Calculus for the Life Sciences Marvin L. Bittinger Solution Manuals.pdf (59)

234

Chapter 4: Exponential and Logarithmic Functions d) 100

2500 2000

80

1500

60

1000

40

500

0

–5

–10

33. G( x) =

20 5

15

10 x

20

100 1 + 43.3 e− 0. 0425x

0

34. C ( t) =

a) When x = 100 G(100)

100

= =

G(150) = =

C (20)

500

G(300) = =

C (50)

100 1 + 43.3 e−0.0425(150)

100 1 + 43.3 e−0.0425(300)

30 = 40.2 = = 43 0 .0 2 −30 1 ln ( 355 ) − 0.092

t

c) = =

0.0425x ] ) [ 0.0425−1.84025e − 0.0425x3 ) (1+43 .3e−

0.0425 1.8 (4025 0.0425x 0.0425 ) = ln − 00425)5 (1.8402 . = ln 1.84025 x −0.0425 = 88.66

40.2 1 + 335 e−0.092(t) 1+335 e−0.092t

74.923 = t

0.0425x

0.0425−1.84025e−0.0425x = 0 e−0.0425x =

40.2

b)

C′ ()t

The possible inflection point occurs when G′′(x) = 0 or undefined, which means the only possible inflection points occurs at

1 + 335 e−0.092(50) = 9.2046

1 + 335 e−0.092(100) = 38.884

c) e− . ( − 184025

40.2

=

C (100) =

b)

(1 + 43.3 e−0.0425x)2

40.2 1 + 335 e−0.092(20)

When t = 100

100 1 + 43.3 e−12.75 = 99.99%

· −1(1 + 43.3 e−0.0425x)−2 × (43.3 · −0.0425 e−0.0425x) 184.025 e−0.0425x

=

When t = 50

When x = 300

=

400

= 0.74164

100 1 + 43.3 e−6.375 = 91.13%

G ′ (x) ′

300

a) When t = 20

When x = 150

=

x

40.2 1 + 335 e−0.092x

100 1 + 43.3 e−0.0425(100)

100 1 + 43.3 e−4.25 = 61.82%

G′(x) = 100

200

t−2

.

4.02 ( − (1+335e −0092) .964 e−0.092t 1238 (1 + 335 e−0.092t)2

(−30.82e−. 0 092t )

d) C ′ (t) ′

=

0.092t

(e− ) [ 1+335e−0.092t ] ×((1+335e−0.092t)3 . t 113.985−76369.741e−0092)

The possible inflection point occurs when C′′(t) = 0 or undefined, which means the only possible inflection points occurs at 113.985−76369.741e−0.092t = 0 e−0.092t =

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113.985 76369.741

Exercise Set 4.3

235 −0.092t = = t

113.985 ln ( 76369.741)

points occurs at 5527.21+32900e−0.04t+ 0 = ( ) 1+32900e−0.04t = − 363689760 363689760 5527.2

ln ( 113.985 ) 73669.741

−0.092 = 70.731

e) 40

363689760 5527.2 − 1

(

32900

5527.2

ln

30

ln

20

(32900 363689760 5527.2

32900

1)

= = =

e−0.04t t−17.33

−1)

− 0.04

t

10

not an acceptable answer or

0

20

40

60

80 100 120 140 160 180 200 x

1+32900e−0.04t = 0 e−0.04t = 0.04t = − = t

105 35. A ( t) = 1 + 32900 e−004t . a) When t = 100 A(100)

= =

105 1 + 32900 e−0.04(100)

−0.04 = 260.031

105 1 + 32900 e−4 = 0.174

We find A(260.031) 105 1 + 32900 e−0.04(260.031) 105 1 + 32900 e−10.40124 = 52.5

When t = 150 A(150) = =

A(200) = =

105 1 + 32900 e−0.04(150)

105 1 + 32900 e−6 = 1.272

1 32 (900 1 ) ln (32901) 0 ln 32900

d) 100

When t = 200 A(200) = =

80

105 1 + 32900 e−0.04(200)

60

105 1 + 32900 e−8 = 8.723

40 20

b) A′(x) = 105 =

· −1(1 + 32900 e−0.04t)−2 × (32900 · −0.04 e−0.04t) 138180 e−0.04t (1 + 32900 e−0.04t)2

0

A ′ (t) ′

=

( e0.04t − ) [ 1+32900 e−0.04t ] × (1+3(2900 e−0.04t3) t+363689760 ] [ − 5527.21+32900e −0.04)

The possible inflection point occurs when A′′(t) = 0 or undefined, which means the possible inflection

200

300x400

500

36. p(t) = 1 − e−0.28t a)

c)

100

600

700

e) The maximum occurs at the limiting value which is 105 p(1)=1 − e−0.28(1) ≈ 0.244 − e−0.28(2) ≈ − ≈ 0.429 0.753 − e−0.28(5) e−0.28(11) ≈ 0.954 −0.989 e−0.28(16) ≈ p(2) = 1 e−0.28(20) ≈ 0.996− p(5) = 1 p(11) = 1 p(16) = 1e−0.28t b) p′(t) = 0.28 p(20) = 1

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236

Chapter 4: Exponential and Logarithmic Functions c) The derivative p′(t) exists for all real numbers.

The equation p′(t) = 0 has no solution. Thus, the function has no critical points and hence no relative extrema. p′(t) > 0 for all real numbers, so p(t) is increasing∞ on [0, ). p′′(t) = −0.0784 e−0.028t, so p′′(t) < 0 for all real numbers and hence is concave down on [0, ∞ ).

100 80 60 40

1 20

0.8 0

0.6

2

4

6

x8

10

12

14

29.47232081 1 + 79.56767122e−0.809743969t b) We round up since the function deals with people. The limiting value appears to be about 30 students.

38. a) N(t) =

0.4 0.2

0

10

20

x

− e−0.4t)

30

40

c) We graph the function in the window [−1, 20, −1, 35], Yscl = 5.

50

30

37. P (t) = 100%(1

25

− e−0.4(0)) = 100%(1 − e0) = a) P(0)=100%(1 100%(1 − 1) = 100%(0) = 0% P (1) = 100%(1 − e−0.4(1)) = 100%(1 − e−0.4) ≈ − e−0.4(2)) 100%(1 e−0.8) ≈ − e−0.4(3)) == = 100%(1 100%(1−− −e−2) e−1.2) − e−0.4(5)) ≈ ≈ 33.0%

20 15 10

P (2) = 100%(1

5

55.1% P (3) = 100%(1

0

69.9%

4

6

x8

10

12

14

d) Use the quotient rule to find N′(t).

P (5) = 100%(1

′

N(t)=

86.5%

− e−0.4(16)) = P(16) = 100%(1 100%(1 − e−6.4) ≈ 99.8% b) P′(t)=100%[

−(−0.4) e−0.4t] = 100%(0.4) e−0.4t (100%=1)

c) The derivative P′(t) exists for all real numbers.

The equation P′(t) = 0 has no solution. Thus, the function has no critical points and hence no relative extrema. P′(t) > 0 for all real numbers, so P(t) is increasing ∞ on [0, ). P′′(t) = −0.16 e−0.4t, so P′′(t) < 0 for all real numbers and hence is concave down on [0,∞).

1898 885181 −0.809743969t . e (1 + 79.56767122e−0.809743969t)2

′ e)limN(t)=0;thisindicatesthateventuallythepop-

− e−0.4(12)) = P(12) = 100%(1 100%(1 − e−4.8) ≈ 99.2%

=0.4e−0.4t

2

t→ ∞

ulation does not change. That is, it reaches and remains at a limiting value.

39. i = e0.073− 1 = 0.0757 = 7.57% 40 i = e0.08− 1 = 0.0833 = 8.33% .

k = ln(0.0924 + 1) = 0.0884 = 8.84%

41. k = ln(0.0661 + 1) = 0.064 = 6.4% 42.

43.

kt 3P0= P0e 3 = ekt ln(3) = t k

44. kt 4P0= P0e 4 = ekt ln(4) = t k

45. Answers vary

237

Exercise Set 4.4 46. After the second year

Exercise Set 4.4

47. 2 = e 24k

1. (d)

l n(2) k= =0.0289 24 y=y kt 48.

0e

2. (f) 3. (a) y2 = y y1 = y

dividing y2 = ( 1yy) = 2 ln ( y1) ln y2y1 (t2 − t1) ln(2) 49. k= T

and

T=

k 1% 2% T 69.315 34.657

kt2

0e

4. (e)

0e

5. (c) k k= = = ln2 6. T k = k k= = 3 ≈0.231≈23.1%permin 7. ln2 ln2 = ≈0.032≈3.2%peryr T 22 8. ln2 ≈7.2days 9.

0.096

kt1

k(t2

−t1)

e 2

t1)

k

ln(2)

10.

k 4.621% 6.931% 14% 15 10 4.951

The graph of T = ln 2/k is not linear since the independent variable appears in the denominator

11.

12.

13.

ln2 ln2 = ≈0.028≈2.8%peryear T 25 ln2 ≈0.00015≈0.015%peryr 4560 ln2 ln2 = ≈0.00003≈0.003%peryr T 23,105

P (100) = 1000 e−0.032(100) ≈ 40.8 g

70

P(t) = P0 e−kt

60

P(20) = 1000 e−0.231(20) Substituting 1000 for P0, 0.231 for k and 20 for t = 1000 e−4.62

50 40

≈9.9

30

Thus 9.9 grams of polonium will remain after 20 minutes.

20

14. The amount of carbon-14 present is 60% P0.

10 2

4

6

8 k

10

12

60%P0=P 0e −0.0001205t 0.6 = e−0.0001205t

14

ln 0.6 = ln e−0.0001205t

50. An exponential function grows indefinitely while a logistic function has a limiting growth value 51. The rule of 69 is used to approximate the doubling time (generation time) of growing things. The name comes from the approximation to ln(2) = 0.6931 which is used to find the doubling time

ln 0.6 = −0.0001205t ln 0.6 − 0.0001205=t ≈ t 4239 The tusk is about 4239 years old.

15.

N(t)=N

0e−0.0001205t

SeeExample3(b).

If a piece of wood has lost 90% of its carbon-14 from an initial amount P0, then 10% P0 is the amount present. To find the age of the wood, we solve the following equation for t:

238

Chapter 4: Exponential and Logarithmic Functions 10%P=Pe−0.0001205t 0 0 Substituting 10% P0 for P(t) 0.1 = e−0.0001205t

19. For carbon-14 the decay rate is 0.0001205 The amount of carbon-14 that remains is N

ln 0.1 = ln e−0.0001205t ln 0.1 = −0.0001205t −2.302585 = −0.0001205t Using a calculator −2.302585 =t − 0.0001205 ≈ t 19, 109

Amount lost is 100 20. k =

Thus, the piece of wood is about 19,109 years old.

ln(1/2) 60.1

= Ne−0.0001205·10000 0 = N e−1.205 0 = 0.2997N0 − 29.97 = 70.03%

= −0.011533

a)

16. A = A0e−kt First find k. The half-life is 60.1 days. ln2 k= 60.1 ≈0.0115 75% of the Iodine-125 remains, so we have:

−0.011533t 0.75N0 = N 0e 0.75 = e−0.011533t

ln(0.75) = −0.011533t t ln(0.75) = − 0.011533 t = 24.94 days

0.75 = 1 · e−0.0115t ln0.75 = lne−0.0115t ln 0.75 = −0.0115t ln 0.75 − 0.0115 =t ≈ t 25

b) N = N0e−0.011533·48 = N0e−0.553584 = 0.5749N0

The sample was on the shelf for about 25 days.

17. If an artifact has lost 60% of its carbon-14 from an initial amount P0, then 40% P0 is the amount present. To find the age t we solve the following equation for t.

After 48 days 57.49% of the iodine 125 will remain 21. k =

40%P0=P 0e −0.0001205t Substituting 40% P0 for P(t) (See Example 3(b).)

0.0884N0 = N0e −5.332×10−10 t 0.0884 = e−5.332×10−10 t ln(0.0884) = −5.332 × 10−10t = ln(0.0884) t −5.332 × 10−10 = 4.55 ×109 years

0.4 = e−0.0001205t ln 0.4 = ln e−0.0001205t ln 0.4 = −0.0001205t −0.916291 = −0.0001205t Using a calculator −0.916291 − 0.0001205 =t ≈ t 7604 The artifact is about 7604 years old.

22.

0.302N0 = N0e −5.332×10−10 t 0.302 = e−5.332×10−10 t ln(0.302) = −5.332 × 10−10t = ln(0.302) t −5.332 × 10−10 = 2.25 ×109 years

18. For carbon-14 the decay rate is 0.0001205 a) The amount of carbon-14 that remains is N = N0e−0.0001205·5300 = 0.528N0 Percentage lost is 100− 52.8 = 47.2% b) N = N0e−0.0001205·5309 = 0.527N0 Percentage lost is 100− 52.7 = 47.3%

ln(1/2) ×19=−5.332×10−10 1.3 0

23.

0.653N0 = 0.653 = ln(0.653) = t =

−5.332×10−10 t

N0e −5.332×10−10 t e 10−10t − 5.332 × ln(0.653)

− 5.332 × 10−10 = 7.99 108 × years

239

Exercise Set 4.4 A 24. a) =Ae

0

b) W=W

−kt

b) First find k. The half-life of A is 3.3 hr. ln2 k = 3.3 ≈0.21

W=W 0e −0.008(30) W=W 0e −0.24

W = 0.786628W0

Then 1 = 10 e−0.21t

Substituting30fort Using a calculator

W ≈ 78.7% W0 Thus, after 30 days, 78.7% of the initial weight remains.

0.1 = e−0.21t ln 0.1 = ln e−0.21t ln 0.1 = −0.21t ln 0.1 =t − 0.21 ≈ t 11

28.

I=I

0e −0.01(100)

I ≈ 0.37 I0, or 37% I0 29. a) I=I

0 e− µ x

There will be 1 g left after about 11 hr.

I=I

25. a) When A decomposes at a rate proportional to the amount of A present, we know that

I=I

dA = −kA. dt ThesolutionofthisequationisA=A

0e

−1.4(1) −1.4

Substituting1.4forµ and 1 for x

0e I = I0(0.246597) Using a calculator I ≈ 25% I0

0e −kt .

b) We first find k. The half-life of A is 3 hr. ln2 k= T 0.693147 k= Substituting0.693147for ln 2 and 3 for T 3 ≈ 0.23, or 23% We now substitute 8 for A0, 1 for A, and 0.23 for k and solve for t.

A=A 0e −kt 1 = 8 e−0.23t Substituting 1 = e−0.23t 8 0.125 = e−0.23t

ln 0.125 = ln e−0.23t −2.079442=−0.23t −2.079442 − 0.23 =t ≈ t 9

−0.008t

0e

Usingacalculator

After 9 hr there will be 1 gram left.

26. W=W 0e −0.009t

a) k = 0.009, or 0.9% The animal loses 0.9% of its weight each day. b) W =W 0 e−0.009(30) W ≈ 0.763 W0, or 76.3% W0

76.3% of the initial weight remains after 30 days.

27. a) W=W 0 e−0.008t k = 0.008, or 0.8%

The starving animal loses 0.8% of its weight each day.

I=I

0e

I=I

−1.4(2) −2.8

Substituting1.4forµ and 2 for x

0e I = I0(0.060810) Using a calculator I ≈ 6.1% I0 I=I

0e

I=I

−1.4(3) −4.2

Substituting1.4forµ and 3 for x

0e I = I0(0.014996) Using a calculator I ≈ 1.5% I0 b) I=I

0 e− µ x

I=I

0e

I=I

−1.4(10) −14

Substituting1.4forµ and 10 for x

0e I = I0(0.00000083) Using a calculator I ≈ 0.00008% I0 30. a) T(t)=ae−kt+C 100 = ae−k·0 + 40 60=a

240

Chapter 4: Exponential and Logarithmic Functions T(t) = 25 e−kt + 75

b)Att=5,T=90◦.

T(t) = 60 e−kt + 40

90 = 25 e−k·10 + 75 Substituting 90 for T and 10 for t 15 = 25 e−10k 15 = e−10k 25 0.6 = e−10k

90 = 60 e−k·5 + 40 50 = 60 e−5k

ln

5 e−5k 6= =lne−5k 5

6 5 ln = 6 −5k 5 ln 6 =k −5 ≈ k

ln 0.6 = e−10k ln 0.6 = −10k ln 0.6 =k =k − 10 ≈ k

−0.510826 − 10

0.04 T (t) = 60 e−0.04t + 40

0.05

c) T(10) = 60 e−0.04(10) + 40 ≈ 80.2◦ d)

41=60e−0.04t+40

1 = 60 e−0.04t 1 = e−0.04t 60 =lne−0.04t −0.04t 1 ln 60 1 = ln 60 1 ln 60 =t − 0.04 ≈ t 102.4 min e) T′(t) = −2.4e−0.04t; the rate of change of the temperature is −2.4e−0.04t degrees per minute.

31. a)T(t)=ae−kt+C

Newton’slawofCooling

At t = 0, T = 100◦. We solve the following equation for a.

100 = ae−k·0 + 75 Substituting 100 for T, 0 for t, and 75 for C 25 = ae0 (e0=1) 25=a The value of the constant is 25◦.

Thus, T(t) = 25 e−kt + 75. b) Now we find k using the fact that at t = 10, T = 90◦.

Thus, T(t) = 25 e−0.05t + 75. c)

T(t)=25e−0.05t+75 T(20) = 25 e−0.05(20) + 75 Substituting 20 for t = 25 e−1 + 75 = 25(0.367879) + 75 Using a calculator = 9.196975 + 75 ≈ 84.2 The temperature after 20 minutes is 84.2◦.

d)

T(t)=25e−0.05t+75 80 = 25 e−0.05t + 75 Substituting 80 for T 5 = 25 e−0.05t 5 = e−0.05t 25 0.2 = e−0.05t ln 0.2 = ln e−0.05t ln 0.2 = −0.05t ln 0.02 − 0.05 =t =t ≈ t

−1.609438 − 0.05 32

It takes 32 minutes for the liquid to cool to 80◦. e) T′(t) = −1.25e−0.05t; the rate of change of the temperature is −1.25e−0.05t degrees per minute.

32. Find a in T(t) = ae−kT + C. Assume T = 98.6◦ at t = 0.

98.6 = ae−k·0 + 10 88.6 = a T (t) = 88.6 e−kt + 10 We want to find the number of hours N since the murder.

We first use the two temperature readings to find k.

61.6 = 88.6 e−kN + 10, or 51.6 = 88.6 e−kN 57.2 = 88.6 e−k(N+1) + 10, or 47.2 = 88.6 e−k(N+1)

Exercise Set 4.4

241

51.6 88.6e−kN = 88.6e−k(N+1) =ek 47.2 51.6 ln =lnek 47.2 ≈ k 0.09 Now we find N. 51.6=88.6e−0.09N 51.6 = e−0.09N 88.6 =lne−0.09N −0.09N 51.6 ln 88.6 51.6 ln = 88.6 51.6 ln 88.6 − 0.09 =N ≈ N 6

34.

a) Let 1995 correspond to t = 0

(150144) = ln150 = 7 −0.005832 = k

Substituting

b) In 2010, t = 15 N = 150e−0.005832·15 =

c) 100 = 150e−0.005832t 100 = e −0.005832t ( 15)02 ln = (2)3 = ln 3 − 0.005832 69.5 = t

98.6◦=ae−k·0+60◦

(Roomtemperatureis60◦.) so a = 38.6◦. Thus T is given by T(t) = 38.6 e−kt + 60. We

35.

ln

Dividing the first equation by the second, we get

−0.399009 − 0.10 =N ≈ N 4hr The coroner arrived at 11 P.M., so the murder was committed about 7 P.M.

51.9e6k

0.010265 = k Thus, N = 51.9 e−0.010265t b) In 2015, t = 20 N = 51.9e−0.010265·20 = 51.9 · 0.8144029589 = 42.27

0.10 ≈ k

25.

38.6 9 ln 0.670984 =

48.)8 = 48.8 (

(51.9 ln 48 ).851.9 =k

83.4 = 38.6 e−k(N+1) + 60, or 23.4 = 38.6 e−k(N+1)

ln

−0.005832t t

The population of Russia will be 100 million in the year 2065. a) Let 1995 correspond to t = 0

85.9 = 38.6 e−kN + 60, or 25.9 = 38.6 e−kN

Now we substitute 0.10 for k in the equation 25.9 = 38.6 e−kN and solve for N. 25.9 = 38.6 e−0.10N 25.

9 e−0.10N 38. = =lne−0.10N 6 −0.10N

150 · 0.9162411337

= 137.44

Assuming the temperature of the body was normal when the murder occurred, we have T = 98.6◦at t = 0. Thus

25.9 38.6e−kN =e−kN+k(N+1)=ek. = 23.4 38.6e−k(N+1) We solve this equation for k: 25.9 ln =lnek 23.4 ln 1.106838 = k

k

Thus, N = 150 e−0.005832t

The coroner arrived at 2 A.M., so the murder was committed at about 8 P.M. 33. We first find a in the equation T(t) = ae−kt + C.

want to find the number of hours N since the murder was committed. To find N we must first determine k.

From the two temperature readings, we have

= 150e7k

14)4

144 ln (

c) 1 = 51.9e−0.010265t 1 = e−0.010265t ( 51.)9 = 1 ln −0.010265t (51.91 t ln 51.9 ) − 0.010265 384.7 = t The population of Ukraine will be 1 in the year 2380.

Chapter 4: Exponential and Logarithmic Functions

242 36.

P = 14.7 e−0.00005a

38.

N=Ne−0.0001205(4000) 0

a) P = 14.7 e−0.00005(1000) ≈140lb/in2 . b) P = 14.7 e−0.00005(20,000) ≈54lb/in2 . c)

1.47=14.7e−0.00005a 0.1 = e−0.00005a

N ≈ N0 (0.62), or 62% N0 Thus, 62% of the carbon-14 remains, so 38% has been lost.

39. a) y = 84.94353992 − 0.5412834098 ln x b) When x = 8, y = 83.8%;

ln 0.1 = ln e0.00005a

when x = 10, y = 83.7%;

ln 0.1 = −0.00005a ln 0.1 − 0.00005 =a ≈ a 46, 052 ft d)P′= −0.000735e−0.00005a ; the rate of change of the pressure is −0.000735e−0.00005a lb/in2 per unit of altitude.

37. a)

P(t) = 50 e−0.004t P(375) = 50 e−0.004(375) Substituting 375 for t = 50 e−1.5 = 50(0.223130) ≈11 After 375 days, 11 watts will be available.

b)

N(t)=Ne−0.0001205t 0

ln2 k 0.693147 = Substituting0.693147for ln 2 and 0.004 for k 0.004

when x = 24, y = 83.2%; when x = 36, y = 83.09% c)

82=84.94353992 − 0.5412834098 ln x −2.94353992 = −0.5412834098 ln x 5.438075261≈lnx ≈ e5.438075261 x ≈230

x The test scores will fall below 82% after about 230 months. d)y=′ −0.5412834098;thetestscoresarechangingat x 0 5412834098 percentpermonth. the rate of − . x 40. a) Let 1980 correspond to t = 0

T=

≈173 The half-life of the power supply is 173 days. c)

=

(72 ln64)72

=

72e20k k

= 20 −0.005889 = k Thus, N = 72 e−0.005889t

P(t) = 50 e−0.004t

b) In 2010, t = 30

10 = 50 e−0.004t Substituting 10 for P ( t) 1 0 = e−0.004t 5 0 0.2 = e−0.004t

N = 72e−0.005889·30 =

150 · 0.8380524814 = 125.71

ln 0.2 = ln e−0.004t ln 0.2 = −0.004t ln 0.2 − 0.004 =t =t ≈ t −1.609438 − 0.004 402

6)4

64 ln (

c) 20 = 72e−0.005889t 20 = e−0.005889t 7)220

Usingacalculator

ln (

= 72 = ln ( 20)72

− 0.005889

The satellite can stay in operation 402 days. d) When t = 0, P = 50 e−0.004(0) Substituting 0 for t

217.5 = t

The population of Russia will be 100 million in the year 2198.

= 50 e0 =50 · 1 =50 At the beginning the power output was 50 watts. e) P′(t) = −0.2e−0.004t; the power is changing at a rate of −0.2e−0.004t watts per day.

−0.005889t t

41.

a) Let 1980 correspond to t = 0 4)8 ( 48

=

(52 ln48)52

=

ln

= 20 −0.004002 = k

52e20k k

243

Exercise Set 4.5 Thus, N =52 e−0.004002t b)

13.

f (x) = x [(6.2)x ′ d ] (6.2)x+] [ d x(6.2)x Product f ()=x x dx dx =x · ( ︸ln6.2︷)︷(6.2)x︸+1·(6.2)x Theorem14 ↑ = (6.2)x[x ln 6.2 + 1]

14.

f (x) = x(5.4)x

In2010,t=30 N

= 52e−0.004002·30 = 52 · 0.8868636211 = 46.12

c) 10 = 52e−0.004002t 10 = e−0.004002t 5210 = ) ln ( −0.004002t 510)2 t ln ( − 0.004002 411.9 = t The population of Russia will be 100 million in the year 2392.

f ′( x) = x(ln 5.4)(5.4)x + (5.4)x = (5.4)x(x ln 5.4 + 1) 15.

16.

42. Answers vary.

17.

Exercise Set 4.5 1.

54=e· 4ln5

Theorem 13: ax = e x ·ln a

4(1 .609438)

≈ e

≈ e .

2. 23= 3.

3 ln 2 e·

y = log5 x 1 dy = l n· 5 dx

19.

f ()=2logx x

2.0794

≈ e

· 3.410 = e10ln3.4

= log4 x y 1 14 d dy Theorem16: logax = = l n· x 1 x dx

18.

Using a calculator

6 4378

y = x310x dy =x3 ProductRule · ︸(ln1︷0︷)10x︸+3x2·10x dx Theorem14 ↑ = 10xx2(x ln 10 + 3) = x45x y dy = x4(ln 5)5x + 4x35x dx = 5xx3(x ln 5 + 4)

f ′()=2 x Theorem13:ax=ex· Using a calculator

ln a

= =2

≈ e

12. 238

≈ e

24ln104.

5. 4k=k·e ln4

20.

≈ e0.941

e ·

′

10.

11.

y =6x dy =(l n6)6x dx

Theorem13:ax=ex·lna

dyax=(ln Theorem14: dx

y =7x dy = (ln 7)7x dx f (x)=10x f ′( x) = (ln 10)10x Theorem 14

12.

logx

1

·

1

1 ·ln10

Theorem16(logx=log

10 x )

1 ·x

5 1 ln 10 x· x f (x)=log 3 f x ( ) = log x log 3 (P2) =

21.

8. 10kR= ekR·ln10 9.

f ()=5logx x f ()=5 x

Theorem13:ax=ex·lna

ln5 6. 5R= R· e ln8 7. 8kT=kT· e

d

ln10 x 2 1 ln 10 x·

10(1 .223775)

= 4. (1.04) 24

· ·

1a l1 n x ·

f (x) = 100x f ′( x) = (ln 100)100x

1 −1 0 Theorem16(logx=log ln10 x − ·1 1 = ln 10 x · x =logx −log5 f ()=log x 5 1 1 f ′ ()= x ln 10 x· ′

f ()= x

a)ax

22.

10 x )

Chapter 4: Exponential and Logarithmic Functions

244 23.

24.

25.

26.

27.

y =x3 (log8x 1︷ ︷ · 1 2 dy =x3 ProductRule ) 8 ln8 x ︸+3x·logx dx =x2 ︸ ↑ Theorem16 =x2 · 1 +3x2·log ln8 )8x 1 ( +3log8x y = x log6 x 1 ·1+log dy =x · ln6 x dx = 1 +log6x ln6

y = cot xlog5 (2x) dy ·1−csc2x·ln = cot x · 1 dx x −ln5 cotx = xln5

y = log2 sec x 1 dy 1 ln2 secx = 1 · dx = ln2

x ln5

31.

ln

4

4

Theorem14

(P2) =250,000(ln1−ln4)( 1 ) t 4 (ln1=0) 1t =−250,000(ln4) ( ) 4 b) The rate of change of the number of cans still in use, when 250, (000)cansareinitiallydistributed,is t 1 cansperyear. −250,000(ln4) 34. a)

V(t)=$5200(0.80)t

V ′(t) = $5200(ln 0.80)(0.80)t b) The value of the office machine is changing at a rate of $5200(ln 0.80)(0.80)t per year.

R ==log log I I0 105· I 0 R I0 = log 105

Substituting105 ·I0forI (P5)

=5 ·secxtanx

y = logx 3 ln3 y = lnx 1 dy = ln3 · −21 · dx − ln x x = − ln3 ln2x

30.

t

35. ·cosx

t

N′ (t)=250,000dx d ( 1 ) (41)(1)t

x ln2

·tanx 29.

N(t)=250,0004 ( 1 )

=250,000·

·cotx 28.

g(x) = 2x log2 x) 1 ln x ′ g(x)= + 2x · ·2xln2 ln2 2xx = +lnx xln2

33. a)

6x

y = csc xlog2 x dy ·1−cscxcotx·ln = csc x · 1 x dx −ln2 cscx xcscxcotxlnx = xln2

y = log10 sin x 1 dy 1 ln10 sinx = 1 · dx = ln10

32.

y = x logx 10 x ln 10 y= ln dy =x ( x −1 1 ) + ln10 · ln(10 · ln ) · dx x 12 ln10 lnx x 1− = g( )x= (logx 10)(log10 x) 1 1 g′x ()=logx10 +log10x ( ln10· −21 · ) · (· ln x ln10· 1 lnx· 1 x xln10 = + ln10 · −21 · ) lnx ln10 ln x =0

The magnitude on the Richter scale is 5.

36. R =log a) 37.

I=I

106.9· I 0 =log106.9=6.9 I0 0 ·10R

·107 Substituting7forR I=I0 =107 · I 0

b) I=I

0 ·10R

·108 Substituting8forR I=I0 =108 · I 0

c) The intensity in (b) is 10 times that in (a).

108 I0 = 10 · 107 I0 d)

I=I1R 0 0

dI = I 0 · d 10R I0isaconstant dR dR =I0 ·(ln 10)10R Theorem 14 =(I0 · ln 10)10R e) The intensity is changing at a rate of (I0 · ln 10)R. 38. I=I100.1L 0 a) I = I 100.1(100) = 1010 0 b)I=I 100.1(10) 0

·I0 =10 · I 0

Exercise Set 4.5

245 e) The pH is changing most rapidly at t = 0 which cor-

c) The intensity in (a) is 109 times that in (b).

responds to a pH of 7

1010 I0 = 109· 10 I0 dI d Ie0.1L·ln10 (100.1L=e0.1L·ln10 = ) 0 dL dL = I 0 ·0.1·ln10·e0.1L·ln10 =0.1 · I 0 ·ln10·100.1L e) The intensity is changing at a rate of 0.1 · I0 · ln 10 · 100.1L. d)

39. a)

R=log

43.

a) When t = 0 [OH−] = x = 10−7 moles/liter

c)Whent=0 d)

([H+][OH−)] = 10−14 log[H+][OH−]) = log(10−14

I

I0 −log I0 (P2) 1 1 dR Theorem16;I0isaconstant. = · I −0 = ln10 dI · 1 1

(0.002t+10−7]) = log[H+][ ( log[H+]+log (0.002t+10−7−7) =

R=logI

14+log0.002t+10

ln10 I b) The magnitude is changing at a rate of 1 ·1. ln10 I 40. a)

I =10(logI I0 1 · −0)

L=10log dL

1 1 I 10 ln10

=10 ( =

Thu s,p H =14+log0.002t+107 ( e)

44.

45.

46.

T)heorem13:ax=exlna (exlnx x·1+1·lnxexlnx

=

0001 . −1 ln10· 0.001t+10− 7

d) When t = 0, x = 10−7 thus, the pH is

x

(eelnx )Theorem13:ax=exlna x·1 f′(x)= ex lnx x +e lnxe e x 1 ex = ex ( +lnxx ) Substituting xex for exelnx x 1 ex = e x ( )

48.

y = af(x)

49.

dy = f′(x)(ln a)af(x) dx y = loga f(x), f(x) > 0

c) d log0.001t+10−7)] ( dt [ −

·d(x2+1) dx

f(x)= exx , x>0 f(x) =

− log[H+]

b) x = [H+] Therefore, [H+] = 0.001t + 10−7

1

2x (ln 3)(x2 + 1) 47.

Substituting xx for ex ln x

y=log x23( +1) 1 d y = ln3 · d x =

−7 = log[H+] 10−7 = [H+]

− log(10−7) = 7

y=xx,x>0

x dx = (1 + ln x)xx

mandbareconstants

7 =

−)

y = 2x4

dy=

b) The response is changing at a rate of m 1 . · ln10 x 42. a)

−log[H+]

0.002 d [OH−]= dt ln10(0.002t+10−7

y=

y=mlogx+b

Theorem16

−14

x4 d x4 d y =( ln2)2 · d x =43x(ln2)2 x4

·

dy =m · d logx+0 dx (dx1 1) =m ln10 · m 1 = ln10 · x

=

−14

f) The pH is changing most rapidly at t = 0 which corresponds to a pH of 7

−log I0)

b) The loudness is changing at a rate of 10 ·1. ln10 I 41. a)

10−14 =10−7moles/liter 10−7 pH = − log107=7 ( −)

b)Whent=0[H+]=

ay=f(x) eylna=f(x)

Exponentialequation Theorem13

Differential implicitly to find dy/dx.

246

Chapter 4: Exponential and Logarithmic Functions d ylna dxe

=

4 6.fx ( )=ex+x4 dy =4ex+4x3 4 dx

d f(x) dx

dy ·lna·eylna=f′(x) dx dy ·lna·f(x)=f′(x) dx dy 50.

=

7.fx ( ) = Substitutingf(x) for ey ln a

f′(x) =

1 f′ (x) lna · f(x)

=

dx y = [f(x)]g(x), f(x) > 0

8. f (x)=ex

ln y = ln [f(x)]g(x) = g(x) ln f(x) 1 1dy ·f′(x)+g′(x)lnf(x) ydx = g(x) ·

=f(x)g(x)

[f(x) g(x)·

f′x() +g′(x)lnf(x) f(x)

·ln 4x f′(x) = =

]

51. Since ax can be written as ex·ln a, we can find the derivative of f(x) = ax using the rule for differentiating an exponen- tial function, base e.

52. Ifwefirstwritelog a x as alogaxwehave alogax=x.Then we can take the natural logarithm on both sides of this ex1 pression, use P3, and write loga x = lna ·lnx.Thisallowsustofind d loga usingtherulesfordifferentiating x dx f(x) = c ·ln x.

Chapter Review Exercises

2

=

Differentia ]tingimplicity f′x()+g′(x)lnf(x)

dy dx =yg(x)· [

lnx tanx

2 4 x2 + 2 xex ·ln4x e · 4x x2 e +2x ex2 ln4x x ex2(2 x2 ln 4 +1)x x

x

9. f (x)= 4xe −

4e4x f ′ ( x )= = 4e4x

1 1 − x ·4 4

1 − x 10. g( )=x8 x g′x ()=8x 11.

y=

−8lnx 7 −8 =x

lnex x = ex ex

8 8(x−1) x ex

dy

= = dx =

1. y=lnx dy 1 dx = x 2. y=ex dy dx =ex

·1−x·ex (ex)2 ex(1 − x ) (ex )2 −x) (1 ex =e−x(1 − x )

12.

3. y = ln (x4 +5)

loga 14 = loga (2 d y d x

= =

1 ·4x3 x4 +5 4x3 x4 + 5

√

4. y = e2 x d y d x 5. y=ln(

1

tanx · x − lnx · sec2 x tan2 x cotx csc2x ln x − x

√

=

e2x ·

=

e2x √x

√

·7) = loga 2+loga 7 = 1.8301+5.0999 = 6.93

13. loga

2 2√ x

sinx + x )

=

1

+1 sinx + x · cos x cosx + 1 sinx + x

= log a2 − loga7 = 1.8301 − 5.0999 = −3.2698

14. loga28

dy = dx

2 7

=

2

loga(2 ·7) a 2 2+loga 7 = log = 2loga 2+loga 7 = 2(1.8301)+5.0999 = 8.7601

247

Chapter Review Exercises 15.

22.

a)

7 2

loga35. = loga = loga 7

dN dt dN N

− loga 2 − 1.8301

= 5.0999

lnN = 0.12t+C N = e0.12t+C

= 3.2698

16.

N = N e0.12t 0

60 = N e0 0

√ 1/2 loga 7 = loga7 1 = = 2 loga7 1 2 ·5.0999

60 = N0 Thus, N = 60 e0.12t b) When t = 10, y = 60 e1.2 = 199

= 2.54995

c) The time to double is T = 23. T =

loga 2− 2

ln(1/2) =5.3years 0.13

24.

−2 loga 2 −2(1.8301) −3.6602

25.

ln(1/2) =0.182=18.2% 3.8

17. loga

1 4

= = =

k=

dA = dt = dA = A

18. dQ

= = kQ kdt dt dQ Q integrating ln Q = kt+C C constant Q = ekt+C Q = ekt Let eC = Q0 Q =

20.

21.

T =

−0.07A −0.07dt −0.07t + C

lnA A = e−0.07t+C A = A e−0.07t 0 800 = A e0 0 800 = A0 Thus, A = 800 e−0.07t b) When t = 20, y = 800 e−1.4 = 197 grams

·eC

Q0e

ln(2) =5.8 years 0.12

a)

=

19. k =

= 0.12N = 0.12dt

kt

ln(2) =0.043=4.3% 16 ln(2) =10.2 0.068

a) C = 4.65 ekt. When t = 43, C = 25.38 25.38 = 4.65e43k 25.38 = e43k ( 4.65) 43k 25.38 = ln k (4.65 4.65

43 ln25.38) 0.0395 = k Therefore C = 4.65 e0.0395t b) When t = 53, y = 4.65 e0.0395(53) = 37.73 When t = 58, y = 4.65 e0.0395(58) = 45.96

ln(1/2) − 0.07 =9.9days

c) The half-life is T = 26.

a)Whent=1,y=1− 4−0.7(1) = 0.5 − 4−0.7(2) == 0.97 0.75 − 4−0.7(5) Whent=2,y=1 Whent=10,y=1 Whent=14,y=1 − Whent=5,y=1 − 4−0.7(10) 4−0.7(14) = = 0.9991 0.99994 p′(t) = 0.7 e−0.7t b) p′(t) represents the rate of change of the probability c) of mastering a certain assembly line task with respect

d) to1the number of learning trials 0.8 0.6 0.4 0.2

0

2

4

6

8

10 t

12

14

16

18

20

Chapter 4: Exponential and Logarithmic Functions

248

27.

dy =3xln3 dx

d) 6200000000 = 984e0.0387t 6200000000 = e0.0387t 984 6200000000) = 0.0387t ln ( (984 ln 6200000000) = t 984 0.0387 404.6 = t

1 28. dy = dx xln5 29. dy

( e2x− e−2x)(2e2x − 2e−2x) − (e2x −−2x2 e ) ( e2x + e−2x)(2e2x + 2e−2x) (e2x − e−2x)2 2e4x + 2e−4x− 4 − 2e4x − 2e−4x − 4 (e2x − e−2x)2 −8 ( e2x− e−2x)2

=

dx

= = =

−8 2x−1)2 22x (e −8e4x (e4x −1)2

=

e) The model obtained does not seem to be realistic.

From the answer of part (d) we would expect 6.2 billion reported cases of SARS within 405 days after March 17, 2003 which corresponds to the middle of February of 2004, and we know that was not the case.

Chapter 4 Test 1. y=ex

30. f′(x) = x4

x

·

1

+4x3·ln

x = x3(1+4lnx) f′(x) = 0 (1+4lnx) = 0 1+4lnx = 0

3

lnx

=

dy

x

−1 4

31.

1/44

)

dy = dx = 3. f (x) = e − x

x = e−1/4 When x = e− 1/4, f(x)=e ( −

=ex dx 2. y = (tan x) ln x

2

f ′ ( x )=

lne−1/4=− 4 1e

0.25

1 sec2x· ln x + x ln x sec2 x+ tanx x

tanx ·

−x2

e

= 4. f (x)=ln

·17 f ′ ( x )=

0.15

= 0.1

–10 –8

–6

–4

–2

5. f (x) = ex − 5x3

2

4 x 6

8

10

e1/x =0 32. lim →0(1+e1 x /x)2 33.

0 0387 t

a) y = 984(1.039)t = 984 e .

Exponential growth rate is k = 3.87% b) When t = 100, y = 984 e . 0 0387(100) = 47110 When t = 200, y = 984 e .0 0387(200) = 2254840 ln(2) c)T= =17.9days 0.0387

x2

x 7

0.2

0.05

· − 2x

2−xe−

7 x 1 x

f ′ x( ) = ex − 15x2 6. f (x) = 3 ex lnx f ′ ( x )= =

3ex · 1 +3ex·ln

x

3ex(xln x+1) x

7. y = ln(e x − sinx ) dy dx

= =

1 · ( e x − cos x) ex − sinx ex− cosx ex− sinx

249

Chapter 4 Test

8.y=

lnx

b) When t = 17, y = 0.748 e0.0331(17) = $1.31 When t = 27, y = 0.748 e0.0331(27) = $1.83

ex dy dx = =

ex · 1 x − l n x · e x (ex)2 (1 −xlnx) xex

c) 16.

Answers vary

a) dA dt = = dA = A

9. logb 18 = logb 2

· 9 = logb 2+logb 9 = 0.2560+0.8114 = 1.0674

10. logb 4.5 = logb = logb 9

11. logb3

9 2

−0.1A −0.1dt −0.1t + C

lnA A = e−0.1t+C A = A e−0.1t 0 3 = A 0 e0 3 =

− 2 − logb 0.2560

A0

Thus, A = 3 e−0.1t

= 0.8114

b) When t = 10, y = 3 e−1 = 1.1 cc

= 0.5554

c) The half-life is T =

√ = logb 9 = log 91/2 b 1 = 2 ·logb 9 = 1 ·0.8114 2

17. T =

ln(1/2) =83.5days 0.0083

18.

k=

19.

ln(1/2) =0.0000003=0.00003%peryear 2300000

a) When t = 0

= 0.4057

100 P(0) = = 1 + 24e(0) = 100 25 4%

12. dM

= = kM kdt dt dM M integrating lnM = kt+C C constant M = ekt+C M = ekt Let eC = M0 M = M ekt0

100 −028=5.22% 1+24e .

100 −028(5)=14.45% .

1+24e 100 When tt = 10, P(10) = −028(10)=40.66% .

When = 15, P(15) = 1+24e When tt = 20, P(20) = When = 30, P(30) = When t = 35, P(35) = 100 −028(15)=73.54% 1+24e .

100 −028(20)=91.85% 1+24e .

100 −028(30)=99.46% 1+24e .

100 −028(35)=99.87% 1+24e .

b) When When tt = = 5, 1, P(5) P(1) = =

·eC

ln(2) =0.173=17.3% 4 14. T = ln(2) =10years 0.6931 15. a) C = 0.748 ekt. When t = 10, C = 1.042 13. k =

1.042 = 0.748e10k 1.042 = e10k 0.74)8 1.042 = 10k ln ( 0 (.748 .0.748 = k ln1042) 0.0331 = k Therefore C = 0.748 e0.0331t

ln(1/2) =6.93hours − 0.1

c) ′ P(t)

= =

e−0.28)t−2·24e−0.28t·−0.28 ( − 1001+24 672e−0.28t (1 + 24e−0.28t)2

d) P′(t) represents the rate of change of the percentage of people who buy products with respect to the number of times the ad is run

250

Chapter 4: Exponential and Logarithmic Functions e) 100

24.

1

80

0.5 60

–4

40

0

–2

2 x

4

–0.5

20

0

20 t

10

30

40

20. f(x) = 20x

ex − e − x =0 25. lim x →0 ex + e−x

f′(x) = 20x ln(20)

26.

21. y = log20 x 1 ·1 dy = = ln(20) x dx 22. y = x (ln x)2

–1

a) y = 0.027(19.2)t = 0.027 e2.96t b) When t = 6, y = 1352605 When t = 7, y = 25970025

1

c) 6300000000 = 0.027 e2.96t 6300000000 = e2.96 t 0.027) 6300000000 ln ( = 2.96t (0.027 ln6300000000) = 0.027 t 2.96 8.86 = t

−2xlnx+2x dy = (lnx)2+2xlnx dx − 2lnx−2x·

· 1

1 x

+2

= (lnx)2 23. f(x) = x4 e−x

d) T = ln(2)/2.96 = 0.2342 hours

f′(x) = 4x3

· e−x + x4 · −e−x = e−x(4x3 − x 4 )

Technology Connection

Solving f′(x) = 0 we get 4x3 x3(4

−x4 −x) = = 0 0 0 x = 4 or = x

Using test values to determine the sign of the first derivative we get: f′(x) < 0 on ( −∞, 0] f′(x) > 0 on [0, 4] f′(x) < 0 on [4, ∞ ) Thus we have a relative minimum at x = 0 and f(0) = (0)4 =

·e0 0

Relative minimum at (0,0) And a relative maximum at x=4 f(4) = (4)4

·e−4 = 256e−4 ≈ 4.689

• Page 265: =156.9925453 1. 5π √ 3 2. 5 =16.24245082 −√2

3. 7

=0.0638044384

π 4. 18−=0.000138793867

• Page 282: f (3) = 1000 f (0 .699) = 5.0003 g(5) = 0.69897 g(1000) = 3 • Page 285: 1.t

= 6.90776

2.t = 4.09435 − 3.t = 74.89331 4.t = 64.21959 5.t = 38.74448 • Page 303: 1.y

= 8.2432

2.y = 9.638

251

Extended Technology Application 3. y = 11.269

4.

a)

4. y = 13.176

800

5. y = 30187.92(1.04686554x) 600

y = 30187.92 e0.0458004996x 6. 75449, 149974, 298111

400

200

Extended Technology Application 1.

0

a) 50

100

200

P 300

400

500

b-c) P0 = 400, H (400) = 400

40

5.

a)

10

30

5 20 10

–10 –8

0

b-c)

2.a)

10

20

P

30

40

f′P ( ) = 1 → P0 = 1 P H (1) = ln(20) − 1 = 1.9957

4 P 6

8

10

–10

6.a) y = −0.0011083333x3 + 0.0715357143x2 0.0338095238x + 4 b)

140

100

300

80

200

40 20

100 –20 0 –100

0

10

20

P

30

40

50

–400

c) P0 = 33.841447

50 40 30 20 10

0

10

20

P

30

40

50

1+ √ 5

2P+1 (→P0=√ 2 √ √ P 1+52 )(P+1) (PH =ln [4+45][ 3+ 2 5 ]) −

)=

–200 –300

f′P ( )= 0.05P+4 →P0=60 − H (60)= 0.025(60)2+4(60) −60=90 −

f′(

2

b-c) P0 = 0.428, H(0.428) = 1.0834

–40

b-c)

–2 –5

60

3.a)

–4

50

120

b-c)

–6

√1+5 2

= 1.905

20

40 P 60

80

100

−

Chapter 5

Integration Exercise Set 5.1 (t2 ∫ − 2 t ∫+3)dt∫ = t2dt− 3dt ∫ = Theintegralofasumisthe t sum of the integrals.

3 2 −2·t t C 3 2[+∫3+ xr+1 xrdx= + ] r+1

9.

1.

x6dx ∫ x6+1 + C = 6+1 = x7 7 + C 7

2. ∫ xdx =

x8 8

( ∫ xrdx =

r +1

+ C)

(Forkaconstant,

+ C

2 dx ∫ + =2x C

3. ∫

rr+1

= 10.

(For k aconstant, kdx = kx + )C. ∫

4.

4 dx = 4 x + C

5.

x1/4 dxC

∫ 1/4+1 x + = 1 4 +1

11.

∫

( ∫ xrdx =

r +1

+ C)

7.

dx =

x

=

=

3

+

x2

4/3

4 3

2 −

b ( ∫ beaxdx= eax+C ) 3 e5x+C

a

5

∫(w3−w8∫/7)dw x3dx− x8/7dx

Theintegralofasumisthe sum of t[he∫integrals. r+1 w4 − w 8/7+1 xrdx= x = +C +C ] 8 r+1 +1 = 4 =

+ C=

( + x−∫ 1) dx−∫ ∫x2 x2dx + xdx = x3

5 e8x+C 8

13.

x5/4 + C 5 4 4 x5/4 + C = 5 x1/3

+t2+3t+C

12. ∫ 3e5xdx=

=

6. ∫

3

(3t2 ∫−4t+7)dt=t3−2t2+7t+C 5e8xdx =

xr+1

t3

∫ Ckdx=kx+C.)

+

3 x4/3 1

+ C

dx The integral of a sum is the sum of the integrals.

DON’TFORGETTHE [x∫C← xr+1 xrdx + C] = r +1

(For k aconstant, 8. ∫ (x2 − x +2) dx =

3

2

∫

kdx = kx + C.

w4 15/7 +C 4 −w 15 7 w4 4 −7w15/7+C 15 14.

t5 t11/5 (t4−t6/5)dt= − 5 +C

∫

C!

=

)

∫ 15.

t5 5 t11/5+C 5 − 11

1000r ∫dr1

x −x +2 x + C 3 2

=1000

r

dr

=1000lnr+C ∫ 16.

Theintegralofaconstant ∫constanttimestheintegral.

1 dx=lnx+C,x>0; wegenerallyconsiderr>0

500dx=500lnx+C x

Chapter 5:

254 17.

18.

19.

1 dx −2 ∫ x2= ∫ 2dx= ∫ x dx x x−2+1 = +C [ ∫ xrdx= = − 2+1 x−1 +C −1 1 = x−1+C,or − x +C −

xr+1 +C ] r+1

dx 1 x− 2 −3 +C=−x−+C,or 2 ∫ x dx= 3 = +C 2 2 x − − 1 2x 2√ ∫ // sds= s12ds ∫ s1 2+1 +C = 1 2 +1 3/2

s

3 2

x5/3 ∫ x2/3dx= 5 +C= 3

20. ∫ √ 3x 2dx=

21.

− 6·

x−2/3+1 +C 2 3 + 1 −

x1/3 = − 6· 1 + C 3 = 6− 3x1/3+C · = 18x1/3+C −

23.

√20 dx= 20x−4/5dx ∫ ∫ 5x4 x1/5 =20 · +C 1 5 = 100x1/5 + C ∫ ∫

24.

5

6 ∫ √ − 6 3dx= ∫ − dx= ∫ −6x−2/3dx x2 x ∫2/3 = − 6 x−2/3dx =

22.

3 x5/3+C

8 e−2x+C −2 = 4e−2x+C −

8e−2xdx=

7 e−0.25x+C 0.25 − = 28e−0.25x+C −

7e−0.25xdx=

x

−4/3

) dx

x3 − 3 x3/2 x−1/3 + C · 3 + 1 3 −3 2 = x3 x3/2 − 3x−1/3 + C − ∫3( x−2/5 x4 + √ 1 − 4 ) dx 8 x 5 ∫ x4 ) x−1/2 x−2/5 dx = ( 5 = x5 1 · x1/2 4 x 3/5 + 8 5· 3 + C − 1 1 5 − 4 2 5 x5 4 x3/5 + C + x1/2 = 3−5 5 5 sin 2θdθ = cos2+ θ C ∫ 2 ∫ 1 1 cos2θdθ = sin 2+ θ C 4 8 2 x −4 cos x dx ∫ ( 5 sin 5 ) =

26.

+C

2 / = s32+C 3

− 3 √ +x 2

x2dx − 3 = ∫ ∫ x1/2dx + ∫ x−4/3dx 2 = / x2+1 3 x1/2+1 x−43+1 + + C − ·1 4 2+1 2 2 +1 +1 3 −

∫

=

∫ ( x2

25.

Integration

27.

28.

29.

2 = −cos 5x − 2 sin x + C

30. ∫ 3 cos2 πx− (

8

sinπx ) dx =

31.

3 xdx= tan3x + C ∫ 3 sec2

32.

∫ 5 csc22xdx = − 5cot 2x + C

33.

∫ 1

34.

3

35.

36.

37.

38.

8 3 sin 2πx + cosπx + C 2π π

x 2 x x sec tan dx =3 sec + C 9 9 9

=2 sec2x∫ ( +C sec xtan x ∫(4 2 2 )dx + secxtanxdx ) + + tanx secx C =∫( cscx 2 + cotxcscxdx = cscx + cscxcotxdx ) ) = −cotx−cotx + C ∫[ 1−1 1 1 1 + + + C dt =ln t− ] t2 et t t et ) dw= 4 e7w+ 10w3/5+ e7w w−2/5+7 w100 − ∫(4 7 w101 + C 101 ∫(

secx + tanxsecxdx =

39. Find the function f such that f′(x) = x − 3 and f(2) = 9.

We first find f(x) by integrating.

Exercise Set 5.1

255

f(x)= (x

−3) ∫ ∫dx = xdx− 3dx ∫ = x2 −3x+C 2 The condition f(2) = 9 allows us to find C. x2 f(x) f(2) = = 2 −3x+C 22

−3·2+C=9 −6+C=9

2 2

Substituting2forx and 9 for f(2)

C=13 Thus, f(x) = 40. f(x)= f(1) =

x2 2

− 3x + 13. x2

∫ (x−5)dx= 12 2

−5·1+C=6 C =

f(x) =

−5x+C

x2 − 21 5x+ 2

21 2

Substituting0forx and 7 for f(0)

−4·0+C=7

Solving for C we get C = 7. x3 Thus, f(x) = 3 −4x+7.

42. f(x)= f(0)=

f(x)=

∫ (x2+1)dx= 03 3 x3 ∫3

43. f(x)= f(0)=

f(x)=

−3 cos8x+C 8

cos(0)+C=3

−38+C=1 11 C = 8 − 3 cos8x+ 11 f(x)= 8 8 5 f(x)= 5e2x dx= e2x+C 45. ∫ 2 5 f(0)= +C= −10 2 C= −25 2 5 e2x−25 f(x)= 2 2 ∫ 46. f(x)= 3e−xdx= f(0) = −3 e−x + C f(x) = −3+C=6 C=9

x3

157t2+1000t+C 2

f(0)=0+0+C 156239 = C f(t)= f(2)=

157t2+1000t+156239 2 (2)2+1000(2)+156239 157 2 ∫=158553

48. f(t)=

500+10t+0.3t3=500t+5t2+0.075t4

f(0)=0+0+C 452937 = C f(t) = 452937 + 500t + 5t2 + 0.075t4 f(4) = 452937 + 500(4) + 5(4)2 + 0.075(4)4

+x+C

+0+C=8 C=8 +x+8

2 sin3x+C 2cos3xdx= sin(0)+C=1 2 C=1 3 2 3 sin3x+1

−3 8

47. f(t)= 157t+1000dt= ∫

f′(x) = x2 − 4 and f(0) = 7.

We first fi ∫ndf(x)byintegrating. f(x)= ∫(x2−4)∫dx = x2dx− 4dx = x3 −4x+C 3 The condition f(0) = 7 allows us to find C. x3 f(x) f(0) = = 3 −4x+C 3

f(0)=

∫ 3sin8xdx=

−3 e−x + 9

41. Find the function f such that

03

44. f(x)=

= 455036 s(0)=4 49. v(t)=3t2, We find ∫s(t)byintegratingv(t). s(t)= ∫v(t)dt 3t2dt = t3 =3 · 3 +C =t3+C The condition s(0) = 4 allows us to find C. s(0) = 03 + C = 4 Substituting 0 for t and 4 for s(0)

+ C

256

Chapter 5: 54. v(t)= (

Solving for C, we get C = 4.

Thus, s(t) = t3 +4.

50. s(t)=

v(0) = v(t) =

2tdt=t2+C ∫ s(0)=02+C=10

−3t2 + 7t + 10) dt ∫ 7+ t2+10t+C =− 2 − t033

v(0)=20

7+

s(0) =

·02+10·0+C

∫v(t)byintegratinga(t).

2 =20

C2 = 20

∫a(t)dt = = 44tdtt2 · 2 +C =2t2+C

s(t) = − t 3 55.

7+ t2+10t+20

a(t) = −32 ft/sec2 v(0) = initial velocity = v0

The condition v(0) = 20 allows us to find C. v(0) = 2 · 02 + C = 20 Substituting 0 for t and 20 for v(0) Solving for C, we get C = 20.

Thus,v(t) =2 t2 +20.

52. v(t)=

−3·02+7·0+C1=10 C1 = 10

−3t2 + 7t + 10

s(t) = t2 + 10

v(t)=

−6t + 7) dt = −3t2 + 7t + C1

s(t)= (

C=10 51. a(t)=4t, We find

∫

Integration

s(0) = initial height = s0 We find ∫v(t)byintegratinga(t). v(t)= ∫a(t)dt (−32)dt = =−32t+C1 The condition v(0) = v0 allows us to find C1.

6tdt=3t2+C v(0)=3 ·02+C=30 v(t) = 3t2C=30 + 30 ∫

v(0) = −32 · 0 + C1 = v0 C1 = v0

Substituting 0 for t and v0 for v(0)

Thus, v(t) = −32t + v0. v(0)=6 53. a(t) = −2t+6, We find ∫v(t)byintegratinga(t). v(t)= ∫a(t)dt = (−2t+6)dt

ands(0)=10

We find ∫s(t)byintegratingv(t). s(t)= ∫v(t)dt = =−16t2+v0t+C2 (−32t+v0)dt v0isconstant The condition s(0) = s0 allows us to find C2.

The condition v(0) = 6 allows us to find C1. v(0) = −02 + 6 · 0 + C1 = 6 Substituting 0 for t and 6 for v(0) Solving for C1, we get C1 = 6.

Thus, v(t) = −t2 + 6t + 6.

We find ∫s(t)byintegratingv(t). s(t)= ∫v(t)dt = (−t2+6t+6)dt = t3 2 − 3 +3t2+6t+C

The condition s(0) = 10 allows us to find C2.

03 s(0) = − 3 +3·02+6·0+C2Substituting0fort and 10 for s(0) Solving for C2, we get C2 = 10.

Thus, s(t) = −1t3+3t2+6t+10.

3

s(0) = −16 · 02 + v0 · 0 + C2 = s0 Substituting 0 for t and s0 for s(0) C2 = s0 Thus,st ∫()=−16

t2 +v0t+s0.

56. v(t)=

−32dt=−32t+C1

v(0)=

−32·0+C1=80 C1 = 80

v(t) = s(t)=

∫−32t+80 (−32t+80)dt=−16t2+80t+C2

s(0)= s(t) =

−16t·02+80·0+C2=10 C2 = 10

−16t2 + 80t + 10 Solve −16t2+ t ±√80+10=0.

10 110 t= 4 √ 10+ 110 Since t > 0, t = ≈5.1sec.

4

257

Exercise Set 5.1 57.

a(t) = v(t)= s(t)=

∫k a(t)dt= ∫

t2

v(t)dt= ∫ ktdt=k·

132= (v(0)=132) =

1 kt2

(s(0) = 0; thus C = 0.)

We know that

= s(t) =

−68.5dt=−68.5t+C

∫0+C→C=∫102.7 v(t)dt=

−68.5t+132dt

The time it takes to go from 132ft/sec to 0 is

1 t= min.

2

−68.5t+132 = 0 68.5t = 132

1 s(t)= kt2 2

·60mph s( 1 min) = 1 1 ·( 2 min = 1 2 = = = 1 ·60mi·1min hr 2 2 ·60mi· 1

1

132 68.5 = 1.927

t =

min) 2

hr

60 mi 240 1 mi 4 1 Thecartravels miduringthattime.

4

Thus s(1.927) = −34.25(1.927)2 + 132(1.927) = 127.182 The car travels almost 127.2 feet during that time.

60. A′(t) = −43.4t−2 a) A(t) = 43.4t−1+C

a(t) = − ∫68.5 a(t)dt= −68.5dt=−68.5t+C ∫ v(t)= (v(0)=102.7) 102.7 = s(t)= ∫0+C→C=∫102.7 = s(t) =

∫

−34.25t2 + 132t + C (s(0)=0;thus C = 0) −34.25t2 + 132t

and that

58.

s(t)= 59.

60 mph 1 2 min

a(t) = k =

Thus

a(t) = − ∫68.5 v(t)= a(t)dt=

C∫onstantacceleration kdt=kt (v(0)=0;thusC=0.)

v(t)dt=

39.7 = 43.4(1)−1+C −3.7 = C Thus, A(t) = 43.4t−1 −3.7 b) A(7) = 43.4(7)−1

−68.5t+102.7dt

−34.25t2 + 102.7t + C (s(0)=0;thus C = 0) −34.25t2 + 102.7t

The time it takes to go from 102.7ft/sec to 0 is

=

2.5

−3.7

61. M′(t) = 0.2t − 0.003t2 a) We integr ∫atetofindM(t).

M(t)= (0.2t−0.003t2)dt =0.1t2−0.001t3+C

−68.5t+102.7 = 0 68.5t = 102.7 102.7 = t 68.5 = 1.499

We use M(0) = 0 to find C. M(0) = 0.1(0)2 − 0.001(0)3 + C = 0 C=0 M(t) = 0.1t2 − 0.001t3 b) M(8)=0.1(8)2

Thus s(1.499) = −34.25(1.499)2 + 102.7(1.499) = 76.98 The car travels almost 77 feet during that time.

− 0.001(8)3 =6.4 − 0.512 = 5.888

≈ 6 words 62. I′(t) = 3.389e0.1049t

258

Chapter 5: 66.

a) I(t)

−32.307 = C I(t) = 32.307e0.1049t

67. − 32.307 32.307 −

2 3 ∫ ( x − 1) x dx = =

I(27) = 32.307e0.1049(27)

∫ (x5 − 2x4 + x3) dx 4 x6 − 2 x5+ x +C 6 5 4 ∫

516 68.

c)

2

√t ∫ (1 − ) tdt =

I(34) = 32.307e0.1049(34) − 32.307 = 1111

3 =

d) 1111− 516 = 595 63. N ′ ( t) = 38.2e0.0376t

∫ N(t)

∫ 38.2e0.0376tdt

=

(t + 3) 2 √t

t3/2−2t5/2+C 5

∫ (t3/2+6t1/2+9t−1/2)dt

dt = =

2 t5/2+4t3/2+18t1/2+C 5

= 1015.957e0.0376t+C 1622 = 1015.957+C

70.

606.043 = C N(t) = 1015.957e0.0376t+606.043

∫

b)

x4 − 6x2 − 7dx = x3 =

N(30) = 1015.957e0.0376(30)+606.043 c) N(60) = 1015.957e0.0376(60)+606.043 64. f ′(t) = t1/2 + t−1/2 2 3 t3/2+2t1/2+C (4)3/2+2(4)1/2+C 2 3

0= =

f(t) =

C 2 3

t3/2+2t1/2

= =

−

∫(t3+3t2+3t+1)dt

=

t4 3 4 +t3+ t2+t+C 2

28 3

∫

7 x−2+ C 2

√ 3+1

= C f(t) = t √ 3+1 √ 3+1 +8

1 dx ∫ ln(x) ln(10

ax ∫ be dx =

b∫ e

=

√

t 3+1 +C √

1 dx = ln(10) x =

+C

73.

∫ t3dt

8 = 0+C 8

dx

72.

√

f(t)

x2 2 −6ln(x)+

∫ (t + 1)3 dt =

= 10294

f(t) =

∫ (x−6x−1−7x−3)

71.

= 3742

−28 3

(t1/2− t3/2) dt

69.

a)

65. f (t)=t3

25 t3+20t2+16t+C 3

=

b) =

∫ (25t2+40t+16)dt

∫ (5t + 4)2 dt =

∫ 3.389e0.1049tdt

=

= 32.307e0.1049t+C 0 = 32.307+C

′

Integration

ax

dx

b ax e + C a

74. ∫ (3x − 5)(2x + 1) dx = ∫ (6x2+3 x − 5) dx = 2x3+ 3 x2 5 + 2 − x C

259

Exercise Set 5.2 75.

Exercise Set 5.2

√3 6 ∫ x4/3dx

∫ √ 364x4dx =

1. a) f(x)=

4 = 4 · x7/3+C 3 12x7/3+C = 77

In the drawing in the text the interval [1,7] has be (endividedinto6subintervals,eachhavingwidth 7−1 1 ∆x= =1. ) Theheightsoftherectanglesshownare 1 f(1)= =1 12 1 1 f(2) = 2= =0.2500 f(3) = 2 4 f(4) = f(5) f(6) == 1 1 2= 9 ≈0.1111 3

76. ∫

x2 − 1dx x+1

= =

∫ (x−1)dx x2 2

−x+C

77. ∫

t3t+8 dt = +2 = =

∫ ∫ t3

(t+2)(t2−2t+4) (t+2) (t2−2t+4)dt −t2+4t+C

dt

3 78. =

∫ cosxtanxdx

=

1 x2

∫ sinxdx −cos x + C

1 1 2= =0.0625 4 16 =0.0400 ≈0.0278 1 1 2= 5 25 1 1 2= 36 6 The area of the region under the curve over [1,7] is approximately the sum of the areas of the 6 rectangles.

Area of each rectangle: 1st rectangle: 1 · 1 = 1 [f(1) = 1 and ∆x = 1] 2nd rectangle: 0.2500· 1 = 0.2500

79.

3

∫ (cosx + cosxsin2x)dx

= =

∫ cosxcos2x + sin2x)] dx ∫[ ( cosxdx

= sinx+C

[f(2) = 0.2500 and ∆x = 1] 3rd rectangle: 0.1111· 1 = 0.1111 ·· 1 1= = 0.0625 0.0400 4th rectangle: 0.0625· 1 = 0.0278 5th rectangle: 0.0400 6th rectangle: 0.0278

80. ∫ tan23xdx = =

∫ (sec23x−1)dx

∫ cot22xdx =

∫ (csc22x−1)dx

tan 3x 3 −x+C

81.

=

−cot2x 2 −x+C

82. Answers could vary. The integral is not unique since the student forgot to mention the constant of integration.

83. Answers could vary. The antiderivative of a function represents the area under the curve of that function.

The total area is 1+0.2500+0.1111+0.0625+ 0.0400+ b) f(x) = 0.0278, or 1.4914.

1 interval [1,7] has been d The (ividedinto12subinter7−1 6 x2 ∆x= = = vals ),eachhavingwidth0.5 12 12 0.5.Theheightsofsixoftherectangleswerecomputed in part (a). The others are computed below:

260

Chapter 5: 1 f(1.5) = 2= f(2.5) = (15) . f(3.5) f(4.5) = = f(5.5) = 1 f(6.5) = 2= (2.5) 1 2= (3.5) 1 2= (4.5) 1 2= (5.5) 1 2= (6.5)

1 2 . ≈0.4444 2 5 ≈0.1600 1 6.25

1st rectangle: 1(0.5) = 0.5 [f(1) = 1 and ∆x = 0.5] 2nd rectangle: 0.4444(0.5) = 0.2222 [f (1.5) ≈ 0.4444 and ∆x = 0.5] 3rd rectangle: 0.2500(0.5) = 0.1250 4th rectangle: 0.1600(0.5) = 0.0800 5th rectangle: 0.1111(0.5) ≈ 0.0556 6th rectangle: 0.0816(0.5) = 0.0408 7th rectangle: 0.0625(0.5) ≈ 0.0313 8th rectangle: 0.0494(0.5) = 0.0247 9th rectangle: 0.0400(0.5) = 0.0200 10th rectangle: 0.0331(0.5) ≈ 0.0166 11th rectangle: 0.0278(0.5) = 0.0139 12th rectangle: 0.0237(0.5) ≈ 0.0119 The total area is 0.5+0.2222+0.1250+0.0800+ 0.0556 + 0.0408 + 0.0313 + 0.0247 + 0.0200 + 0.0166 + 0.0139 + 0.0119, or 1.1420. (Answers may vary slightly depending on when rounding was done.)

2. a)∆

5− 0

b)∆x=

=0.5 10 10f(xi)∆x=f(0)·0.5+f(0.5)·0.5+

∑

f(1)·0.5+f(1.5)·0.5+f(2)·0.5+ f (2.5) · 0.5 + f(3) · 0.5+ · 0.5 + f(4) · 0.5+ · + f(5) · 0.5 f (3.5) f0.5 (4.5)

1 ≈0.0494 12.25≈0.0816 ≈0.0331 ≈0.0237 1 20.25 1 30.25 1 42.25

The area of the region under the curve over [1, 7] is approximately the sum of the areas of the 12 rectangles.

Area of each rectangle:

= (02 + 1) · 0.5 + (0.52 + 1) · 0.5+ · 0.5 + + 1) · 0.5+ · 0.5 + (1.52 (2.52 1) 0.5+ 0.5 (3.52 ++ +1) 1)· ··0.5 0.5+ ·· 0.5 ++ (4.52 (12 + 1) (22 + 1) (32 + 1) = 40.625 (42 + 1) an antiderivative. It also 3. The shaded region represents represents velocity, the antiderivative of acceleration.

4. The shaded region represents an antiderivative. It also represents distance, the antiderivative of velocity.

The shaded region represents an antiderivative. It also 6. represents total energy used in time t.

The shaded region represents an antiderivative. It also 7. represents the total number of divorces in time t.

The shaded region represents an antiderivative. It also 8. represents the amount of the drug in the blood.

9. The shaded region represents an antiderivative. It also represents total sales.

10. The shaded region represents an antiderivative. It also represents the number of words memorized in time t.

11. The shaded region represents an antiderivative. It also represents the number of pages typed in t hours. ∆x=2−0=1 5.

∫

4

0

f (2)(1/2) = (1/4)(1/2)+(1)(1/2)+(9/4)(1/2)+(4)(1/2) = 1/8+1/2+9/8+2 = 3.75

f(3) ·1+f(4)·1

= (02 + 1) ·1+(12+1)·1+ ·1+ (32+1)·1+ ·1

=1+2+5+10+17 (42 + 1) =35

2

2x2dx = f(1/2)(1/2)+f(1)(1/2)+f(3/2)(1/2)+

∑x=15f(xi)∆x=f(0)·1+f(1)·1+f(2)·1+

(22 + 1)

Integration

∆x = 1−(−1) =1 4

12. ∫

1x2dx −1

2

=

f(−1/2)(1/2)+f(0)(1/2)+f(1/2)(1/2)+ f (1)(1/2)

= (1/4)(1/2)+(0)(1/2)+(1/4)(1/2)+(1)(1/2) = 1/8+0+1/8+1/2 = 0.75

261

Exercise Set 5.2 13.∆ =5−4 16 6= ∫x 5xdx =

20. P ′(t) = 10et

4

2

∫

= (25/6)(1/6)+(26/6)(1/6)+(27/6)(1/6)+

P ′(t) dt

P′(1/2)(1/2)+P′(1)(1/2)+P′(3/2)(1/2) +P ′(2)(1/2) + P ′(5/2)(1/2) + P ′(3)(1/2)

0

(28/6)(1/6) + (29/6)(1/6) + (5)(1/6) = 25/36+26/36+27/36+28/36+29/36+5/6

= 8.24+13.59+22.41+36.95+ 60.92 + 100.43

= 4.58333

= 242.54 ≈243

∆x=1−0=1 6

14. ∫

5xdx 4

=

6

3− 0 1 = 6 2

∆t =

f(25/6)(1/6)+f(26/6)(1/6)+f(27/6)(1/6)+ f(28/6)(1/6) + f(29/6)(1/6) + f(5)(1/6)

21. P ′(t) = −500(20 − t) f(1/6)(1/6)+f(2/6)(1/6)+f(3/6)(1/6)+ f(4/6)(1/6) + f(5/6)(1/6) + f(1)(1/6)

∆t =

= (1/6)(1/6)+(2/6)(1/6)+(3/6)(1/6)+ (4/6)(1/6) + (5/6)(1/6) + (1)(1/6)

∫

=

0

=

= 1/36+2/36+3/36+4/36+5/36+1/6 = 0.58333

∫

πsinxdx

4

=

0

( −4000)(4) + (−2000)(4) + 0 −80000 22. P ′(t) = −50t2 ∆t

sin(π/4)(π/4)+sin(2π/4)(π/4)+

sin√(3π/4)(π/4)+sin(π)(π/4)√ (1/2)(π/4)+(1)(π/4)+(1/ = 1.89612

=

=

10P′(t)dt

2)(π/4)+0

∫

=

0

=

∆x=π−0=π 4

16. ∫

πcosxdx 4 0

= =

cos(π/4)(π/4)+cos(2π/4)(π/4)+ cos( √3π/4)(π/4)+cos(π)√(π/4) (1/2)(π/4)+0+( −1/ 2)(π/4)+

bf(x)dx=0,becausethereisthesameareaabove ∫a) a ∫thex-axisasbelow.Thatis,theareaisA−A,or0. b) bf(x)dx<0,becausethereismoreareabelowthe ax a) ∫-axisthanabove.TheareaisA−2A,or−A. 18. bf(x)dx>0,becausethereismoreareaabovethe ax b) ∫-axisthanbelow.Theareais3A−A,or2A. bf(x)dx<0,becausetheareaisentirelybelowthe ax-axis.Theareais−3A.

∆t = ∫

17.

19. P′(t) = 200e−t

∫

0

2− 0 1 = 6 3 P′(1/3)(1/3)+P′(2/3)(1/3)+P′(1)(1/3)+ P ′(4/3)(1/3) + P ′(5/3)(1/3) + P ′(2)(1/3)

= 143.31(1/3)+102.68(1/3)+73.58(1/3)+ 52.72(1/3) + 37.78(1/3) + 27.07(1/3) = 145.71333 ≈146

10 − 0 =2 5 P′(2)(2)+P′(4)(2)+P′(6)(2)+ P ′(8)(2) + P ′(10)(2) −400 − 1600 − 3600 − 6400 − 10000 −22000

23. v(t) = 3t2 + 2t

(1)(π /4) = 1.11072

∆t =

P′(4)(4)+P′(8)4+P′(12)(4)+ P ′(16)(4) + P ′(20)(4) ( −8000)(4) + (−6000)(4) +

∆x=π−0=π 4

15.

20 − 0 =4 5

5− 1 =1 4 v(1)(1)+v(2)(1)+v(3)(1)+v(4)(1)

1

= 5+16+33+56 = 110

v(t) = 4t3 + 2t

24.

∆t ∫

1

= 3− 0 = 1 6 2 = v(1/2)(1/2)+v(1)(1/2)+v(3/2)(1/2)+ v(2)(1/2) + v(5/2)(1/2) + v(3)(1/2) = 0.75+3+8.25+18+33.75+57 = 120.75

25. f(x) = x. Since we are integrating over [0, 2] the length of the subintervals is given by ∆x=

2 2−0 = n n

Now, we need an expression for xi: x0=0

262

Chapter 5: x1=0+2/n=2/n x2 = 2/n + 2/n = 4/n = 2(2/n) x3 = 2(2/n) + 2/n = 3(2/n) In general, we can write xi=i(n)· ∑ n f(xi)∆x=

=

4

∑ni 2 ni=1

=

4 ·n(n+1) n2 2 2(n + 1) n

2

∫

∆x=

∑ n f(xi)∆x=

=

n

= = =

→im∞

=

n 2 →im∞2+nl n

= =

2 26. f(x) = 2x. Since we are integrating over [0,1] the length of the subintervals is given by ∆x=

1 1−0 = n n

Now, we need an expression for xi: x0=0 x1=0+1/n=1/n x2 = 1/n + 1/n = 1/n = 2(1/n) x3 = 2(1/n) + 1/n = 3(1/n) Ingeneral,wecanwritex i=i · ( n1 )

= ∑i=1n ∑i=1n 1 2i·n2 = 2 = n (n + 1)

∑n

n2i=1 i +1)n 22· ( n 2

1n

∫

0

n

∑ n f(xi)∆x

2xdx=l→im∞ (n + 1) l n = →im∞ n

11

∑i=1n 3 3 i2·n3 ∑ n i2 n3i=1 3 n(n+1)(2n+1) n3· 6 2n3 + 3n2 + n ∑2n3 n→im∞ 2n3fx(i)∆ + 3n2x+ n l →im∞ n →im∞1+ 2n3 ln 1 3 2n+ 2n2 1

28. f(x) = x2. Since we are integrating over [0,3] the length of the subintervals is given by ∆x=

11

=

=

0

3x2dx=l

∑i=1n

= = =

11 f(xi)∆x= ∑ n f(i·)nn

∑n

1

∫

11

∑n =

∑ n f(xi)∆x xdx=l →im∞ 2(n + 1) l

0

1 1−0 = n n

Now, we need an expression for xi: x0=0 x1=0+1/n=1/n x2 = 1/n + 1/n = 1/n = 2(1/n) x3 = 2(1/n) + 1/n = 3(1/n) In general, we can write xi = i ·(1n)

i=1

=

1

27. f(x) = 3x2. Since we are integrating over [0, 1] the length of the subintervals is given by

22

∑i=1n 4

=

n

2

∑i=1n

=

1 l→im∞1+n

= =

22 ∑ n f(i·)nn

Integration

3 3−0 = n n

Now, we need an expression for xi: x0=0 x1=0+3/n=3/n x2 = 3/n + 3/n = 6/n = 2(3/n) x3 = 2(3/n) + 3/n = 3(3/n) In general, we can write xi = i (3 · n) n

∑n f(xi)∆x=

33 f(i · 3 n 27 n n3 3 i2( )2 · n · ∑ ∑i=1

= =

i=1 n

∑

i=1

i2 ·

Exercise Set 5.2

263

= = =

3

∫

= = = =

x2dx

0

n

27 i2 n3 ∑ 27 i=1 3·n(n+1)(2n+1) 6 n 9(2n3 + 3n2 + n)

x3 = 2(1/n) + 1/n = 3(1/n) In general, we can write xi = i ·(1n) ∑n

11

∑ n f(xi)∆x= =

n2n3

f(xi)∆x n→im∞ l ∑i=1 3 n2 + n lim9 ·2n+3 n→ ∞ 2n3 →im∞9+27 9 l 2n + 2n2 n

=

∑i=1n 1 4i3·n4

=

i=1

= =

9 29. f(x) = x3. Since we are integrating over [0,4] the length of the subintervals is given by ∆x=

1

4 4−0 = n n

∫

Now, we need an expression for xi: x0=0 x1=0+4/n=4/n x2 = 4/n + 4/n = 8/n = 2(4/n) x3 = 2(4/n) + 4/n = 3(4/n) In general, we can write xi = i ·(4n) ∑ n f(xi)∆x= =

4 0

44

=

∑i=1n

=

i=1

=

∑ n i3 n4i=1

=

∫

∑i=1n

x3dx=l = = =

256

256

256 n2(n+1)2 n4 · 4 64(n4 + 2n3 + n2) ∑ n n4 n→im∞ 64(n4 + 2n3 fx(i)∆ x + n2) l →im∞ n

n4 128 64 ln→im∞64+ +n n2 64

30. f(x) = 4x3. Since we are integrating over [0, 1] the length of the subintervals is given by ∆x=

1 1−0 = n n

Now, we need an expression for xi: x0=0 x1=0+1/n=1/n x2 = 1/n + 1/n = 2/n = 2(1/n)

11

4 ∑ n i3 n4i=1 4 n2(n+1)2 n4· 4 (n4 + 2n3 + n2)

4x3dx=l→im∞ n

= = =

44

∑n

0

∑i=1n

n4 ∑ n f(xi)∆x

(n4 + 2n3 + n2) l n ln→im∞ →im∞1+ n4 2 1 1 +n n2

31. f(x) = x2. Since we are integrating over [1,3] the length of the subintervals is given by 2 3−1 = n n Now, we need an expression for xi: x0=1 x1=1+2/n x2 = 1 + 2/n + 2/n = 1 + 2(2/n) x3 = 1 + 2(2/n) + 2/n = 1 + 3(2/n) Ingeneral,wecanwritex i=1+( i2n ) ∆x=

∑ n f(xi)∆x=

∑n = =

∑i=1n

2i2 2i 2

= = ∑i=1n 2 4i2] i4 + [ 1+ = ∑ni∑ 1+ n4i2 4 = 2 2[ ∑[n ] + = n2 n i=1 i [ n n+ 4 ∑i=1 4n∑=1] i2 i+ i=1 n2 2 [n n n(n+1) ]i=1 n+ · 4 + 2 4 nn n 2 ] n 6 +1)(2 +1) n 8 n2+n 8 2n3+3n2+n n2· +n3· 2 6 2 + 4 4 8 4 2 + 4 + n + 3 + n + 3n2

264

Chapter 5: 26 8 3∑n n ++

= 3

∫

1

x2dx=lim

4

x0=0 x1=0+2/n=2/n x2 = 2/n + 2/n = 2(2/n) x3 = 2(2/n) + 2/n = 3(2/n) Ingeneral,wecanwritex i=i · 2n

3n2 f(x)∆x

n→ ∞

[i=1 26 8 4 = lim ++ ] n → ∞ 3 n 3n2 26 = 3

= =

3 4−1 = n n

=

Now, we need an expression for xi: x0=1 x1=1+3/n x2 = 1 + 3/n + 3/n = 1 + 2(3/n) x3 = 1 + 2(3/n) + 2/n = 1 + 3(3/n) Ingeneral,wecanwritex=1+i(3 n) i ∑ n f(xi)∆x=

= = = =

3i3

∑n

= =

∑i=1n 3i3·3 (1+ )n n == ∑i=1n 9i 27i2 27i3] [ 3 1+ + + n3 n n n = 1+ n n27i3 ] ∑9i∑27i2∑ [3 + + ∑ n2 n3 n n i27∑=1 i=27∑1 ∑i=1 n i2+ n i=1 9 i=1 i+ 3 [ n+ i=1 i3] n2] i=1 n3 n n [ n(n+1) 3 + · 9 [ n + 2 n 27 ] + [ n 2 ·n 3 27·n2(n+1)2 n(n+1)(2n+1) ] n3 4 n 81 27 27 + 6 = 3+ +27+ + 2 3 2n 2n 81 81 81 + + 4 2n 4n2 n 255 54 81 = + + n2

∫

= =

4

∫

1

x3dx=l→im∞ = =

n

f(xi)∆x n [i=1 255 54 81 + + lim ] 255 ∫ dx 4 x2

2

2

33. ∫ 02(x+x2) dx= 0 xdx + 0 ∫ ThefirstintegralwasevaluatedinExercise25,andyielded a2 value of 2. The second integral is now computed: ∆x = −0=2 n n

2x2 0

22

∑n

∑ n f(xi)∆x=

32. f(x) = x3. Since we are integrating over [1,4] the length of the subintervals is given by ∆x=

Integration

dx

∑i=1n 22 8 ∑i=1n 8 i2·n3 2

n n3 ∑ii=1 8 n3 ·n(n+1)(2n+1) 42n3+3n2+n63 43 n3 8 1 ] 3 [l 2+ + +

83 4 4 + 3n2 n 8 4 4 n→im∞+ +

Thus, ∫ 02 (x+x2)dx=2+

8

=

14

34. - 42. Left to the student π

43. 0sinxdx=2 ∫ π 44 0cosxdx=0 ∫ 4√ .

0 xdx=5.33334 ∫ 45. 3 √ 2x+1dx=4 ∫0 46 4 ∫ 2 ln(x) dx = 2.15888 . e 1 ∫1 x d x = 1 47

Exercise Set 5.3 .

48 1. Find the .

∫areaunderthecurvey=4ontheinterval[1,3].

4dx

A(x)= =4x+C

Since we know that A(1) = 0 (there is no area above the number 1), we can substitute for x and A(x) to determine C. A(1) = 4 · 1 + C = 0 Substituting 1 for x and 0 for A(1) Solving for C we get:

Exercise Set 5.3

265

4+C=0 C = −4 Thus, A(x) = 4x−4.

Then the area on the interval [1, 3] is A(3).

A(3) = 4 ·3−4 Substituting3forx =12 − 4 2. A(x)=

∫=8 5dx=5x+C

A(1) = 5 · 1 + C = 0 C = −5

A(x) = 5x − 5 The area on [1,3] is A(3).

A(3) = 5 ·3−5=10 3. Find the a ∫reaunderthecurvey=2xontheinterval[1,3].

A(x)= 2xdx =x2+C Since we know that A(1) = 0 (there is no area above the number 1), we can substitute for x and A(x) to determine C. A(1) = 12 + C = 0 Substituting 1 for x and 0 for A(1) Solving for C we get: 1+C=0 C = −1 Thus, A(x) = x2

−1.

Then the area on the interval [1, 3] is A(3).

A(3) = 32 −1 Substituting3forx = 9− 1

∫=8 4. A(x)=

x2dx=

03 +C=0 Substituting0forxand 3 0 for A(0) Solving for C, we get C = 0: x3 Thus,A(x)= .

3 Then the area on the interval [0, 5] is A(5).

53 A(5)= Substituting5forx 3 2 125 ,or41 = 3 ∫ 3 x4 x3dx= 6. A(x)= +C 4 Use A(0) = 0 to find C. 04 A(0)= +C=0 4 C=0 A(0)=

x3

+C

Use A(0) = 0 to find C. 03 A(0)= +C=0 3 C=0 x3 A(x) = 3 The area on [0,3] is A(3).

33 A(3)= =9 3 5. Find the a ∫reaunderthecurvey=x2ontheinterval[0,5].

A(x)= x2dx = x3 +C 3 Since we know that A(0) = 0 (there is no area above the number 0), we can substitute for x and A(x) to determine C.

A(x) =

x4 4

The area on [0,2] is A(2).

24 A(2)= =4 4 7. Find the a ∫reaunderthecurvey=x3ontheinterval[0,1].

A(x)= x3dx = x4 +C 4 Since we know that A(0) = 0, we can substitute for x and A(x) to determine C. 04 A(0)= +C=0 Substituting0forxand 4 0 for A(0) Solving for C, we get C = 0. x4 Thus,A(x)= .

4 Then the area on the interval [0, 1] is A(1).

14 A(1)= Substituting1forx 4 1 = 4 8. A(x)= ∫ (1−x2)dx=x− Use A( −1) = 0 to find C. A ( −1)=−1−(−1)

3

3

x3

+C

+C=0

−2+C=0 3 2 C = 3 A(x)=x

x3 2 − 3 + 3

The area on [−1,1] is A(1).

13 2 4 1 A(1)=1 − 3 + = ,or1 3 3 3

Chapter 5:

266 9. Find the area under the curve y = 4 − x2 on the interval [−2, 2].

A(x)= (4−x2)dx ∫ x3 +C =4x − 3

Since we know that A( −2) = 0, (there is no area above the −2),we can substitute for x and A(x) to determine number C. −2)=4(−2)−(−2) A(

The area on the interval [0, 3] is A(3).

A(3) = e3 − 1 = 20.085537 −1 Usingacalculator ≈∫19.086 2 dx=2lnx+C 12. A(x)= Use A(1) = 0 to find C. A(1)=2ln1+C=0 2 ·0+C=0

3

+C=0 3 Substituting −2 for x and 0 for A(−2)

Solving for C, we get: 8 +C=0 − 8+ 8+ −24 +C=0 8 3 −16+C=0 3

C=0

A(x) = 2 ln x The area on [1,4] is A(4).

A(4) = 2 ln 4 ≈ 2.773 Find the 13.

16 C = 3 +

3 ontheinterval[1,6]. ∫areaundert∫hecurvey= x 3 1 A(x)= dx=3 dx = 3ln x+C

Since we know that A(1) = 0, (there is no area above the number 1), we can substitute for x and A(x) to determine C. A(1) = 3ln 1 + C = 0 Substituting 1 for x and 0 for A(1) 3 ·0+C=0 (ln1=0) C=0

x3 16 .

Thus, A(x) = 4x − 3 The area on the interval [−2,2] is A(2).

23 16 A(2) = 4 · 2 − + Substituting2forx 3 3 =8 − 8+3 16 16 24 − 8 + = 3 3 32 2 ,or10 = 3 ∫3 exdx=ex+C 10. A(x)=

Thus, A(x) = 3 ln x.

The area on the interval [1, 6] is A(6).

A(6) = 3lnSubstituting 6 6 for x Usingacalculator ≈∫5.375 14. A(x)= A(0) =

1+C=0 C = −1

A(x) = −3cos x + 3

The area on [0,2] is A(2).

A(2) = e2 − 1 ≈ 6.3891

=ex+C Since we know that A(0) = 0, (there is no area above the number 0),we can substitute for x and A(x) to determine C. A(0) = e0 + C = 0 Substituting 0 for x and 0 for A(0) 1+C=0 (e0=1) C = −1

Thus, A(x) = ex 1.−

−3cos(0) + C = 0 −3·0+C=0 C=3

A(x) = ex − 1

Find the a ∫reaunderthecurvey=exontheinterval[0,3].

A(x)= exdx

3sinxdx=−3cosx+C

Use A(0) = 0 to find C.

A(0)=e0+C=0

11.

Integration

A4) 4 Theareaon[0,π]is (π A(π 4)= −cos(π 4)+3 3 15.

∫≈0.8787 1.5 x−x2 )dx 0 ( x2 − x 3]1.5 = [ 3 0 . = ( 2(1.5)2 3 2 ↑− = 2.25−3.375

Usingacalculator

03) 2− − ( 02↑ 3

2 3 ) −(0−0) = 1.125 − 1.125 =0 (

Substituting0forx Substituting1.5forx

Exercise Set 5.3

267

The area above the x-axis is equal to the area below the x-axis.

16.

39.

2

(x2 ∫ 0 3 − x )]dx x2 2 x = [ −3 2 )0 = ( 23 02) 22 (03 = = 3 − 2 − 3 − 2 8 3 −2 2 3 The area above the -axis is greater than the area below x the x-axis. ∫

0

40.

=16 = −60 (9 41.

2 42.

1

x)

dx

]e 1

= e2 ( = = 2

−lne) − ( 122 −ln1 ) 1 +0 e2 2 −1− 2 e2

43.

-36 ∫.]Lefttothestudent 37. betdt a

=[et]ba =eb ∫−ea

3 −2 2 π/6 5 2 sin 2x dx ∫ 0 cos2x ]π/6 = [ − 5 4 0 = = = −5 cos(π/6) − (−5cos(0)) −

5

5 − ( − )4

8 5 2π/3 8

a 0

)dx [(ax−x2 x3]a = − 3 0 (2 03) = a·a2 · − 3 − ( a 02 2 − 3 = = a3 a3 3 −2

∫

1 + (2 e x−1

x2 = [ 2 −lnx

19.

6

dx

)1 2 e +lne − ( 122 +ln1) 2 (ln e2 1 e=1, = +1 − = 2 2

b

(0 2 e3·0) = − e23 b − − 2 3 3 The area x-axis.

2 is below the 2 3 (e3b−1) = − e 3 b + ,or− 3

a3

1

1)

e2

= This means that the area between [0, π 2 ] (area above the 2 ] (area below x-axis) is less than that area between [ π2 ,3π thex ∫-axis)by1unitofarea. b−2e3xdx 18.

38.

x+

= (

= −1−0 −1

=

e

∫

x2 = [ 2 +lnx ] e

cos x dx

2 e3x ] −

3

2

−54t dt ∫ = [t4]2− 5

=24 − (−5)4 −625

= [ sinx ]3π/2 0 = 2) − sin(0) sin(3 π/

[0

3t2 dt t3 = [ 3·3a ] b = [t3]ba =b3 −a3

3 π/2

17.

b

∫a

44.

1 4cos xdx ]2 2π/3 1 = [ 8sin x2 ∫

π/3

=8sin(π/3) 8sin(π/6) − =8 ·

√

1 3 8 2 − ·2

√ = 4 3− 4

ln1=0)

Chapter 5:

268 45.

∫14 10 17t3dt −

3

2(x2 +x+1dx ])3 ∫ x3 x2 +x = [ 3 + 2 = 33 + )2 32 22 ( +3 − ( 2 3 + +2 ) 3 = 3 ( 92 ) =12+ 9+ +3 − ( 8 +2+2 ) 2 3 8 93 2− − 4 16 27 6 = =8 8+ + −6

1

10 = ∫ t3dt 17 − 4 1 −4 10 4 = = 17[ t4 ] = = = = 10 14 −(−4) 4 4 ) 17( 4 10 1 ( −64 ) 17 4 −255 (· ) 4 17 2550 − 68 1275 − 34 10

46.

∫ 1 12t2 dt 13 12 [ t3 ]1 = 13 303−

12 1 =13 ( 3 12 1 = = 13 · 3 4 13

50.

11 6 5 =8+1 6 5 =9 ∫61 −(2 − x − x 2 )dx 2 [2 − x −x31 == 2 3 ]− 2 =

03 3)

2 − 1 3) 21 2(2) · 2 ( 2 3 − [ − = 4 8 − 1 − 1 +4+2− 2 1

47. Find the area under y = x 3 on [0, 2]. x3 ∫ 02 dx x4]20 = [ 4 = 4 04 2 − 4 4 =4 −0 =4

48.

Integration

∫[0x]4dx 1 x5 10 = 5 = = 15 05 5 − 5 1 5

−(−2) 2

2 51. Find ∫theareaundery=5−x2on[−1,2].

2 1(5 − x ] 2 )dx 32 [− = 5x − x 3 −)13 3] = 5(1) −(−1) · ( 52 3 − [ − 3 = 1) 8) −5+ ( 10− − ( 3 = 3 30 1+ − 8 −15 ( ) −( ) 3 = −14 ) = 22 = 3 − ( 3 = 22 + 14 3 3 36 3 ∫123

2

49. Find the area under y = x +x+1on[2,3].

52.

x

−2edx =[ex]3− 2

=e3 − e − 2,ore3−e2

1

53. Find the area under y = ex on [ −1, 5].

2

− (− 2)3 3 ]

Exercise Set 5.3

269

5

58.

ex dx ∫ − x1 = [e ]5− 1

=e5 − e − 1 =e5−1 54. ∫

4 1

1

1 [ [81(x+2)4 ] −0] 0 1 4 81 4

∫10(x + 2)3 dx = = =

4 e1 x−2)dx ( 2x+x2 ) dx = = ∫ 1(2 x + = x−1 ]4 [ x2+− = ]11 14 [ x2−x

(

∫81

√3x2 √ − 31x dx =

)11 1) 12− 4 − (

=

1 4 −1+1

=

42−

=16 − =15 3

59.

= =

∫ 8 (x1/3−x−1/3)dx [1 3 x4/3−3x2/3

[84/3− 4

∫

=

∫

=

21

=

3 3 + 4 12 − 6 2−

3(x−1)(x+1) 3

dx

x− 1 (x+1)dx

x2 +]x ]

60.

3

∫10

2

x3 + 8 x + 2 dx = = = =

2 x5 − x − 1 = x2 dx =

[1x4 +4

]1

x3−x2+4x

1 3 −1+4−0 10 3

1

]

0

∫21(4x + 3)(5x − 2) dx =

∫ 2 (20x2+7x−6)dx

=

[1 20x3 7+ x2−6x

= =

3888 25 57. =

∫16 (x3/2−x1/2)dx

=

[42 x5/2

=

5

= =

1 [0

61.

5

625 2x2]1 1 +] [ 50 − 4 [ 1 1+4

=

4

∫ 1 (x2−2x+4)dx

∫ 5 (x3−x−3)dx

=

∫16( x − √1) xdx

2

27 4

[2 [ +3−[2+2] 279 2

56. ∫51

]1 2 82/3]−

14/3− 3

x2 ∫32 − 1 dx = x− 1

=

2

[4

4

55.

8

]

2 x3/2 ] −

160 +14 −12− 3 307 6

1

20 3

−

62.

16

∫52(t+ √ 3)(t − √ 3)dt =

4

=

3 3 5/2− (16) 3/2 ] −

= =

[ [(16) 5 2 2 (4)5/2− (4)3/2]

16 2048 5 −128−64 + 5392 3 5 15

2

]

∫5 (t2−3)dt [2t ]5 3 −3t 3125 3 −15− 8 3 +6 30

63.

1

∫

0

(t + 1) 3 dt =

∫ 1 (t3+3t2+3t+1)dt

7 2 +6

270

Chapter 5: 4

t +t3+ 3t 2 +t ] 1 4 3 1 4 +1+ + 1 −0 15 2 4

=

[

= =

69. The average is give ∫nby 1 x3dx = 1 − − 1) − 1 = = =

64. ∫31 ( x −

1 ) x

2

3

dx = =

∫

1

(x2 −2+x−2)dx

3

x − 2 x − x − 1] 3 [ 3 1 1 9 − 6 − 3 − 3 +2+1 16 3

= =

∫

1

3

t − t dt = t3

∫

1

= = =

1

−1

1x4 ] ·[

1

2 −1 2 1 4 (14 −(−1)4)

70. The average i ∫sgivenby 2(4− 1 x 2)dx = 2− − −2 = = =

2

1 4 1

∫

−2

(t2 −t−2)dt

(4 −x2)dx 3 2

x ] 3 −2

[ 4x−

4 1 4( 8 8 3

−

8 3 +8−

8 3)

71. The average is given by

]3 +t−1 1 1 3 3 1 3 9+ − −1 8

[t

1 1 2 ∫ 2x3dx

0

65.

3 5

Integration

3

1

1

1

exdx = 1− 0 0 = = ∫

∫

0

exdx 1

[ ex]

0

e− 1

66.

9

∫94 t √+1 dt = t =

∫

4

72. The average is given by

(t1/2+t−1/2)dt

1

∫ e dx = 1 −0 0 =

9

3/2

1

t +2t1/2 ] [2 3 4 = 54+6 −16−4 = 40

−x

= =

5

∫ ∫

= =

3

5

3

(x −2)(x+2) x− 2

25 22 +10

−

73. The avera ∫geisgivenby 2(x2 1 −x+1) 2 0 0

(x+2)dx

[ x2 +2x ]

dx

5 3

9 2 −6

12 68.

1

3

∫10 t + 1 dt = t+1

∫

0 3

(t2 −t+1)dt 2

=

[t − t

= =

1 33 1 2 +1−0 5− 6

2

+t ] 1

74. The avera ∫geisgivenby 4(x2 1 +x−2) 4 0 0

∫

0

e− x dx 1

[ e− x ]

0

e− 1 − 1 1 − 1 e

67.

2 4 ∫ 5 x − dx = = x− 2 =

1

dx = = = =

1

2

( 2 − x +1) dx 2 0 x 2 2 3 x 1 x + x] [ − 2 3 2 0 1 8 ) 2 ( 3 −2+2−0 4 3∫ ∫

1 4 4 (x2+x −2) dx = dx = == 1 0 4 x3 x2 1[ + −2 ] 4 3 2 )x0 +8 4 64 16 ( −8 3 3

271

Exercise Set 5.3 75. The average is given by 2

1

1 6 4 ∫ (3x+1) dx 2 1 6 32 4[ + x ] 2 2 1 (54+6 6 4 − − 2) 13

∫ (3x+1) dx = = 6 −2 0 = =

81. The populat ∫ionincreaseisgivenby 2

0

3

0 7

1 1 ∫ (4x+1) dx = = = ∫ 3 (4x+1) dx 7 −3 0 = 4 7 1 2x2+x [ ] 4 3 1 4 (98+7 −18 − 3) 21 77. The average is given ∫by111−00

1

xndx =

∫

=

78. The average is giv ∫enby12n2−11 xdx

2

=

∫

n

xdx

1

n+1

=

2

x ] n+ 1 1 n+1 1 2 +1 − n +1 nn+1 2 − 1 n+ 1

[

= =

10

0

1

=

[ t3+t2 ]

= 125+25 =

85. The work done

− 1 − 1

[ t4+t2 ]

= 81+9 = 90

[

−500(200 − 50) −75000

3 0

t3]6 − 50[ 3

0

−50(72 −3600 − 0)

∫isgivenby 1171.3x−4.15x2+0.434x3dx

11710 .x −1032+09863 .x .

x dx

[2 35x2−34333+246542 = . x . x ]11 = [3274.2 − 116.48] = 3157.72 Nmm ·

1

148

=

−

10

−t2 ] 2 0

[2 35.65x2−1.383x3+.1085x42 ]11 = = [4061 − 133.27] = 3927.73 N ·mm ∫isgivenby 86. The work done

5

80. The distance is given by ∫3 (4t3+2t)dt

= = =

−50 t2dt = 0 = =

79. The distance is given by ∫5 (3t2+2t)dt

(−500(20−) tdt

6

x [ n + 1 ]10 1 n+1 − 0 1 n+1

=

0

10e (3+10 10e3 )+1

83. The po ∫pulationdecreaseisgivenby

n+1

=

[ 10et] 3

10etdt = = =

84. The populatio ∫ndecreaseisgivenby

n

xdx

0

−20(0e−2+2)00 200 1−e21

82. The population ∫increaseisgivenby

76. The average is given by 2

[ −200e−t ] 2

200e−tdt = = =

87.

a) The initial dosage is 42 03 . µg/ml b) The aberage amoount i ∫sgivenby 1204203−0.01050t 1 . e dt 120 10 120 [−10 −0.01050t = ] e 1 110−0.01050 10 = [−1135−(−3604)] 110 = 22.45 µg/ml

Chapter 5:

272 88.

1 0 (x − x 3 dx ] 1) ∫ 2

The avera ∫getemperatureisgivenby 1 1243.5−18.4x+8.57x2−0.996 = = =

x −x4 2 4 02 04 = 12 14 = ( 2 − 4) − ( 2 − 4) = 1 2 1 − 24 1 = [

3 x + 0.0338x4dx

[10 43.5x−9.2x2+2.8567x3−0.249 12 −0] [652.36 12

4 x + 0.25576]12x50

54o 89.

4

a)

2. ∫ 0b 2πrdr

= = =

r 2π[ 2 ] b 2π 2 0 b2 2 πb2

1.8 1.6 1.4 1.2 1 0.8

b) The circle could be thought of as a rectangle with length πb and width b 90.

0.6 0.4 0.2 0

a) ∫ b 4πr2 dr = 0

= =

r 4π[ 3 ] b 3 0 4πb3 3

0.2

b) The volume could be thought of as a stacking of surface areas 91. The lower limit of the integral, x = 1, was not evaluated 1

Exercise Set 5.4 1. First graph the system of equations and shade the region bounded by the graphs.

0.4

0.6

0.8

1

We find the boundaries by looking at the graph or by solving the following: x4=x − x x4− 1 ) = =0 0 x(x3

4 πb3 3

92. ∫

Integration

x ∫=0orx=1 1 0( x−x4 = [

])dx x2 − x 51 5

0

= ( 212 15) ( 0 2 05) − 5 − 2 − 5 = 3 10 3. First graph the system of equations and shade the region bounded by the graphs.

4

2

3

1.8

2

1.6 1.4

1

1.2 1 0.8

–2

0.6

–1

1

2

3

–1

0.4 0.2 0

–2 0.2

0.4

0.6

0.8

1

Here the boundaries are easily determined by looking at the graph. Note which is the upper graph. Here it is x ≥ x3 over the interval [0, 1].

Compute the area as follows:

Here the boundaries are easily determined by the graph.

Note which is the upper graph. Here it is (x+2) ≥ x2 over the interval [−1, 2].

Compute the area as follows:

Exercise Set 5.4 2

∫ −21

273 Here the boundaries are easily determined by the graph. −x2)≥x

[(x + 2) − x2] dx

Note which is the upper graph. Here it is (6x

− 1( x2+x+2)dx − 2 3 x + x2 +2x = [ − ] 2 3 ] − 1 [ ( 1)3 ( + −1)2 = [ − + +2( −1) ] 23 22 +22 · − − − 2 ) 3 = ( − 1 3 +2+42 ) −( 1 + − 2 8 = −8 1 1 3 +3 2 + 4 − 3 − 2 + 2 − = 9 1 1 9 3 − + 2 + 4 + 2 = 4 , or 2 2 2 =

4.

∫

over the interval [0,5].

Com∫putetheareaasfollows: ∫50[(6x−x2)−x]dx 5 +5x)d]x5 = 0 [(−x2 2 = −x3 +5·x 3 2 5)0 3 2 03 02) = 5 +5· ( − 3 +5· 2 − ( − 2 3 = 125 125) + ( − 2 −0 = 375 + = −250 6 125 6

4 3 2

6.

1 0

0.5

1

1.5

2

2.5

3

4

5

–4

–2

We find the boundaries by looking at the graph or by solving the following: x2 −2x=x −3x=0 −3)=0 x2 x x(x ∫=0orx=3 2 ∫ 0[x−(x−2x)]dx 3 3

2 0(−x+3x)]dx

= [ −x

3

3

+

–6 –8 –10

We find the boundaries by looking at the graph or by solving the following: x2 −6x=−x −5x=0 −5)=0 x2

23

3x

x x(x ∫=0orx=5 ∫ 0[ 5 − x − x( 2 +5x)dx

0

2 3 = ( − 3 3 + 3·32) − 0 + 3·02) − ( 3 2 3 2 = 9 2

2 0(−x+5x)dx 5x2)5 = ( −x3 + 3 2 0 3 3 = ( − 5 + 5·52) − 0 + 5·02) − ( 3 2 3 2 = 125 6

=

5. First graph the system of equations and shade the region bounded by the graphs.

10

7. First graph the system of equations and shade the region bounded by the graphs.

8 6 4 2 0

2

–2

3

–1

=

1 0

1

2

3

4

5

274

Chapter 5: 1 √ x2) 0( x− ∫ ]dx 3 1

2

2 x = [ x3/2− 2 3 = (3 13 2· 03) 13/2− 03/2− · = 3) − ( 3

1 0.5

1

1.5

2

2.5

3

0 –1

1 3

–2 –3

9. First graph the system of equations and shade the region bounded by the graphs.

–4

1.4

–5

The boundaries are easily determined by looking at the graph. Note which is the upper graph over the shaded region. Here it is (2x − x2) ≥ −x over the interval [0, 3].

1.2 1 0.8

Com∫putetheareaasfollows: 2 ∫ 0[(2x−x)−(−x)]dx 3

0.6 0.4

3

=

2 0(3x−x)3]dx

=

3 x2−x [

0.2

3

3

0

0

33) 03) = (2 ( 3·02− ·32− − 3 3 = = = 3 −27 = 3 2 −54 27 6 2

0.2

0.4

0.6

0.8

1

The boundaries are easily determined by looking at the graph. Note whic √histheuppergraphovertheshaded region.Hereitis4x ≥ x over the interval [0, 1].

Com∫putetheareaasfollows: ∫ 01 √ x−x)dx (4 1 = 0 [(x1/4 −x)d]x 1 4 x5/4− 1 x2 = 0) (5 21 4 1 = 4 · 1 2 − ( ·05/4− · 0 2) ·15/4− = 4 2 −1 −0 = 5 = 3 8 5 1 −0 1 0 3

81

6 927 26 8.

Integration

1.4 1.2 1 0.8 0.6

10

0.4

10.

0.2 0

0.2

0.4

0.6

0.8

1

We find the boundaries by looking at the graph or by solving the following: 2 = √x x x4=x x4 − x = 0 −1)=0 x(x3 x=0orx=1

2.5 2 1.5 1 0.5 0

0.5

1

1.5

2

2.5

3

From the graph we see that the boundaries are 0 and 3.

Exercise Set 5.4

275 x2 = x3

3

0(3−x) ]dx3 x2 = [ 3x−2 ) 2) 32 = 3·3− 30 − 0 2 − ( · 2 = ( ∫

0=x3 − x 2 0 = x2(x −1) x=0orx=1 Note that x2 ≥ x3 on [0, 1].

9 2

(Note that this is the area of a triangle with base 3 and height 3.)

11.

Graph the system of equations and shade the region bounded by the graphs.

5

x4])1 ∫[0(x2−x3dx 1 x3 = − 4 0 (3 4) 13 0 − 0 4) = −1 − ( 33 4 4 = 1 12 13. First graph the system of equations and shade the region bounded by the graphs.

4 3

6 4

2

2

0

5

10

15

20

25

The boundaries are easily determined by looking at the graph. Here 5 ≥ √x over the interval [0, 25].

Com ∫putetheareaasfollows: ∫250(5−√x)dx 25 = [0(5−x1/]2)dx x3/225 = 5x− 3/2 ] 25 0 = 2 [ 5x− x3/2 0 =

3 2 5·25− ·253/2) − ( 5·0− (

2503 =125 − 3 − 0

2

·03/2)

[253/2=(52)3/2=53=125]

375 250 3 − 3 125 2 ,or41 = 3 3

=

12.

1

0 –2 –4

1

4

–6 –8 –10 –12

Then determine the first coordinates of possible points of intersection by solving a system of equations as follows. −x2andy=4−4x, At the points of intersection, y = 4 so 4 −x2=4−4x 0=x2 − 4 x 0 = x(x −4) x=0orx=4 Thus the interval with which we are concerned is [0,4].

Note that 4 − x2 ≥ 4 − 4x over the interval [0, 4].

=

0.8 0.6 0.4 0.2

0.2

3

Com∫putetheareaasfollows: 2 −x)−(4−4x)]dx 0 ∫4[(4 4 2 = 0−x+4x) [( ]dx4 3 = − x +2x2 0 3

1

0

2

0.4

0.6

0.8

1

We find the boundaries by looking at the graph or by solving the following:

− = ( = = 64 − +32 3

+2·42

64 − +3

96

32 3

3

) − ( −

03

+2·02

)

Chapter 5:

276 14.

Integration

From ∫thegraphweseethattheboundariesare0and

10

π/4

8

[cosx−s]inx]dx

[0 = sinx+cosx0

π .4

π/4

√1 + √ 12 ] −[0+1] = [ 2 √ 2− 1 = 17. First graph the system of equations and shade the region bounded by the graphs.

6

4

2

6 1.2 1.4 1.6 1.8

2

2.2 2.4 2.6 2.8

3

From ∫ the graph we see that the boundaries are 1 and 3. ∫31[(x2+1)−x2]dx 3 = = [1]dx

5 4 3

=3−1

2

=2

1

15. First graph the system of equations and shade the region bounded by the graphs.

0

0.4

0.6

0.8

1

1.2

1.4

From the graph we can easily determine the interval with which we are concerned. Here x + 3 ≥ sin x + cos x over the interval [0, π 2].

7 6 5 4 3 2 1

1.2

1.4

1.6

1.8

2

From the graph we can easily determine the interval with which we are concerned. Here x2 +3 ≥ x2 over the interval [1, 2].

18.

Compute the area as follows: π 02 [x+3 sinx−cosx] − ∫ ]dxπ = 122x+3x+c [ ]osx−sinx20 π2 +3π2 − 1−[1] [ = π2 +3π2 − 2 8 = 2.5 2 1.5

Com ∫putetheareaasfollows: +3)−x2]dx 1 ∫2[(x2 2 = [13]dx =3

1 0.5

0

=6− 3

0.1

π/4

1

=

0.3

0.4

0.5

0.6

0.7

0

2 [secx]−1]dx

π .4

π/4

tanx x0 ]− −0 [ =1−π4

0.8 0.6

[

19. First graph the system of equations and shade the region bounded by the graphs.

0.4 0.2

0

0.2

From ∫thegraphweseethattheboundariesare0and

=3 16.

0.2

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Exercise Set 5.4

277 From the graph we can easily determine the interval with

2.5

which we are concerned. Here x2 + 6x−15≥2x2−6x+5 over the interval [2,10].

2

Com∫putetheareaasfollows: 2 2 2[x+6x − 15−2x+6x−5]dx ∫10

1.5 1 0.5

0 0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

From the graph we can easily determine the interval with which we are concerned. Here x + 3 ≥ sin x + cos x over the interval [ 4π ,2]. π Com ∫putetheareaasfollows: π sinx]dx π [2[csc2x − π = − co ] [tx+cosx2π4 =[0] − − 1+√12 ] =1 − √ 12 20.

2[x+12x−20] −2

=

− 13x+6x−20x2

=

−02−0 + 0 3 4

]dx ]10 [ 8 = −1000 3 +600−200−−3+24−40 [ [ ] =256 ∫30 2 3 −∫1[0 − (2x + x − x )] dx+ 1 2 3 0 0[(2x+x−x)−0]d ∫ 1x 3 2 3 4]x)dx = −1(−2x−x+x2 4])dx+ [∫ [0(2x+x− 0 1 3 3 x −x +x2+ = − x2− x + x 3 4)−1 3 4 03 ( 3

2

3

(

12+

12

=

10 8

13

2−(−1) 3

− [ − −(1)

+

(−1) 4] + 4

14) 03 04) − 4 − ( 02+ − 4

18 12

=

3 2

23. f(x) ≥ g(x) on [−5,−1], and g(x) ≥ f(x) on [−1,3]. We

6

uset∫wointegralstofindthetotalarea. −1 3 2

4

− ∫5[(x+3x−9x−12)−(4x+3)]dx+ 3 3 2

2 0 –2 –4

1

2

−1[(4x+3) − (x+3x−9x−12)]dx

3

=

From ∫ the graph we see that the boundaries are -1 and 3. x−x−3)]dx (22 ∫ −1[x2+x− 3 3 == [−1[−x2+2x+3] ]dx 3 2 =[9+9+ − 13x+x+3 3x−1 9]− − 13+1−3 − [ ] =223 21. First graph the system of equations and shade the region bounded by the graphs.

140 120 100

=

−1 3 2 ∫ − ∫5(x+3x−13x−15)dx+ 3

−1(x−3x2+13x −3

4

2 x +x3− 13x [ 2 −15x = 4 4

[ [ −x

4 −

(− 1) 4

6

8

10

13(1) −

2

2

−15(−1)− ] 2

−15(−5)+ ]

− 33+ 13(3) +15·3− ] 4 2

(

4

−

4

−3

40

2

+

1

2 3−13(−5) 2

60

20

2

−

[4 4 (5) − +(5)− 4 [

80

]+15)dx

x3 + 13x +15x ] 3 2 5 −1

+(−1)3−

[ −(−1) 4 1 13 = ( −1− = 2

0

4

5 1 = 0 + 12 + 3 1 2 −0

14

–1

22.

10

=

4

13(−1) 2 +15(−1) ] 2 325 +15 ) − (625 −125− +75+) 4) 2 117 13 27+ +45 − ( − 1 +1+ −15 ) − −(1)+

4 −81− 31 225 225 31 + + + 4 4 4 4 =128

4

Chapter 5:

278 24.

[( x+28) − (x4 − 8x3 + 18x2)] dx ∫4−1 ∫4 = (xx − 4+8x3−18x2+x+28)dx [−1 5 4 x2 3 = +2x 4− 6x+ +28x ] − 5 −1 = 45 42 − +2·44−6·43+ +28·4− ) 5 = ( 2 = [ −(−1) 5 (−1) 2 3 +2(1)− 4− 6(1)+ +28(−1) ] − 5 2 216 193 + 10 5 625 1 =62 10 2 ≥ g(x) on [1,4]. We find the area.

25. f(x)

3

−128 +80−32 ) − ( − 3 16 11 27 = + 3 3 3 =9 (

26.

8+

3

) +(12 − 0)

) +12

16 36 + 3 3 1 52 ,or17 = 3 3 28. 25 20 15 10 5

)

+5−8 )

2 ∫ −23]1 1 xdx+11·dx ∫3 x +[x]3 [ 1 =3 − 2 ] −31 3 [ 1 −(−2) +(3−1) = = 3 3 1 8 + +3 3 3 =5

27. Find the area under f(x)=

=− ( −24 3

8

=

1

2

( −8+

3 =− ( −16 ) +12 3

[(4x − x2) − (x2 − 6x + 8)] dx ∫∫414 = ( −2x2 + 10x − 8) dx [1 4 3 = − 2x +5x2−8x ] 3 = − 2 · 4 +54 · 2 − 84 − 2 · 1 +5·12−8·1 · − = ) ( 3 3 = (

=(0−0)−

Integration

−x2,ifx<0 { 4 4, ifx ≥0 on[−2,3]

We have to break the integral into two parts in order to complete this problem f (x) dx ∫∫3−2 f(x)dx+ ∫30f(x)dx = 3 ∫ 0− = (420 − x] 2 dx+04dx ) ∫ [ − 2 30 −x +[4x]3 = 4x 3 −2 3 3 = ·0−0 −(−2) 4(2) +(4·3−4·0) − ] [ − {[4 3 3 ]}

0 1

2

3

4

5

From ∫thegraphweseethattheboundariesare1and5.

5 [ 1( x 2 −x−2)dx x

3

+ ]15 (3 x1 1) = 53 + 1) − ( 133 + = 608 15 =

29. First graph the system of equations and shade the region bounded by the graph.

5

4 3 2 1

0

0.2

0.4

0.6

0.8

1

From the graph we can easily determine the interval with which we are concerned. Here ex≥ e−x over the interval [0, 1].

Compute the area as follows:

Exercise Set 5.4

279

1

x 0(e−e−

1

(x2 − x3 ])dx ∫ 1 [− x3 x4 1 = 13 − 14 − = 3 − )4 − − (− 1)4 ) ( ((1)33 ( 4 ) = 1 1 1 1) − − − = ( − = 1 3 1 1 4 1 3 4 2 3 − + +

]x)dx

∫ −x1 x = e+e 0 [ =(e1+e−) ( + e− 0 ) )1−e0 1 = ( e+ e −(1+1) 1 = e + e −2 = =

30.

2

e −2e+1 e (e −1)2 ≈1.086 e

32.

6

60

4

50

2 40 30

0

20

–2

10

1

2

3

4

5

x +1 ≥ − x +1,andon[2,4], 2 ≥ − x +1. − 2x ∫+7 2 On[02], ,

0 1

2

3

4

5

6

7

8

From ∫thegraphweseethattheboundariesare1and8. √ 8 [1(x2−3x2dx8 ]) x3 3 x5/3 = − 1 (3 5 83 3 = − ·85/3 ) − ( 133 − ·15/3) 3 (3 = 512 96) 3) 5 ( 1− − − 5 = 5 2276 15 31. First graph the system of equations and shade the region bounded by the graph.

2

0

( x +1) − ( − x +1 ) ] dx+ 2( ∫4[ − x +1 ) ] dx (2x+7) − − 22

2 3 xdx+ 4( 3 x+6 ) dx ∫ − 2] 2 2 4 2 x22 = [0 3 − 3 x +6x + [ ] = 0 2 3 ( 4 2 2 3·22 = 322 · −3·0 + −3·4 ) − ( − 4 +6·2 ) 4 4 4 6 −0+12−9 = =∫

2

33. On[ 1

–1 –0.8–0.6–0.4–0.20

0.2 0.4 0.6 0.8

1

–1

–2

From the graph we can easily determine the interval with which we are concerned. Here x2 ≥ x3 over the interval [−1, 1].

Compute the area as follows:

∫0,11],f(x)≥g(x).

11[f(x)−g(x)]dx

∫011 [0.3x+6.15x2−0.552x3]dx == [ 0 =[726.24 0] 11 14.45 x3−0.138x4 ] 0.15x2+ 3 0 − =726.24 N mm ·

Chapter 5:

280 34.

T(t)

0

∫=5.76+24t−16t2,f(t)=8 1[T(t)−f(t)]dt

∫

∫01 [−2.24+24t−16t2]dt [0 = − 16 2.24t+12t2− t33 ] 1 −2.24+12− 163 −0] =[ = 4.43

0

38.

10 8 6 y 4

∫=25+3e−t−20(t−0.5)2,f(t)=8 1[T(t)−f(t)]dt

2 –4

–8

From the graph we see that the first coordinate of the relative maximum is x = −1 and the first coordinate of ther ∫elativeminimumisx=1 0

36. a) −0.003t2 ≥ −0.009t2 rate ∫ofmemorization.

2 0[(0.003t+02) .t−−. (0009 t2 +0.2t)]dt ∫10− 10 = 0 0.006tdt ]2 310 .002t 0 [ =0.002·103−0.002·03

−1

3 −1 =[0 − 14+2−2]−[4+2−2]

39. V=

=2

p (R2 −r2) 4Lv = Q

Subject B memorizes 2 more words then subject A during the first 10 minutes.

= =

100

=

80

=

60 y 40 20 2 x

4

–60

0

−

1 x4+ 3 x2−2x

3

= −32

=2− 0

0 –20 –40

∫ 1 [0−x3+3x−2]dx

[x3−3x+2−0]dx+

[−1 1 x4− 3 x2 = +2x ] 0 + [

Thus, M ′(t) ≥ m′(t), so subject B has the higher

–2

4

–10

= 17.23 degree days

–4

2 x

–6

− 3e− 5 6− − [3+6] 5

37.

0 –2 –4

–2

∫01 = = [17+3e−t−20(t−0.5)2]dt [0 =[17 17t− 20 (t−0.5)3 ] 1 3e−t−

b)

∫ 2 [0 − 3x 5 +20x 3 ]dx

[3x5−20x3−0]dx+

−2

1 1 = [ x6− 5x4] 0 + [ − x6+5x ]240 −2 =[ =96−32 + 80] − [32 − 80]

=

35. T(t)

Integration

∫ R [2π · p (R2−r2)r]dr πp∫

R 4Lv

[R2r−r3]dr

πp[02 R r2− 1 r4] R 2Lv[ 2 4 1 4]0 R − πp 2 R4

2Lv πpR4 8Lv 40 The area bounded in the graph is 4 The area bounded in the graph is 24.961

–80

.

–100

41. The area bounded in the graph is 5.886

From the graph we see that the first coordinate of the relative maximum is x = −2 and the first coordinate of the relative minimum is x = 2

42. The area bounded in the graph is 416.708 43. The area bounded in the graph is 0.238 44 .

1

]

0

281

Exercise Set 5.5 45. a)

4.

60 y40 20

–4

–3

–2

–1

0

1

2 x

3

4

3x

e dx ∫ Let u =3,then x du=3 .dx 1 = 3 ∫ e3x3 · dx = = = 1 3 ∫ eu du 1 3 eu + C 1 e3x+ C 3

–20 –40 –60

b) x = −1.8623, x = 0, and x = 1.4594

5.

c) The area is 64.5239 d) The area is 17.683

Exercise Set 5.5 3x2 dx ∫ 7+x3

1.

Let u = 7 + x3, then du = 3x2 dx. = ∫

du u

Substitutingufor7+x3and du for 3x2 dx

1 = ∫ u du =lnu+C

UsingFormulaC

=ln(7+x3)+C 3x2 2. ∫ 1 + x 3dx Let u = 1 + x3, then du = 3x2 dx. = ∫

(1/2)x

dx e dx = ∫ e ∫ x/2 1 1 Let u = x ,then du = dx.

2 2 1 Wedonothave dx.Weonlyhavedxandneedtosupply 2 1 1 a bym asfollows. ∫ultiplyingby2· 2 2 1 x/2 2 Multiplyingby1 e dx ∫· 21 =2 e dx 2 =2 x/21 ∫ ( 2 dx) ∫e 1 =2 eu Substitutinguforx/2anddufor du dx =2eu+C = 2 x/2 +C ∫e ex/3dx

6.

x ,thendu= ∫= 3x/3 ·1dx =3 3 ∫e =3 eudu

Letu

du u

=lnu+C ∫=ln(1+ 3. e4xdx

=

1

eu+C

= e4x+C 4

3

= 3 x/3 e +C 7.

We do not have 4 dx. We only have dx and need to supply 1 a4.Wed ∫othisbymultiplyingby 4 ·4asfollows.

1 4x Multiplyingby1 ∫ · 4 e dx 41= dx ∫4e 41= 4x (4dx) 41= ∫ e eudu

dx.

=3eu+C

x3)+C

Letu=4x,thendu=4dx.

4 1 4

1

Substitutingufor4xand UsingFormulaB

∫

x3ex4dx

We do not have 4x3 dx. We only have x3 dx and need to 1 supply a 4 ·4asfollows. ∫.Wedothisbymultiplyingby 4 1·4 x3ex4dx Multiplyingby1 41= ∫ 4x3ex4dx 1= ∫ 41= ex4 (4 x3 dx) ∫ 4 Substitutinguforx4and eudu 1 4 UsingFormulaB ·eu+C 1 = = 4 x4e +C

Chapter 5:

282 8.

9.

x4 x5edx ∫ Let u = x5, then du = 5x4 dx.

1 = 5 ∫ x5e·5x4dx 1 = ∫ eudu 5 1 = eu+C 5 1 5 = ex+C 5 ∫

− 3 t2etdt

12.

= =

= udu ∫ 2 = = u +C 2 (ln 5x)2 2 13.

= = =

3−1∫ e 3−

−1 3

e−t3+C

Letu=

Substitutingufor−t3and dufor −3t2 dt UsingFormulaB

15.

−t2 = −1 ·(−2t)dt 2−1∫ e = eudu 2 = −eu+C 1 = 2 e−t2+C −1 ∫2ln4xdx

x

(ln 4x)2 2

∫ cos(2x + 3) dx

∫

3 sin(4x + 2) dx

∫

2 sec2(2x + 3) dx

Let =2x+3,thendu=2dx. u 2 = ∫ sec2(u)duSubstitutingufor2x+3and 2

1 dx. x

+C

UsingFormulaC

uLet =4x+2,thendu=4dx.

3 Substitutingufor4x+2and = ∫ sin(u)du 43 1 4 3 4du for dx 4 = − cos(u)+C cos(4x+2)+C = − 16.

Letu ∫=ln4(x,then)du= 1 = dx ∫ln4x x Substitutinguforln4xand = 1 dufor dx x u2 = +C UsingFormulaA 2 =

Substitutingufor1+xand du for dx

Let =2x+3,thendu=2dx. u 1 = ∫ cos(u)du Substitutingufor2x+3and 2 1 1 2du for dx = sin(u)+C 2 1 = sin(2x+3)+C 2

∫−t2,thendu=−2tdt.

11.

du u

=lnu+C

−1 ∫3te−t2dt

10.

dx 1+x Let u = 1 + x, then du = dx.

1 = ∫ u du

14.

(−3t2dt)

eu+C

+C

=ln(1+x)+C

−t3

eudu

13

1 dx. x

∫

= ∫

∫−3t2e−t3dt

1

ln 5x dx x

Let u = ln 5x, then du =

We do not have −3t2dt.Weonlyhavet2dt.Weneedto supply a 1 −3by ∫multiplyingby− ·(−3)asfollows.

3 −1 ∫·(−3) t2e−t3dt Multiplyingby1 3−

∫

Integration

1

= tan(u) + C

2du for dx

= tan(2x + 3) + C 17.

∫ csc(2x + 3) cot(2x + 3) dx Let =2x+3,thendu=2dx. u 1 = ∫ csc(u)cot(u)duSubstitutingufor2x+3and 2 1

= csc(u)+C −

= csc(2x+3)+C −

2du for dx

283

Exercise Set 5.5 2 ∫ cot(2−x)dx=

18.

∫ csc2(2−x)−1

22.

Letu=2 − x, then du = − dx. = ∫ [−csc2(u)+1]du Substitutingufor2−xand −du for dx = cot(u) + u + C = cot(2 −x)+2−x+C

19.

∫=cot(2−x)−x+K dx4−x

(t −1)6 +C 12

Letu=4 − x, then du = −dx.

We do not have−dx. We only have dx and need to supply a −1 by multip ∫lyingby−1·(−1)asfollows. ∫(1) dx Multiplyingby1 −1 ·− = −

1

=− ∫

−1 · 1

4−x

=

20.

23.

4−x dx

4 x − (−dx) =− ∫ 1du −lnu+C = − ∫ln(4−x)+C dx1−x

24.

(x + 1) sec2(x2 + 2x + 3) dx ∫ Letu=x2+2x+3,thendu=2x+2dx.

1 = ∫ sec2udu 2 1 = tan(u)+C 2 1 = tan(x2+2x+3)+C 2

26.

(x2 + 2x + 2) csc(x3 + 3x2 + 6x − 7) cot(x3 + 3x2 + ∫ 6x − 7) dx Letu=x3+3x2+6x − 7, then du = 3x2 + 6x + 6 dx.

1 = csc(u)cot(u)du 3

∫=−ln(1−x)+C t2(t3−1)7dt We do not have 3t2 dt. We only have t2 dt. We need to 1 supply a 3 ∫bymultiplyingby 3 ·3asfollows.

1 t2(t3−1)7dt Multiplyingby1 ∫·3 31=

= −1cscu+C 3

∫3t2(t3−1)7dt 31=

= −1csc(x3+3x2+6x−7)+C 3

∫(t3−1)73t2dt 31= 3u7du 1 3 u8 8 +C (t3 ·

= = 1

24

−1)8+C

= −1 2 = −1 2 ∫ x2 cos x3 dx

25.

= −lnu+C

Letu=t3−1,thendu=3t2dt.

x sin x2 dx ∫ Let u = x2, then du = 2x dx.

1 = ∫ sinudu 2 cosu+C cosx2+C

Let u = x3, then du = 3x2 dx.

1 = ∫ cosudu 3 1 = sinu+C 3 1 = sinx3+C 3

Substitutingufor4−xand dufor −dx UsingFormulaC

Let u = ∫1−x,thendu=−dx.

1 =− (−dx) 1 x − =− 1dx ∫

21.

t(t2 − 1)5 dt − 1, then du = 2t dt. ∫ Let u = t2 1 = ∫ (t2 −1)5·2tdt 2 1 = ∫ u5du 2 · 1 2 u6 = = 2 6 +C

27.

Substitutingufort3−1and UsingFormulaA

exdx 4+ex Let u = 4 + ex, then du = ex dx. du = ∫ u Substitutingufor4+exand du for ex dx ∫

1 = ∫ u du =lnu+C =ln(4+ex)+C

UsingFormulaC

Chapter 5:

284

28.

etdt 3+et Let u = 3 + et, then du = et dt. ∫

= ∫

du u

=lnu+C

29.

+ e)+C t ∫=ln(3 lnx2dx x

1) 2 dx. ( 2x· x 2 dx= 2 1 We do not have dx. We only have dx and need to x x 1 supply a 2 ∫bymultiplyingby 2 ·2asfollows.

1 lnx2 d Multiplyingby1 · 2 x x = 2 2· 1 lnx2 2d = ∫ 2 x ∫ xx Substitutinguforlnx2and 1 lnx2· 2 2 ∫ d dufor dx x 2 1 u2udu 2 +C UsingFormulaA = x · 2 = 1= u2 4 +C Letu=lnx2,thendu=

1 = (lnx2)2+C, 4 1 or (2lnx)2+C= 4 ∫ 30.

x

dx x ln x2

2 dx. x

1x2·2 1 = dx = 2 ∫ ln x ∫ 1 1 du 2 u 1 = lnu+C 2 1 = ln(lnx2)+C 2√ ax+bdx,or (ax+b)1/2dx 33. ∫ ∫ We do not have a dx. We only have dx and need to supply 1 ∫ultiplyingby a ·aasfollows.

1√ ax+bdx Multiplyingby1 a ∫· a1 √ ∫aax+bdx a1 √ ax+b(adx) a1 √ Substitutinguforax+band ∫ a

anabym

= = =

∫ u1/2du 1=

duforadx UsingFormulaA

a +C

= 3 = a1·u3/2 2 2 3a ·u3/2+C 2

1 ·4(lnx)2+C 4 =(lnx)2+C

= 34.

1 dx. x

1 dx.

Letu ∫=lnx(,then)du= x = 1 1 dx = lnx x Substitutinguforlnxand du ∫ 1 1 dufor dx x =lnu+C UsingFormulaC u

(ax+b)3/2+C 3 ∫a√xax2+bdx

Letu= ∫ax√2+b,thendu=2axdx.

1= ax2 + 2b·axdx 2a ∫ 1= u1/2du 2a 1 u3/2 3 +C = = 2a · 2 1 3a (ax2+b)3/2+C

∫ xlnx

=ln(lnx)+C

∫

Let =lnx2,thendu= u

(lnx)2dx

Letu ∫=lnx,thendu= = u2du = = u3 +C 3 (ln x)3 +C 3dx 31.

32.

Integration

35.

∫ ∫beaxdx eaxdx =b Letu=ax,thendu=adx.

We do not have a dx. We only have dx and need to supply 1 an a by multiplying by a ·aasfollows.

Exercise Set 5.5

285

1 =b· · a ∫ eaxdx ab= ∫ aeaxdx ( eaxadx ∫b=

40.

)

eudu b= ∫ eu+C a = b a

Substitutinguforaxand UsingFormulaB

= = =

b+Pekt = eax C a 36.

∫ Letu= 1= k1= k P= =

37.

0

dt

∫kt,thendu=kdt.

Pekt 0 ·kdt

41.

∫ Peu0du 0eu+C k P0 e k t +C kasin(bx+c)dx

∫ Letu= 1=

∫

∫bx+c,thendu=bdx. cosudu

= =

sin(u)+D b= b sin(bx+c)+D 1 1 = b

44.

∫ 3x2dx 39.

(1+x3)5 Letu ∫=1+x3,thendu=3x2dx.

1 = · x3 2 dx = =

∫

1(1+x3)5 du

Substitutingufor1+x UsingFormulaA

u5 ∫ u−5du

u−4 +C +C −14 1 + x )4 4(1 = = − 4u4 =

−

3

sinx )+ C ∫n( √ 12x 51+62 x dx

3

45.

− 5 3/2+ C u 12 −5 (1 − 4t23/2 ) + C 12t t

e sine dt ∫ Let u e du=t e dt. ∫=t,then ==− sin u du cosu + C = − coset + C ∫ t t e sine dt Let =u 1=

√ ,then w du= u

∫ e du 2 eu+ C = 1 = √ +C

du=12 xdx.

= (1+6 2)6/5+ C x 6 ∫ √ 5 t1−42t dt 43.

Letu= t du=−8 tdt. ∫1−42,then =−5 1/2 u du

38.

1=

cosx

dx sinx Let u sinx ,then du= cosxdx. ∫= du = u =ln u + C

Let u x ∫=1+62,then = u1/5 du 6 u6/5+ C 5

bacos(bx+c)dx

Letu=

∫ cotxdx ∫ =

42.

1 b cos(u)+D 1 cos(bx+c)+D

b= = − −

− 1 · u− 6 + C 4 −6 1 + C 24 u6 1 + C 24(2 − x 4)6

=l

∫bx+c,thendu=bdx. sinudu

3

x dx (2− x 4)7 Let u = ∫ 2 −x 4,then du=−43 x dx.

3 =−1 − 4x dx 4 x 4)7 ∫(2− =−14 du u7 =−1 −7 ∫ u du 4 ∫

2

ew+C

dw .

2√ w

Chapter 5:

286 x2 e3dx ∫ x x3 Let = u ,then du=3 x2dx.

1 = ∫ e−u du = = 3 = 1 − 3 e−u+ C 1 − e− x 3 + C 3 1 − 3ex3 ∫ sin2x cos x dx 47.

2 = ∫ secudu

46.

Let =u

=tanu+C ∫=tanlnx+C x2 53. 102xe dx Firstfindtheindefiniteintegral. x2

2xe dx ∫ Let u = x ,2then du = 2x dx. x2uu

= e (2xdx) ∫ = edu

sinx ,then du= cosxdx.

=e+C

= ∫ u2du = = 1 u3 3 + C 1 sin3x+ C 3 48.

∫ Let =u

=e

cos2x sin x dx cosx,then du= −sin x dx.

54.

2

uu edu ∫ =e+C x3

+C

Then evaluate the definite integral on [0,1]. x3

∫ 1 3x2e dx [ 0 x 13 ] = e 0 3 13 =e −e0

sint dt ∫ ecos t cost,then du= −sin t dt.

= − ∫ e−udu

∫=e − 1 2 55. 1x(x+1)5

dx 0 Firstfindtheindefiniteintegral. x(x 22 +1)5dx ∫ Letu=x +1,thendu=2xdx.

= e−u+ C + = 1 C ecost

r3+7) ∫ r2sin (3

dr

r3 Let u =3 ,then du=9 r2dr.

1 = = 9 ∫ sin u du 1 + − 9cosu C 1 = −cos (3 r3 +7)+ C 9 1 sec2 ln xdx x Let =ln u ,then x

=

=e

= −esint+ C

52.

x3

3x e dx ∫ Let u = x ,3then du = 3x2 dx.

= − e u+ C

51.

∫ 1 2xe dx [ 0 x 12 ] = ee120 2 = −e0 1 = e ∫−1 2x3 dx 0 3x e First find the indefinite integral.

et sinetdt ∫ Let u = −sint ,then du= −costdt.

Let =u

+C x2

= − ∫ eudu

50.

x2

Substitutinguforx2and du for 2x dx

Then evaluate the definite integral on [0,1].

= − ∫ u2du = = 1 −u3 + C 3 1 + − cos3x C 3 49.

Integration

du=

dx . x

We only have x dx and need to supply a 2 by multiplying 1 by 2 · 2.

1 2 x(x2+1)5dx Multiplyingby1 2∫ ∫ 1 2x(x2+1)5dx = 2· 1 = ∫ (x2+1)5 ·2xdx 2 1 Substitutinguforx2+1and = ∫ u5du du for 2x dx 2

Exercise Set 5.5

287

1 6 = 2 ·u+CUsingFormulaA = 6 (x2 + 1)6 +C 12

Then ∫evaluatethedefiniteintegralon[0,4].

4 dt 1+t] 4 [0 = ln(1+

Then ∫evaluatethedefiniteintegralon[0,1].

1 +16])5dx 0 ( [xx2 1 (x2 +1) = = 2 (12+1)6 −(0+1) = = = 12 −1 12 64 12 12

First

Letu=∫x2−1,thendu=2xdx.

21= 2

58.

∫(x

7 2− 1)·2xdx

dt

dt1+t Letu ∫=1+t,thendu=dt. = du Substitutingufor1+tand =

∫

1u

du

=lnu+C =ln(1+t)+C

dufordt UsingFormulaC

[ 1e 2 x]dx 1 e2x

1 = e2·2 −1e2·1 2 2 e4 − e1 1 −1e1 e4 ,or 4 22 2

= 59. ∫

dx

2

1

First

∫findtheindefiniteintegral.

2x+1 dx 2 x+x−1

Letu ∫=x2+x=1,thendu=(2x+1)dx. = 1 (2x+1)dx 2 =

First∫findtheindefiniteintegral.

∫ eudu

1 = e4x+C 2 Then ∫evaluatethedefiniteintegralon[1,3].

3 =

0

∫2x,thendu=2dx. e2x·2dx

2 1 = e2u + C

1 u8 = 2 · 8 +C =

57. ∫

∫findtheindefiniteintegral. e2xdx

21=

u7du

(x2 −1)8 +C 16 Then ∫evaluatethedefiniteintegralon[1,2].

2 − 8])7dx 1 1 ( [xx2 2 (2− 1) x = 16 8]1[ 2 8] (1−1) 1) = [ − 16 = 4 38 6561 = 16 16

dx

1

First

1=

∫findtheindefiniteintegral. x(x2−1)7dx

1=

=l ∫n53e2x

Letu=

∫42 7 1x(x 2− 1)dx

−ln1 −0

=ln5

6

63 12 21 56.

=ln5

∫

1 x+x−1 Substitutinguforx2+x−1and du

=lnu+C =ln(x2+x

−1)+C Then ∫evaluatethedefiniteintegralon[1,4].

4 2x+1 dx 1 x2 +x−1] 4

= ln(x2+ x−1)1 [ =ln(44+4−1)−ln(12+1−1) = ln 19 −ln1 =ln19

(ln1=0)

Chapter 5:

288 3

60.

2x+3 dx 2 1 x +3x First find the indefinite integral.

2x+3 ∫ 2x+3x dx ∫

Let u = x2 + 3x, then du = 2x + 3. du u

= ∫

=lnu+C =ln(x2+3x)+C Then evaluate the definite integral on [1,3].

3

2x+3 ∫ 2 1 x +3x

dx

= ln(x2+3)3 x 1 [ ] =ln(32+3·3)−ln(12+3·1) = ln 18 −ln4 18

=ln =ln

61.

∫

4 9

2 b0e−xdx

We only have dx and need to supply a −1 by multiplying by 1 ( 1). · − ∫ e−xdx − · −−1(1) e−xdx = − ∫− = − e−x(dx) − = −

∫

eudu

= − eu+C

Substitutingufor −x and −dx du for UsingFormulaB

= e−x+C − Then evaluate the definite integral on [0, b]. b

=

∫

0 e−x d ]x e−xb0

[−− =(e−b)

− −(e−0) = e−b+e0 − = e−b+1 − 1 = 1 − eb 62.

b

2e−2xdx ∫0 First find the indefinite integral.

2e−2x dx ∫ Let u = 2x, −2 dx. − then du = = − e−2x(2dx) − ∫ = − eudu ∫− = eu+C = e−2x+C −

Then ∫evaluatethedefiniteintegralon[0,b]. b 2e−2x 0− ]dx e−2xb0 = [ =−e−2b−(−e−2·0) 1 b = − e − 2 +1,or1− e2b ∫ 63. b0me−mxdx ∫Firstfindtheindefiniteintegral. misaconstant me−mxdx Letu=−mx,thendu=−mdx.

We only have mdx and need to supply −1 a by multiplying by −1 · (−1). − 1∫(1) · − ∫ me−mxdx =− ∫−me−mxdx =− ∫e−mx(−mdx) =− eudu Substitutingufor−mxand UsingFormulaB = −eu+C −e−mx + C = ∫evaluatethedefiniteintegralon[0,b].

Then

∫Firstfindtheindefiniteintegral. e−xdx Letu= − x,thendu=−dx.

Integration

b

=

−mx 0me−− ]dx emxb0

[ =(−e−mb)−(−e−m·0) (e−m·0=e0=1)

= −e−mb+1

= 1 − e−mb 1 =1 − b emb −kx dx 64.

0ke ∫ First ∫findtheindefiniteintegral. ke−kxdx Letu= ∫−kx,thendu=−kdx. =− ∫e−kx(−kdx) =− eudu =−eu+C = −e−kx + C ∫evaluatethedefiniteintegralon[0,b].

Then b

−kx 0ke−− ]dx ekxb0

= [ =−e−kb−(−e−k·0) = −e−kb+1,or1−ekb

1

289

Exercise Set 5.5 65.

4x ( −6) 2dx ∫0 First ∫findtheindefiniteintegral. (x−6)dx2 Letu ∫=x−6,thendu=dx. u2du Substitutinguforx−6and = = u3 3 +C = Then(x −6)3 3

+C

∫evaluatethedefiniteintegralon[0,4].

4 [0(x−6)2]dx

=

3

208 ∫33 0(x −5)2 dx First ∫findtheindefiniteintegral. (x−5)2dx Letu ∫=x−5,thendu=dx. = u2du u3 = = +C Then3 − 5 ) 3 (x

3

+C

∫evaluatethedefiniteintegralon[0,3].

3 0(x −5)2 ]dx ( x −5)33 = [ 3 0 = (3 −5)3 (0 5)3 − − 3 3 = = 8 125 − +3 3 117 3

=39

First

∫findtheindefiniteintegral. cos(πx+π/3)2dx

Letu ∫=πx+π/3,thendu=πdx.

1 Substitutinguforπx+ cosudu = du for dx π 1 = sinu+C π 1 sin(πx+π/3)+C = π Then ∫evaluatethedefiniteintegralon[−1/3,0].

0 [−13cos(πx+π/]3)dx /1 0

0

(4 −6)3 (0 6)3 − = =− − 3 3 = = 8 216) 3 − (− 3 8 216 3 − +3

66.

67. ∫0−13cos(πx+π/3)dx /

sin(πx+π/3) = √ π = = 3 − 0 √2π3

−1/3

2π √π/6 2 68. 0 2xsin3xdx ∫

First

∫findtheindefiniteintegral.

2xsin3x2dx

2 Letu ∫=3x,thendu=6xdx.

1 Substitutingufor3x2and sinudu = 1dufordx = 3 −1cosu+C = 3 −1cos3x2+C π π/6].

Then ∫evaluatethedefiniteintegralon[0, √ √π/6 x2 2xsin]3√dx [0 π/6 = −1cos3x2 3 0 =0 1 − (−)3

=

1 3

π/3

Chapter 5: Integration √

290 3x2dx (1+ x35 ) From Exercise 39 we know that the indefinite integral is 3x2dx35 1 = − +C. ∫ 4(1+x3)4 (1+ ) x

Then

69. ∫ 2

Now we evaluate the definite integral on [0,2].

3x2dx ∫2 x35 (1+ ) 0 1 ]2 = [ 4(1+ ) x34]0 = 1 [ = [− 4(1+2) 34 − − = = 1 − 1 − 26244 4+ , 6560 26, 244 1640 6561 70.

21 =42 − 8

]

= 72.

First

(2 −x47) −1 From Exercise 40 we know that x3dx 1 = +C. ∫ (2 −x47) 24(2 − x4)6

= = = =

[−

0

x3dx

1 (2

−x47)

1 ]0 ) 24(2−x46−1 1 1 24(2 −0)4 6 − 24[2− (−1)4]6 1 1 1536 −24 63 21 − 1536 = − 512 2 ∫ 0√ 77x √ 3 1+xdx=7

7 2 = ∫ √ 3 1+x(2xdx) 2 7 = ∫ √ 3udu 2 7 1/3 = 2 ∫ u du

∫81 √x5 012 1 − x 2 dx ∫findt√heindefiniteintegral.

12x51−x2dx

√atethedefiniteintegralon[0,1].

Then∫evalu 1 012 x5 1 − x 2 d ]x = 5(1 − x2)6/5 − [10 = [ − 5(1 − 12)6/5− −5(1−02)6/5 ] =5 (H−x)5/2

∫ 0√ x √ 1+x2dx

Firstfindtheindefiniteintegral. √3 7 x 1+x dx 2 ∫ Let =u 1+x , then du = 2xdx. ∫2√ =7 1 2 2 3 · ∫ · x √1+x2dx 1 2 3 · =7 2 2x 1+xdx

315

Letu= (1−x)2∫,th√endu=−2xdx.

1 5 1 − x 2 (−2xdx) =12− ∫ √2 =− 6 ∫5udu u1/5du =−6 6/5 = − u 6 +C 6/5 = −5 u6/5 +C +C −5(1 − x2)6/5 =

x3dx

∫0

Then ∫

71.

1

7].

∫evaluatethedefiniteintegralon[0, √7 √3 x 1+ x2 0 [7 ]dx√ 7 21 (1+x2)4/3 − = ·14/3 8 0 = 21 21 = √ 8 (1+(7)2)4/3 = 8 (1+02)4/3 21 21 8 · 84/3− 21 · 21 8 16− ·1

7

73.

a)K

∫ H (H−x)3/2dx

7 = 2 ·u +C = 4/3 = 21 8 u4/3+C 21 8 (1+x2)4/3+C

∫ b)K

]H

[

=

− − 2 K H5/2 0 5 2 KH5/2 5

= 4/3

−2K 5

=

0

H/2 0

(H−x)3/2dx −2K(H−x) 5

H/2 5/2

=

[

=

− 2 K (H/2) 5/2 − − 2 K 5/2 H 5 5/2 − √ 52KH 5/2 + 20 5 − 1 √ 5/2 2)KH 20(8

= =

]

0

291

Exercise Set 5.5 c) We divide the answers from the previous two parts 1 √ 2)KH5/2 (8 20 − 2 K H 5/2 −√5 (8 2) 5 · 20 (8 − √ 2) 2

= =

8 = 0.8232

d) Upper half proportion is given by 1−

74. a)

(8

−√

2)

8

√

2 8 = 0.1768 =

∫Firstfindtheindefiniteintegral.

100,000e0.025tdt Letu=0.02

(5t,then)d∫u=0.025dt.

1 0.025t (0.025)dt e 0.025 eudu =4,000,000 ∫ =4,000,000eu+C =100,000

= 4, 000, 000 e0.025t + C Then evaluate [thedefi]niteintegralon[0,99].

0.025t99 4,000,000 e 0 =4,000,000(e0.025(99)−e0.025(0 −e0) −1) )) = 4,000,000(e2.475 = 4,000,000(e2.475 ≈ b) From ∫parta)weknowthat 43,526,828 D(t)dt=4,000,000e0.025t+C Nowevaluatet

[hedefi]niteintegralon[80,99].

0.025t99 4,000,000 e 80

=4,000,000(e0.025(99)−e0.025(80)) = 4,000,000(e2.475− e 2 ) ≈ 17,970,604

75. We have to break the integral into two parts ∫ 0 [−x √ 4−x 2 ∫ 2 [0−(− x 4− x2)] dx √ ]−0] [dx+ −2 0 2 0 1 (4−x2)3/2 −2 + − (4−x2)3/2 ] = [ 0 8 0+0− − 8 = 3− 3 = 16 3 76. We have to break the integral into two parts

0

∫

−4

[0−(x16−x2)]dx √ 0

+ [ − = [1 3(16−x2)3/2 ] −4 64 −−64 = 3 − 0+0 3 = 128 3

+ ∫ 4 [√x 0

16 − − 0] dx ]x242320

(16−x)/

77. Using the hint t2 + 2t ∫ 2dt (+1) t 1 = ∫ 1− dt (t+1)2 = 1 ∫ − (t + 1)−2 dt =t+ 1 +C (t + 1) 78. Using the hint from the previous problem x2 + 6x ∫ ( x +3) 2dx 9 = ∫ 1− (x+3) 2dx = 1 ∫ − (x + 3)−2 dx =x+ 1 +C (x + 3) 79. Using the hint x+ +3 1 + 1 dx ∫ x 2 = ∫ dx (x + 1) = x+2ln | x + 1 | + C 80. Using the hint from the previous problem t − 5 dt t− 4 = ∫ 1+ t− 1 dt − 4 =t −ln|t−4|+C ∫

81. Let u = ln x then du = dx ∫ x(ln x)n = ∫ u−ndu u−n+1 = − n+1 +C = = (lnx)−n+1 +C 1 − n1 (1 −n)(lnx)n−1 82. Using the hint ∫ = ∫

dx ex+1 e−x 1+ex dx −

= ln(1 + e−x) + C

dx x

+C

Chapter 5:

292 83. Let u =

x du= ln ∫ln(ln )then x dx x ln x [ln(ln)]x du = ∫ u =ln u + C =ln |ln(ln x)|+ C

dx

Then

x

d = dxandv=e2x. ∫u x (2 e2xdx)=xe2x−

3.

ae−mxthen du=− ame−mx dx e−mx dx 1+ ae−mx =−1 du ∫ u am = −1ln u+ C am − 1 ln|1+ ae−mx|+ C = am 85. Using th ∫e hint 84. Let u =

4.

u=xanddv=cosxdx.

Then du ∫=dxandv=sinx

5.

u=xanddv=e2xdx

∫ xe2xdx

6.

Then du = dx and v = e5x.

2.

2x e2x dx ∫ Let u = x and dv = 2e2x dx.

7.

1 e2x 2

=

1 2x ·12 2x x e ∫ − 2 e dx

=

1 2x −1e24x 2 xe

∫xe3xdx

∫ xe3xdx

1 =xe5x −e5x+C 5

.

Then du = dx and v =

u = x and dv = 5e5x dx.

uv−∫7: ∫UsingTheorem udv= vdu

− ∫ sinxdx

xe2x dx ∫ Let

Then du = dx and v =

∫

x·sinx

= xsinx cosx+C −

5xe5xdx= x(5e5xdx) ∫ ∫ Let

∫

=

xcosxdx

Exercise Set 5.6

v du e5x·dx

∫ cosxdx

x cos x dx ∫ Let

u=xanddv=e3xdx

u dv u v x (5 e5xdx )= x · e5x−

x·−cosx+

−xcosx+sinx+C = sinx −xcosx+C

csc x dx

1.

= =

86. Using th ∫e hint from the previous problem

87. Left to the student

xsinxdx Let

xsinxdx

2

2 csc x cot x+ csc x = ∫ dx cotx + cscx 2 Let u = cotx + cscx then du=−( cscx + cotxcscx)dx du =− ∫ u = −ln u + C = −ln| cscx + cotx | + C

∫

Then du ∫=dxandv=−cosx

sec x dx sec x tan x+ sec x dx tanx + secx 2 Let u = tanx + secx then du= secx + secxtanxdx du = ∫ u =ln u+ C =ln | secx + tanx | +C

e2xdx ∫ =xe2x − 1e2x + C 2

u=xanddv=sinxdx.

∫1+

= ∫

Integration

.

1 e2x 3

=

1 3x ·13 3x x e ∫ − 3 e dx

=

1 3x −1e39x 3 xe

xe−2x dx ∫ Let u=xanddv

=e−2x .dx

1 − 2x Then du = dx and v e (=−21 ) . x 1 2x −2 x e−2x dx= x · ∫ ( − 2 e− ) dx e − − 2 ∫

Exercise Set 5.6

293

−1xe−2x− = 2

−1 2−2x+ −e

12.

C

b ( ∫ beaxdx= eax+C )

8.

∫

= −1xe−2x − 1e−2x + C 2 4

2

3

x ln x dx .

Let =ln x 3and dv = x 2 dx Thenu 3 3 x du= dx and v = .

3 x ∫

∫

xe−xdx

3 3 3 x ln x x 3 ∫ dx − 3 3 ·x 3 ln 3 2 x = x − ∫ x dx 3 = 3 3 3 x ln x 3 x − 3 + C, or +

2 3 x ln x dx =

Letu=xanddv=e−xdx.

Then d ∫u=dxandv=−e−x. − − xexdx=−xex−

∫ −e−xdx +C =−∫xe − x − e − x

9.

∫

x2lnxdx=

(lnx)x2dx

u=lnxanddv=x2dx.

1 x3 dxandv= .

Thendu= 3 x u u v dv ∫

(lnx)x2

dx=lnx

· 3

10.

∫

x3lnxdx

13.

∫

ln( x +3)

dx

Let u=ln( +3)and x

v

dv = .dx 1 Then du= dx and v= +3x x +3 ∫

du

3 · 1 dx − ∫x3 x 1 ∫IntegrationbyParts 3 x2dx

x = lnx− 3 1 = = 3 x3 x3 lnx − · 3 +C lnx 3 − x3 x3 +C 3 9

ln( x +3) dx = =

14.

x4 1 du= dxandv= .

4 x 4 x4 lnx x ·1 3 dx x ln xdx = ∫ − 4 ∫ x = x4lnx dx − ∫ = x4 ln x x4 +C − 4 16 ∫ xlnx2dx

∫

2

x ∫ (ln x2) xdx =(ln x 2 · x − 2 ∫ 2 x2 lnx2 − ∫ xdx = x2 x2 lnx2 − 2 +C = 2

∫

− x+ C

x +1.(Choosing

ln(x+1)dx=(x+1)ln(

(x+2)lnxdx=

x

C =1)

x +1 dx ∫ x

x +1)−

=(∫x+1)ln(

− ∫ dx +1)−x + C

(ln x )( x +2) dx

u=lnxanddv=(x+2)

dx.

Then du=

(x+2)2 1 dxandv= .

2 x u=lnxanddv=(x+2)dx.

Let

2

dxandv=

=( +1)ln( +1) x x 15.

∫

dv = dx.

Then du=

Then

dx ∫ x +3· x +3) dx

x +3)−

( x +3)ln( x +3)−

∫ln(x+1)dx 1 x+1

u=lnx2anddv=xdx.

2 1 dxandv= Then du= 2 · 2xdx= x )

ln( x +3)·(

= ( +3)ln( x +3) x

Letu=lnxanddv=x3dx.

11.

3

3

x ·3ln x x C − 3 3 3 x + = x 3ln x − C 3

x2 . · 2 x dx

Then 1 dxandv= ∫du= x (lnx)(x+2)dx( = (lnx) =

·

(x+2)2

x+2)2

2

− ∫

2 x2+4x+4 lnx − ∫ 2

.

(x+2)2 2

·

1 x dx

x2+4x+4 2x

dx

294

Chapter 5:

= = = = = =

16.

x2+4 x +4

2 x ln x − ∫ ( +2+ x ) dx ln 2 2 x2 +4 x +4 1 1 x − ∫ xdx − 2 ∫ dx − 2 ∫ x dx 1 C 2 C x2+4 x +4 2 x2 + ln x − · − 2 · x x 1 2 x −2·ln x x 2 2 −2· + x2+4x +4 −2·ln · x ln − 2 2 2 2 x2+4 x +4 x x ) −2ln ( − 4 −2x + C 2 2 2 x ( x +2 x ) ln x − x C 2 4 −2

x +1)ln ∫( Let

xdx

u =ln and x dv =( +1) x . dx Then 1 ( x +1)2 . du= dx and v = ∫x 2 ( x +1)ln xdx ( x +1) 2 · 1 ( x +1)2 ln x = dx ∫ − = 2 x = +2 = ∫ 2 ∫ 2 = ( 2 x +1)ln x x +2 x +1 = x dx − 2 2x − − 1 ( x 2 +2 x +1)ln x x ( 2 +1+ 2x ) dx 2 (2+2 x

x +1)ln 2

18.

sinxdx

dv = sinxdx .

=

( x − 1 ) · − cosx + ∫ cosxdx

=

(1 − x ) cosx + sinx + C

(x −2) cosxdx Let u = x −2and dv = cosxdx . ∫ Then du= dx and v = sinx ∫ ( x −2) cosxdx

19.

x

( x −1) sinxdx Let u = x ∫ Then du= dx and =v − cosx ∫ ( x −1)

√xx+2dx ∫ Let u=xanddv=

du= dx and v = u ∫

=

( x −2)· sinx − ∫ sinxdx

=

( x −2) sinx + cosx + C

√ x+2dx=(x+2)1/2dx.

( x+2)3/2 3/ 2

2 = 3 ( x +2)/ 32.

dv

x √ +x 2

dx

2 · 2 ( +2)3/2− ( x +2)3/2 dx x x ∫ = 3 3 2 2 +2)3/2− +2)3/2 dx ( ( xx x ∫ =2 2 3· ( x +2)5/2 = x( x +2)3/2 + C +2)3/2 − 3 52 / 2 4 = x( x − ( +2)5/2+ x C 3 15 +4 20. x√+4 x dx ∫ Let √ u = x and dv = x dx.

Then 2 3/2 . du= dx and v = (+4) x 3 √ x x +4 dx ∫ 2 ( +4)3/2− ∫ ( x +4)3/2 dx = xx 2 2 3 2 = x( x +4)3/2 − · 5 ( x +4)5/2+ C +4)3/2 3 =

2

x −1ln x x+ C 4 − 2 x 2 2 x x + 2x + 1 − 1 ln x+ C − − ) ( 4 − )2 C 2 2 −1and ( x + x ln x x − x + 2 4 17.

Then

Integration

21.

2 x( x 3

4 − ( +4)5/2+ x 15

C

x3ln2 xdx = (ln2)(3 x x dx) ∫ ∫ Let u=ln2 and x

. dv = x 3 dx 4 1 x Then du= =. dx and v 4 x ∫ 4 4 1 x x (ln2)(3 dx x x dx)=(ln2) x · − 4 41 ∫ ∫ 4 · x x 3 = ln2 x − x dx 4 4 = 4 − 1 4 − = x x + C ln2 x 4 · 4 4 + 4 4 x x ln2 x C 4 16 ∫ 22. x 2ln5 xdx Let . u =ln5 x and dv = x 2 dx Then 3 1 x du= dx and v = .

33 x 2 x x ln5 x 2 x ln5 xdx = dx ∫ − 3 3 ∫ =

3 3 x ln5 x − x + C 3 9