×(1π)=1.352,L10×(1π)=−0.1837,L100×(1π)=−0.2956.L1×(1π)=1.352,L10×(1π)=−0.1837,L100×(1π)=−0.2956. The exact answer 0.303,0.303, so L100 is not accurate to first decimal.

125.

Use tan2θ+1=sec2θ.tan2θ+1=sec2θ. Then, BA=π/4π/41dx=π2.BA=π/4π/41dx=π2.

127.

02πcos2tdt=π,02πcos2tdt=π, so divide by the length 2π of the interval. cos2tcos2t has period π, so yes, it is true.

129.

The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, A=bb24ac2aA=bb24ac2a and B=b+b24ac2a.B=b+b24ac2a.

131.

If f(t0)>g(t0)f(t0)>g(t0) for some t0[a,b],t0[a,b], then since fgfg is continuous, there is an interval containing t0 such that f(t)>g(t)f(t)>g(t) over the interval [c,d],[c,d], and then ddf(t)dt>cdg(t)dtddf(t)dt>cdg(t)dt over this interval.

133.

The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, abf(t)dt=a0a1f(t)dt+a1a2f(t)dt++aN+1aNf(t)dt=a0a11dt+a1a21dt++aN+1aN1dt=(a1a0)+(a2a1)++(aNaN1)=aNa0=ba.abf(t)dt=a0a1f(t)dt+a1a2f(t)dt++aN+1aNf(t)dt=a0a11dt+a1a21dt++aN+1aN1dt=(a1a0)+(a2a1)++(aNaN1)=aNa0=ba. Dividing through by baba gives the desired identity.

135.

0Nf(t)dt=i=1Ni1if(t)dt=i=1Ni2=N(N+1)(2N+1)60Nf(t)dt=i=1Ni1if(t)dt=i=1Ni2=N(N+1)(2N+1)6

137.

L10=1.815,R10=1.515,L10+R102=1.665,L10=1.815,R10=1.515,L10+R102=1.665, so the estimate is accurate to two decimal places.

139.

The average is 1/2,1/2, which is equal to the integral in this case.

141.

a. The graph is antisymmetric with respect to t=12t=12 over [0,1],[0,1], so the average value is zero. b. For any value of a, the graph between [a,a+1][a,a+1] is a shift of the graph over [0,1],[0,1], so the net areas above and below the axis do not change and the average remains zero.

143.

Yes, the integral over any interval of length 1 is the same.

Section 1.3 Exercises

145.

Yes. It is implied by the Mean Value Theorem for Integrals.

147.

F(2)=−1;F(2)=−1; average value of FF over [1,2][1,2] is −1/2.−1/2.

149.

ecostecost

151.

116x2116x2

153.

xddxx=12xddxx=12

155.

1cos2xddxcosx=|sinx|sinx1cos2xddxcosx=|sinx|sinx

157.

2x|x|1+x22x|x|1+x2

159.

ln(e2x)ddxex=2xexln(e2x)ddxex=2xex

161.

a. f is positive over [1,2][1,2] and [5,6],[5,6], negative over [0,1][0,1] and [3,4],[3,4], and zero over [2,3][2,3] and [4,5].[4,5]. b. The maximum value is 2 and the minimum is −3. c. The average value is 0.

163.

a. is positive over [0,1][0,1] and [3,6],[3,6], and negative over [1,3].[1,3]. b. It is increasing over [0,1][0,1] and [3,5],[3,5], and it is constant over [1,3][1,3] and [5,6].[5,6]. c. Its average value is 13.13.

165.

T10=49.08,−23x3+6x2+x5dx=48T10=49.08,−23x3+6x2+x5dx=48

167.

T10=260.836,19(x+x2)dx=260T10=260.836,19(x+x2)dx=260

169.

T10=3.058,144x2dx=3T10=3.058,144x2dx=3

171.

F(x)=x33+3x225x,F(3)F(−2)=356F(x)=x33+3x225x,F(3)F(−2)=356

173.

F(x)=t55+13t3336t,F(3)F(2)=6215F(x)=t55+13t3336t,F(3)F(2)=6215

175.

F(x)=x100100,F(1)F(0)=1100F(x)=x100100,F(1)F(0)=1100

177.

F(x)=x33+1x,F(4)F(14)=112564F(x)=x33+1x,F(4)F(14)=112564

179.

F(x)=x,F(4)F(1)=1F(x)=x,F(4)F(1)=1

181.

F(x)=43t3/4,F(16)F(1)=283F(x)=43t3/4,F(16)F(1)=283

183.

F(x)=cosx,F(π2)F(0)=1F(x)=cosx,F(π2)F(0)=1

185.

F(x)=secx,F(π4)F(0)=21F(x)=secx,F(π4)F(0)=21

187.

F(x)=cot(x),F(π2)F(π4)=1F(x)=cot(x),F(π2)F(π4)=1

189.

F(x)=1x+12x2,F(−1)F(−2)=78F(x)=1x+12x2,F(−1)F(−2)=78

191.

F(x)=exeF(x)=exe

193.

F(x)=0F(x)=0

195.

−2−1(t22t3)dt−13(t22t3)dt+34(t22t3)dt=463−2−1(t22t3)dt−13(t22t3)dt+34(t22t3)dt=463

197.

π/20sintdt+0π/2sintdt=2π/20sintdt+0π/2sintdt=2

199.

a. The average is 11.21×10911.21×109 since cos(πt6)cos(πt6) has period 12 and integral 0 over any period. Consumption is equal to the average when cos(πt6)=0,cos(πt6)=0, when t=3,t=3, and when t=9.t=9. b. Total consumption is the average rate times duration: 11.21×12×109=1.35×101111.21×12×109=1.35×1011 c. 109(11.211639cos(πt6)dt)=109(11.21+2π)=11.84x109109(11.211639cos(πt6)dt)=109(11.21+2π)=11.84x109

201.

If f is not constant, then its average is strictly smaller than the maximum and larger than the minimum, which are attained over [a,b][a,b] by the extreme value theorem.

203.

a. d2θ=(acosθ+c)2+b2sin2θ=a2+c2cos2θ+2accosθ=(a+ccosθ)2;d2θ=(acosθ+c)2+b2sin2θ=a2+c2cos2θ+2accosθ=(a+ccosθ)2; b. d=12π02π(a+2ccosθ)dθ=ad=12π02π(a+2ccosθ)dθ=a

205.

Mean gravitational force = GmM202π1(a+2a2b2cosθ)2dθ.GmM202π1(a+2a2b2cosθ)2dθ.

Section 1.4 Exercises

207.

(x1x)dx=x1/2dxx−1/2dx=23x3/2+C12x1/2+C2=23x3/22x1/2+C(x1x)dx=x1/2dxx−1/2dx=23x3/2+C12x1/2+C2=23x3/22x1/2+C

209.

dx2x=12ln|x|+Cdx2x=12ln|x|+C

211.

0πsinxdx0πcosxdx=cosx|0π(sinx)|0π=((−1)+1)(00)=20πsinxdx0πcosxdx=cosx|0π(sinx)|0π=((−1)+1)(00)=2

213.

P(s)=4s,P(s)=4s, so dPds=4dPds=4 and 244ds=8.244ds=8.

215.

12Nds=N12Nds=N

217.

With p as in the previous exercise, each of the 12 pentagons increases in area from 2p to 4p units so the net increase in the area of the dodecahedron is 36p units.

219.

18s2=6s2s2xdx18s2=6s2s2xdx

221.

12πR2=8πR2Rrdr12πR2=8πR2Rrdr

223.

d(t)=0tv(s)ds=4tt2.d(t)=0tv(s)ds=4tt2. The total distance is d(2)=4m.d(2)=4m.

225.

d(t)=0tv(s)ds.d(t)=0tv(s)ds. For t<3,d(t)=0t(62t)dt=6tt2.t<3,d(t)=0t(62t)dt=6tt2. For t>3,d(t)=d(3)+3t(2t6)dt=9+(t26t)|36.t>3,d(t)=d(3)+3t(2t6)dt=9+(t26t)|36. The total distance is d(6)=18m.d(6)=18m.

227.

v(t)=409.8t;h(t)=1.5+40t4.9t2v(t)=409.8t;h(t)=1.5+40t4.9t2 m/s

229.

The net increase is 1 unit.

231.

At t=5,t=5, the height of water is x=(15π)1/3m..x=(15π)1/3m.. The net change in height from t=5t=5 to t=10t=10 is (30π)1/3(15π)1/3(30π)1/3(15π)1/3 m.

233.

The total daily power consumption is estimated as the sum of the hourly power rates, or 911 gW-h.

235.

17 kJ

237.

a. 54.3%; b. 27.00%; c. The curve in the following plot is 2.35(t+3)e−0.15(t+3).2.35(t+3)e−0.15(t+3).

A graph of the data and a function approximating the data. The function is a very close approximation.
239.

In dry conditions, with initial velocity v0=30v0=30 m/s, D=64.3D=64.3 and, if v0=25,D=44.64.v0=25,D=44.64. In wet conditions, if v0=30,v0=30, and D=180D=180 and if v0=25,D=125.v0=25,D=125.

241.

225 cal

243.

E(150)=28,E(300)=22,E(450)=16E(150)=28,E(300)=22,E(450)=16

245.

a.

A plot of the given data, which decreases in a roughly concave down manner from 600 to 2200.


b. Between 600 and 1000 the average decrease in vehicles per hour per lane is −0.0075. Between 1000 and 1500 it is −0.006 per vehicles per hour per lane, and between 1500 and 2100 it is −0.04 vehicles per hour per lane. c.

A graph of given data, showing that minutes per mile increases dramatically as wehicles per hour reaches 2000.


The graph is nonlinear, with minutes per mile increasing dramatically as vehicles per hour per lane reach 2000.

247.

137037p(t)dt=0.07(37)34+2.42(37)2325.63(37)2+521.232037137037p(t)dt=0.07(37)34+2.42(37)2325.63(37)2+521.232037

249.

Average acceleration is A=1505a(t)dt=0.7(52)3+1.44(5)2+10.448.2A=1505a(t)dt=0.7(52)3+1.44(5)2+10.448.2 mph/s

251.

d(t)=01|v(t)|dt=0t(730t30.72t210.44t+41.033)dt=7120t40.24t35.22t3+41.033t.d(t)=01|v(t)|dt=0t(730t30.72t210.44t+41.033)dt=7120t40.24t35.22t3+41.033t. Then, d(5)81.12d(5)81.12 mph ×sec119×sec119 feet.

253.

140040(−0.068t+5.14)dt=0.068(40)2+5.14=3.78140040(−0.068t+5.14)dt=0.068(40)2+5.14=3.78

Section 1.5 Exercises

255.

u=h(x)u=h(x)

257.

f(u)=(u+1)2uf(u)=(u+1)2u

259.

du=8xdx;f(u)=18udu=8xdx;f(u)=18u

261.

15(x+1)5+C15(x+1)5+C

263.

112(32x)6+C112(32x)6+C

265.

x2+1+Cx2+1+C

267.

18(x22x)4+C18(x22x)4+C

269.

sinθsin3θ3+Csinθsin3θ3+C

271.

(1x)101101(1x)100100+C(1x)101101(1x)100100+C

273.

122(711x2)+C122(711x2)+C

275.

cos4θ4+Ccos4θ4+C

277.

cos3(πt)3π+Ccos3(πt)3π+C

279.

14cos2(t2)+C14cos2(t2)+C

281.

13(x33)+C13(x33)+C

283.

2(y32)31y32(y32)31y3

285.

133(1cos3θ)11+C133(1cos3θ)11+C

287.

112(sin3θ3sin2θ)4+C112(sin3θ3sin2θ)4+C

289.

L50=−8.5779.L50=−8.5779. The exact area is −818−818

291.

L50=−0.006399L50=−0.006399 … The exact area is 0.

293.

u=1+x2,du=2xdx,1212u−1/2du=21u=1+x2,du=2xdx,1212u−1/2du=21

295.

u=1+t3,du=3t2dt,1312u−1/2du=23(21)u=1+t3,du=3t2dt,1312u−1/2du=23(21)

297.

u=cosθ,du=sinθdθ,1/21u−4du=13(221)u=cosθ,du=sinθdθ,1/21u−4du=13(221)

299.


Two graphs. The first shows the function f(x) = cos(ln(2x)) / x, which increases sharply over the approximate interval (0,.25) and then decreases gradually to the x axis. The second shows the function f(x) = sin(ln(2x)), which decreases sharply on the approximate interval (0, .25) and then increases in a gently curve into the first quadrant.


The antiderivative is y=sin(ln(2x)).y=sin(ln(2x)). Since the antiderivative is not continuous at x=0,x=0, one cannot find a value of C that would make y=sin(ln(2x))Cy=sin(ln(2x))C work as a definite integral.

301.


Two graphs. The first is the function f(x) = sin(x) / cos(x)^3 over [-5pi/16, 5pi/16]. It is an increasing concave down function for values less than zero and an increasing concave up function for values greater than zero. The second is the fuction f(x) = ½ sec(x)^2 over the same interval. It is a wide, concave up curve which decreases for values less than zero and increases for values greater than zero.


The antiderivative is y=12sec2x.y=12sec2x. You should take C=−2C=−2 so that F(π3)=0.F(π3)=0.

303.


Two graphs. The first shows the function f(x) = 3x^2 * sqrt(2x^3 + 1). It is an increasing concave up curve starting at the origin. The second shows the function f(x) = 1/3 * (2x^3 + 1)^(1/3). It is an increasing concave up curve starting at about 0.3.


The antiderivative is y=13(2x3+1)3/2.y=13(2x3+1)3/2. One should take C=13.C=13.

305.

No, because the integrand is discontinuous at x=1.x=1.

307.

u=sin(t2);u=sin(t2); the integral becomes 1200udu.1200udu.

309.

u=(1+(t12)2);u=(1+(t12)2); the integral becomes 5/45/41udu.5/45/41udu.

311.

u=1t;u=1t; the integral becomes
1−1ucos(π(1u))du=1−1u[cosπcosusinπsinu]du=1−1ucosudu=−11ucosudu=01−1ucos(π(1u))du=1−1u[cosπcosusinπsinu]du=1−1ucosudu=−11ucosudu=0
since the integrand is odd.

313.

Setting u=cxu=cx and du=cdxdu=cdx gets you 1bcaca/cb/cf(cx)dx=cbau=au=bf(u)duc=1baabf(u)du.1bcaca/cb/cf(cx)dx=cbau=au=bf(u)duc=1baabf(u)du.

315.

0xg(t)dt=12u=1x21duua=12(1a)u1a|u=1x21=12(1a)(1(1x2)1a).0xg(t)dt=12u=1x21duua=12(1a)u1a|u=1x21=12(1a)(1(1x2)1a). As x1x1 the limit is 12(1a)12(1a) if a<1,a<1, and the limit diverges to +∞ if a>1.a>1.

317.

t=π0b1cos2t×(asint)dt=t=0πabsin2tdtt=π0b1cos2t×(asint)dt=t=0πabsin2tdt

319.

f(t)=2cos(3t)cos(2t);0π/2(2cos(3t)cos(2t))=23f(t)=2cos(3t)cos(2t);0π/2(2cos(3t)cos(2t))=23

Section 1.6 Exercises

321.

−13e−3x+C−13e−3x+C

323.

3xln3+C3xln3+C

325.

ln(x2)+Cln(x2)+C

327.

2x+C2x+C

329.

1lnx+C1lnx+C

331.

ln(ln(lnx))+Cln(ln(lnx))+C

333.

ln(xcosx)+Cln(xcosx)+C

335.

12(ln(cos(x)))2+C12(ln(cos(x)))2+C

337.

ex33+Cex33+C

339.

etanx+Cetanx+C

341.

t+Ct+C

343.

19x3(ln(x3)1)+C19x3(ln(x3)1)+C

345.

2x(lnx2)+C2x(lnx2)+C

347.

0lnxetdt=et|0lnx=elnxe0=x10lnxetdt=et|0lnx=elnxe0=x1

349.

13ln(sin(3x)+cos(3x))13ln(sin(3x)+cos(3x))

351.

12ln|csc(x2)+cot(x2)|+C12ln|csc(x2)+cot(x2)|+C

353.

12(ln(cscx))2+C12(ln(cscx))2+C

355.

13ln(267)13ln(267)

357.

ln(31)ln(31)

359.

12ln3212ln32

361.

y2ln|y+1|+Cy2ln|y+1|+C

363.

ln|sinxcosx|+Cln|sinxcosx|+C

365.

13(1(lnx2))3/2+C13(1(lnx2))3/2+C

367.

Exact solution: e1e,R50=0.6258.e1e,R50=0.6258. Since f is decreasing, the right endpoint estimate underestimates the area.

369.

Exact solution: 2ln(3)ln(6)2,R50=0.2033.2ln(3)ln(6)2,R50=0.2033. Since f is increasing, the right endpoint estimate overestimates the area.

371.

Exact solution: 1ln(4),R50=−0.7164.1ln(4),R50=−0.7164. Since f is increasing, the right endpoint estimate overestimates the area (the actual area is a larger negative number).

373.

112ln2112ln2

375.

1ln(65,536)1ln(65,536)

377.

NN+1xex2dx=12(eN2e(N+1)2).NN+1xex2dx=12(eN2e(N+1)2). The quantity is less than 0.01 when N=2.N=2.

379.

abdxx=ln(b)ln(a)=ln(1a)ln(1b)=1/b1/adxxabdxx=ln(b)ln(a)=ln(1a)ln(1b)=1/b1/adxx

381.

23

383.

We may assume that x>1,so1x<1.x>1,so1x<1. Then, 11/xdtt.11/xdtt. Now make the substitution u=1t,u=1t, so du=dtt2du=dtt2 and duu=dtt,duu=dtt, and change endpoints: 11/xdtt=1xduu=lnx.11/xdtt=1xduu=lnx.

387.

x=E(ln(x)).x=E(ln(x)). Then, 1=E'(lnx)xorx=E'(lnx).1=E'(lnx)xorx=E'(lnx). Since any number t can be written t=lnxt=lnx for some x, and for such t we have x=E(t),x=E(t), it follows that for any t,E'(t)=E(t).t,E'(t)=E(t).

389.

R10=0.6811,R100=0.6827R10=0.6811,R100=0.6827

A graph of the function f(x) = .5 * ( sqrt(2)*e^(-.5x^2)) / sqrt(pi). It is a downward opening curve that is symmetric across the y axis, crossing at about (0, .4). It approaches 0 as x goes to positive and negative infinity. Between 1 and -1, ten rectangles are drawn for a right endpoint estimate of the area under the curve.

Section 1.7 Exercises

391.

sin−1x|03/2=π3sin−1x|03/2=π3

393.

tan−1x|31=π12tan−1x|31=π12

395.

sec−1x|12=π4sec−1x|12=π4

397.

sin−1(x3)+Csin−1(x3)+C

399.

13tan−1(x3)+C13tan−1(x3)+C

401.

13sec−1(x3)+C13sec−1(x3)+C

403.

cos(π2θ)=sinθ.cos(π2θ)=sinθ. So, sin−1t=π2cos−1t.sin−1t=π2cos−1t. They differ by a constant.

405.

1t21t2 is not defined as a real number when t>1.t>1.

407.


Two graphs. The first shows the function f(x) = 1 / sqrt(9 – x^2). It is an upward opening curve symmetric about the y axis, crossing at (0, 1/3). The second shows the function F(x) = arcsin(1/3 x). It is an increasing curve going through the origin.


The antiderivative is sin−1(x3)+C.sin−1(x3)+C. Taking C=π2C=π2 recovers the definite integral.

409.


Two graphs. The first shows the function f(x) = cos(x) / (4 + sin(x)^2). It is an oscillating function over [-6, 6] with turning points at roughly (-3, -2.5), (0, .25), and (3, -2.5), where (0,.25) is a local max and the others are local mins. The second shows the function F(x) = .5 * arctan(.5*sin(x)), which also oscillates over [-6,6]. It has turning points at roughly (-4.5, .25), (-1.5, -.25), (1.5, .25), and (4.5, -.25).


The antiderivative is 12tan−1(sinx2)+C.12tan−1(sinx2)+C. Taking C=12tan−1(sin(6)2)C=12tan−1(sin(6)2) recovers the definite integral.

411.

12(sin−1t)2+C12(sin−1t)2+C

413.

14(tan−1(2t))214(tan−1(2t))2

415.

14(sec−1(t2)2)+C14(sec−1(t2)2)+C

417.


A graph of the function f(x) = -.5 * arctan(2 / ( sqrt(x^2 – 4) ) ) in quadrant four. It is an increasing concave down curve with a vertical asymptote at x=2.


The antiderivative is 12sec−1(x2)+C.12sec−1(x2)+C. Taking C=0C=0 recovers the definite integral over [2,6].[2,6].

419.


The graph of f(x) = arctan(x sin(x)) over [-6,6]. It has five turning points at roughly (-5, -1.5), (-2,1), (0,0), (2,1), and (5,-1.5).


The general antiderivative is tan−1(xsinx)+C.tan−1(xsinx)+C. Taking C=tan−1(6sin(6))C=tan−1(6sin(6)) recovers the definite integral.

421.


A graph of the function f(x) = arctan(ln(x)) over (0, 2]. It is an increasing curve with x-intercept at (1,0).


The general antiderivative is tan−1(lnx)+C.tan−1(lnx)+C. Taking C=π2=tan−1C=π2=tan−1 recovers the definite integral.

423.

sin−1(et)+Csin−1(et)+C

425.

sin−1(lnt)+Csin−1(lnt)+C

427.

12(cos−1(2t))2+C12(cos−1(2t))2+C

429.

12ln(43)12ln(43)

431.

125125

433.

2tan−1(A)π2tan−1(A)π as AA

435.

Using the hint, one has csc2xcsc2x+cot2xdx=csc2x1+2cot2xdx.csc2xcsc2x+cot2xdx=csc2x1+2cot2xdx. Set u=2cotx.u=2cotx. Then, du=2csc2xdu=2csc2x and the integral is 12du1+u2=12tan−1u+C=12tan−1(2cotx)+C.12du1+u2=12tan−1u+C=12tan−1(2cotx)+C. If one uses the identity tan−1s+tan−1(1s)=π2,tan−1s+tan−1(1s)=π2, then this can also be written 12tan−1(tanx2)+C.12tan−1(tanx2)+C.

437.

x±1.13525.x±1.13525. The left endpoint estimate with N=100N=100 is 2.796 and these decimals persist for N=500.N=500.

Chapter Review Exercises

439.

False

441.

True

443.

L4=5.25,R4=3.25,L4=5.25,R4=3.25, exact answer: 4

445.

L4=5.364,R4=5.364,L4=5.364,R4=5.364, exact answer: 5.870

447.

4343

449.

1

451.

12(x+4)2+C12(x+4)2+C

453.

43sin−1(x3)+C43sin−1(x3)+C

455.

sint1+t2sint1+t2

457.

4lnxx+14lnxx+1

459.

$6,328,113

461.

$73.36

463.

1911712ft/sec,or1593ft/sec1911712ft/sec,or1593ft/sec

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