Problemas Impares - Copia

8/12/2019 Problemas Impares - Copia

1/33

C H A P T E R 4

Integration


2/33

177

1. d

dx3

x3 C ddx3x3 C 9x4

9

x4 3.

d

dx1

3x3 4x C x2 4 x 2x 2

5.

Check: d

dtt3 C 3t2

y t3 C

dy

dt 3t2 7.

Check: d

dx2

5x52 C x32

y 2

5x52 C

dy

dxx32

Given Rewrite Integrate Simplify

9.3

4x 43 C

x43

43 Cx13dx3xdx

11. 2

x C

x12

12 Cx32dx 1

xxdx

13. 1

4x2 C

1

2x2

2 C1

2x3dx 12x3dx

15.

Check: d

dxx2

2 3x C x 3

x 3dx x2

2 3x C 17.

Check: d

dxx2 x3 C 2x 3x2

2x 3x2dx x2 x3 C

19.

Check: d

dx1

4x 4 2x C x3 2

x3 2dx 14x4 2x C 21.Check:

d

dx2

5x52 x2 x C x32 2x 1

x32 2x 1dx 25x52 x2 x C

23.

Check: d

dx3

5x53 C x23 3x2

3

x2

dx x

23

dx

x53

53 C3

5x53

C 25.

Check: d

dx1

2x2 C 1x3

1

x3dx x3

dx x2

2 C 1

2x2 C


3/33

178 Chapter 4 Integration

27.

Check: x2 x 1

x

d

dx2

5x52

2

3x32 2x12 C x32 x12 x12

2

15x123x2 5x 15 C

2

5x52

2

3x32 2x12 Cx2 x 1

xdx x32 x12 x12dx

29.

Check:

x 13x 2

d

dxx3 1

2x2 2x C 3x2 x 2

x3 1

2x2 2x C

x 13x 2dx 3x2 x 2dx 31.Check:

d

dy2

7y72 C y52 y2y

y2ydy y52dy 27y72 C

33.

Check: d

dxx C 1

dx 1 dx x C 35.Check:

d

dx2 cosx 3 sinx C 2 sinx 3 cosx

2 sinx 3 cosxdx 2 cosx 3 sinx C

37.

Check: d

dtt csc t C 1 csc tcot t

1 csc tcot tdt t csc t C 39.Check:

d

dtan cos C sec2 sin

sec2 sin d tan cos C

41.

Check: d

dytany C sec2

y tan2

y 1

tan2y 1dy sec2ydy tany C 43.

2

2

3

3

2 2

x

3C

2C

0C

y

fx cosx

45.

Answers will vary.

5

4

3

32123

y

x

2x)x)

22x) )xf

ff

fx 2x C

fx 2 47.

Answers will vary.

3

4

3

2

231

2

x

y3

3

3

xxx)f

xx3

x)f )3

)

f

fx x x3

3 C

fx 1 x2 49.

y

x

2

x

1

1 12 1 C C 1

y 2x 1dx x2 x C

dy

dx 2x 1, 1, 1


4/33

Section 4.1 Antiderivatives and Indefinite Integration 179

51.

y

sinx

4

4 sin 0 C C 4

y cosxdx sinx C

dy

dx cosx, 0, 4

53. (a) Answers will vary.

(b)

y x2

4

x 2

2 C

2 42

4 4 C

y x2

4 x C

4 8

2

6dy

dx

1

2x 1, 4, 2

x

y

3 5

3

5

55.

fx 2x2 6

f0 6 202 CC 6

fx 4xdx 2x2 Cfx 4x,f0 6 57.

ht 2t4 5t 11

h1 4 2 5 CC 11

ht 8t3 5dt 2t4 5t Cht 8t3 5, h1 4

59.

fx x2 x 4

f2 6 C2 10 C

2 4

fx 2x 1dx x2 x C2fx 2x 1

f2 4 C1 5 C

1 1

fx 2 dx 2x C1f2 10

f2 5

fx 2 61.

fx 4x12 3x 4x 3x

f0 0 0 C2 0 C

2 0

fx 2x12 3dx 4x12 3x C2fx

2

x 3

f4 2

2 C

1 2 C

1 3

fx x32dx 2x12 C1 2x

C1

f0 0

f4 2

fx x32

63. (a)

(b) h6 0.7562 56 12 69 cm

ht 0.75t2 5t 12

h0 0 0 C 12 C 12

ht 1.5t 5dt 0.75t2 5t C


5/33


65. Graph of is given.

(a)

(b) No. The slopes of the tangent lines are greater than 2

on Therefore,fmust increase more than 4 units

on

(c) No, becausefis decreasing on

(d) fis an maximum at because and

the first derivative test.

(e) fis concave upward when is increasing on

and fis concave downward on Points

of inflection atx 1, 5.

1, 5.5,., 1f

f3.5 0x 3.5

4, 5.f5 < f4

0, 4.0, 2.

f4 1.0

ff0 4.

(f ) is a minimum at

(g)

x

2

2

4

4

6

6 82

6

y

x 3.f

67.

The ball reaches its maximim height when

s15

18

1615

8

2

6015

8

6 62.26 feet

t15

8seconds

32t 60

vt 32t 60 0

st 16t2 60t 6 Position function

s0 6 C2

st 32t 60dt 16t2 60t C

2

v0 60 C1

vt 32 dt 32t C1at 32 ftsec2 69. From Exercise 68, we have:

when time to reach

maximum height.

v0 187.617 ftsec

v02 35,200

v02

64

v02

32 550

s v032 16v0

322

v0 v032 550

tv0

32st 32t v

0 0

st 16t2 v0t

71.

f0 s0 C

2ft 4.9t2 v

0t s

0

ft 9.8t v0dt 4.9t2 v0t C2v0 v

0 C

1 vt 9.8t v

0

vt 9.8 dt 9.8t C1at 9.8 73. From Exercise 71,

(Maximum height when )

f 109.8 7.1 m

t10

9.8

9.8t 10

v0.v t 9.8t 10 0

ft 4.9t2 10t 2.

75.

since the stone was dropped,

Thus, the height of the cliff is 320 meters.

v20 32 msec

vt 1.6t

s0 320

s20 0 0.8202 s0 0

st 1.6tdt 0.8t2 s0v0 0.vt 1.6 dt 1.6t v0 1.6t,

a 1.6


6/33

Section 4.1 Antiderivatives and Indefinite Integration 181

77.

(a)

(b) when or(c) when

v2 311 3

t 2.at 6t 2 03 < t < 5.0 < t < 1vt > 0

at vt 6t 12 6t 2

3t2 4t 3 3t 1t 3

vt xt 3t2 12t 9

0 t 5xt t3 6t2 9t 2 79.

at vt 1

2t32

1

2t32acceleration

xt 2t12 2 position function

x1 4 21 C C 2

xt vtdt 2t12 Ct > 0vt

1

t t12

81. (a)

(b)

s13 275

234132

2

250

3613 189.58 m

st at2

2

250

36t s0 0

a 550

468

275

234 1.175 msec2

550

36 13a

v13 800

36 13a

250

36

v0 250

36 vt at250

36

vt at C

at a constant acceleration

v13 80 kmhr 80 1000

3600

800

36msec

v0 25 kmhr 25 1000

3600

250

36msec 83. Truck:

Automobile:

At the point where the automobile overtakes the truck:

when sec.

(a)

(b) v10 610 60 ftsec 41 mph

s10 3102 300 ft

t 100 3tt 10

0 3t2 30t

30t 3t2

st 3t2Let s0 0.

vt 6tLet v0 0.

at 6

st 30tLet s0 0.

vt 30

85.

(a) (b)

(c)

The second car was going faster than the first until the end.

S230 1970.3 feet

S130 953.5 feet

In both cases, the constant of integration is 0 because S10 S

20 0

S2t V2tdt

0.1208t3

3 6.7991t2

2 0.0707t

S1t V1tdt 0.10683 t3 0.04162 t2 0.3679t

V2t 0.1208t2 6.7991t 0.0707

V1

t 0.1068t2 0.0416t 0.3679

1 mihr5280 ftmi

3600 sechr

22

15ftsec

t 0 5 10 15 20 25 30

0 3.67 10.27 23.47 42.53 66 95.33

0 30.8 55.73 74.8 88 93.87 95.33V2ftsec

V1ftsec


7/33

87.

since

At the time of lift-off, and Since

7.45 ftsec2.

18,285.714 mihr2

1.4k 1602 k1602

1.4

v1.4k k1.4k 160t

1.4

k

k2t2 0.7,k2t2 0.7.kt 160

v0 s0 0.st k

2t2

vt kt

at k

89. True 91. True

93. False. For example, becausex3

3 C x

2

2 C

1x2

2 C

2x xdx xdx xdx

95.

is continuous: Values must agree at

The left and right hand derivatives at do not agree. Hence is not differentiable atx 2.fx 2

fx x 2,

3x2 2,

2

0 x < 2

2 x 5

4 6 C2C2 2

x 2:f

f1 31 C1 3C

1 2

fx x C

1,

3x2 C

22,

0 x < 2

2 x 5

fx 1,3x,0 x < 2

2 x 5



8/33

Section 4.2 Area 183

19.

12,040

14,400 2,480 120

152162

4 2

151631

6

1516

2

15

i1

ii 12 15

i1

i 3 215

i1

i 2 15

i1

i 21. sum seq (TI-82)

202141

6 60 2930

20

i1

i2 3 2020 1220 1

6 320

2 3,x, 1, 20, 1 2930x

23.

s 1 3 4 921 252 12.5

S 3 4 92 51 332 16.5 25.

s 2 2 31 7

S 3 3 51 11

27.

s4 014 1

41

4 1

21

4 3

41

4 1 2 3

8 0.518

S4 141

4 1

21

4 3

41

4 11

4 1 2 3 2

8 0.768

29.

s5 1

651

5 1

751

5 1

851

5 1

951

5 1

21

5 1

6

1

7

1

8

1

9

1

10 0.646

S5 115 16515 17515 18515 19515 15 16 17 18 19 0.746

31. limn

81n4n2n 12

4 81

4limn

n4 2n3 n2

n4 81

41

81

4

33. limn

18n2nn 1

2 18

2limn

n2 n

n2 18

21 9

39. 8 limn

1 1n 8 limn8n2 n

n2 limn16

n2nn 1

2 limnn

i1

16in2 limn16

n2

n

i1

i

35.

S10,000 1.0002

S1000 1.002

S100 1.02

S10 12

10 1.2

n

i1

2i 1n2

1n2

n

i1

2i 1 1n22nn 12 n n 2n Sn

37.

S10,000 1.99999998

S1000 1.999998

S100 1.9998

S10 1.98

6n22n

2 3n

1

3n

36 1n22n2 2 Sn

n

k1

6kk 1

n3

6

n3

n

k1

k2 k 6

n3nn 12n 1

6

nn 1

2

>


9/33


41.

limn

162 3n 1n2

1 1

3 lim

n

1

62n3 3n2 n

n3

limn

1

n3n 1n2n 1

6 limnn

i1

1

n3i 12 lim

n

1

n3n1

i1

i 2

43. 2 limn1 n2 n

2n2 21 1

2 3 2 limn1

nn 1

nnn 1

2 limnn

i11 i

n2

n 2 limn1

nn

i11

1

n

n

i1i

45. (a)

(b)

Endpoints:

(c) Since is increasing, on

(d) on

Sn

n

i1

fxix

n

i1

f2i

n 2

n

n

i1i2

n2

n

xi1

,xifM

i fx

i

n

i1

f2i 2n

2

n n

i1

i 12n2

n

sn n

i1

fx

i1x

xi1

,xi.fm

i fx

i1y x

0 < 12n< 22

n< . . .< n 12

n< n2

n 2

x 2 0

n

2

n

x31

3

2

1

y (e)

(f)

limn

2n 1

n 2

limn

4n2nn 1

2

limn

n

i1

i2n2

n limn4

n2n

i1

i

limn

2n 1n 4

n 2

limn

4

n2nn 1

2 n

limn

n

i1

i 12n2

n limn4

n2n

i1

i 1

x 5 10 50 100

1.6 1.8 1.96 1.98

2.4 2.2 2.04 2.02Sn

sn

47. on

Area limn

sn 2

3 2

n2n

i1

i 3 2n 1n

2n2 2

1

n

sn n

i1

f in

1

n n

i1

2 in 31

n

321x

2

1

yNote: x 1 0n 1

n0, 1.y 2x 3

49. on

Area limn

Sn 7

3

1n3n

i1

i 2 2 nn 12n 16n3 2 1

62 3

n

1

n2 2

Sn n

i1

f in1

n n

i1

in2

21n

x2 3

3

1

1

y

Note: x

1

n0, 1.y x2 2


10/33


51. on

Area limn

sn 30 8

3 4

70

3 23

1

3

30 8

6n2n 12n 1

4

nn 1

2

n15n 4n2

nn 12n 1

6

4

nnn 1

2

2

n

n

i1

15 4i2

n2

4i

n

sn n

i1

f1 2in 2

n n

i1

16 1 2in 2

2n

x

y

11

2

4

6

8

10

12

14

18

2 3

1, 3. Note:x 2ny 16 x2

53. on

Area limn

sn 189 81

4 27

27

2

513

4 128.25

189 81

4n2n 12

81

6n2n 12n 1

27

2n 1

n

3

n63n 27

n3n2n 12

4

27

n2nn 12n 1

6

9

nnn 1

2

3

n

n

i1

63 27i3

n3

27i2

n2

9i

n

sn n

i1

f1 3in 3

n n

i1

64 1 3in 3

3n

x5 6

30

y

40

50

60

70

20

10

1 21 3

1, 4. Note:x 4 1n 3

ny 64 x3

55. on

Again, is neither an upper nor a lower sum.

Area limn

Tn 4 10 32

3 4

2

3

4 101 1

n 16

32 3

n

1

n2 41 2

n

1

n2

4

nn

20

n2

nn 1

2

32

n3

nn 12n 1

6

16

n4

n2n 12

4

n

i1

2 10in 16i 2

n2

8i3

n3 2

n 4

nn

i1

1 20

n2n

i1

i 32

n3n

i1

i 2 16

n4n

i1

i 3

n

i1

1 4in 4i2

n2 1 6i

n

12i 2

n2

8i3

n3 2

n

Tn n

i1

f1 2in

2

n n

i1

1 2in 2

1 2in 3

2n

Tn

x1

2

1

1

yNote: x 1 1n 2

n1, 1.y x2 x3


11/33


57.

Area limn

Sn limn

6 6n 6

6n 1

n 6

6

n

12

n2

n

i1

i 12n2 nn 1

2

n

i1

f2in

2

n n

i1

32in 2

nSn n

i1

fm

iy

64x

4

2

2

2

yNote:y 2 0n 2

n0 y 2fy 3y,

59.

Area limn

Sn limn

9 272n

9

2n2 9

9

n22n2 3n 1

2 9 27

2n

9

2n2

27

n3

nn 12n 1

6

n

i1

3in 2

3n 27

n3

n

i1

i2Sn n

i1

f3in

3

nx

2 4 6 8 10

2

4

6

2

4

yNote:y 3 0n 3

n0 y 3fy y2,

61.

Area limn

Sn 6 10 8

3 4

44

3

2

n

n

i1

3 10in 4i2

n2

8i3

n3 2

n3n 10

nnn 1

2

4

n2nn 12n 1

6

8

n2n2n 12

4

2

n

n

i1

41 4in 4i2

n2 1 6i

n

12i2

n2

8i3

n3

n

i1

41 2in 2

1 2in 3

2n

Sn n

i1

g1 2in 2

n

x

6

y

8

10

2

2

4

24

gy 4y2 y3, 1 y 3. Note: y 3 1n 2

n

63.

Let

69

8

1

21

16 3 9

16 3 25

16 3 49

16 3

Area n

i1

fcix

4

i1

ci

2 312

c4

7

4c3

5

4,c

2

3

4,c

1

1

4,x

1

2,

ci

xix

i1

2.

n 40 x 2,fx x2 3, 65.

Let

16tan 32 tan 332 tan 532 tan 732 0.345

Area n

i1

fcix

4

i1

tan ci16

c4

7

32c3

5

32,c

2

3

32,c

1

32,x

16,

ci

xix

i1

2.

n 40 x

4,fx tanx,

67. on

Exact value is 163

0, 4.fx x

n 4 8 12 16 20

5.3838 5.3523 5.3439 5.3403 5.3384Approximate area


12/33


69. on 1, 3.fx tanx8n 4 8 12 16 20

2.2223 2.2387 2.2418 2.2430 2.2435Approximate area

71. We can use the line bounded by and

The sum of the areas of these inscribed rectangles is the

lower sum.

x

y

a b

x b.x ay x The sum of the areas of these circumscribed rectangles is the

upper sum.

We can see that the rectangles do not contain all of the area in

the first graph and the rectangles in the second graph covermore than the area of the region.

The exact value of the area lies between these two sums.

x

y

a b

n 4 8 20 100 200

15.333 17.368 18.459 18.995 19.06

21.733 20.568 19.739 19.251 19.188

19.403 19.201 19.137 19.125 19.125Mn

Sn

sn

73. (a)

Lower sum:

(c)

Midpoint Rule:

(e)

(f) increases because the lower sum approaches the exact value as n increases. decreases because the upper sum

approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the

exact value, whereas the upper sum is always larger.

Snsn

M4 223 4

45 5

57 6

29

6112315 19.403

x

1

2

2 3

4

4

6

8

y

s4 0 4 513 6 15

13

463 15.333

x

1

2

2 3

4

4

6

8

y (b)

Upper sum:

(d) In each case, The lower sum uses left end-

points, The upper sum uses right endpoints,

The Midpoint Rule uses midpoints,i 124n.i4n.i 14n.

x 4n.

S4 4 513 6 6

25 21

1115

32615 21.733

x

1

2

2 3

4

4

6

8

y


13/33

188 Chapter 4 Differentiation

Section 4.3 Riemann Sums and Definite Integrals

75.

b. square unitsA 6

x1 2 3 4

1

2

3

4

y 77. True. (Theorem 4.2 (2))

79.

Let area bounded by thex-axis, and Let area of the rect-

angle bounded by and Thus,

In this program, the computer is generating pairs of random points in the rectangle whose

area is represented by It is keeping track of how many of these points, lie in the region

whose area is represented by Since the points are randomly generated, we assume that

The larger is the better the approximation toA1.N

2

A1

A2

N

1

N2

A1

N1

N2

A2.

A1.

N1,A

2.

N2

A2 21 1.570796.x 2.x 0,y 0,y 1,

A2x 2.x 0fx sinx,A

1

x

0.25

0.5

0.75

1.0

4

2

2( ), 1f x x( ) sin( )=

y0, 2fx sinx,

81. Suppose there are n rows in the figure. The stars on the left total as do the stars on the right. There are

stars in total, hence

1 2 . . . n 12nn 1.

21 2 . . . n nn 1

nn 11 2 . . . n,

83. (a)

(c) Using the integration capability of a graphing utility,

you obtain

A 76,897.5 ft2.

y 4.09 105x3 0.016x2 2.67x 452.9 (b)

0

0 350

500

1.

3323 0 23 3.464

limn

33n 12n 13n2 n 1

2n2

limn

33

n32nn 12n 1

6

nn 1

2

limn

33

n3 n

i1

2i2 i

limn

n

i1

fcix

i lim

n

n

i1

3i2

n2

3

n22i 1

xi

3i2

n2

3i 12

n2

3

n22i 1

3n2

3(2)2

n23(n1)2

n2. . .

. . .

3

1

2

3

y

x

y= x

fx x,y 0,x 0,x 3, ci

3i2

n2


14/33

Section 4.3 Riemann Sums and Definite Integrals 189

3. on

104

6 dx limn

36 36

n

i1

66n n

i1

36

n 36

n

i1

fcix

i

n

i1

f4 6in 6

n

Note: x 10 4n 6

n, 0 as n4, 10.y 6

5. on

1

1

x3dx limn

2

n

0

2 61 1n 42 3

n

1

n2 41 2

n

1

n2 2

n

2 12

n2n

i1

i 24

n3

n

i1

i2 16

n4

n

i1

i3

n

i1

fc

ix

i

n

i1

f1 2in 2

n n

i11 2in

3

2n n

i11 6in

12i 2

n2

8i3

n32

n

Note:x 1 1n 2

n, 0 as n1, 1.y x3

7. on

21

x2 1dx limn

103 3

2n

1

6n2 10

3

2 2

n2

n

i1

i 1

n3

n

i1

i2 2 1 1n 1

62 3

n

1

n2 10

3

3

2n

1

6n2

n

i1

fc

ix

i

n

i1

f1 in1

n n

i11 in

2

11n n

i11 2in

i2

n2 11n

Note:x 2 1n 1

n, 0 as n1, 2.y x2 1

9.

on the interval 1, 5.

lim0

n

i1

3ci 10x

i 5

1

3x 10dx 11.

on the interval 0, 3.

lim0

n

i1

ci2 4 x

i 3

0

x2 4 dx

13.50

3 dx 15.44

4 xdx 17.22

4 x2dx

19.0

sinxdx 21.20

y3dy 23. Rectangle

x5

5

3

2

42 31

1

Rectangle

y

A 3

0

4 dx 12

A bh 34


15/33


25. Triangle

A 40

xdx 8

A 1

2bh

1

244

x

Triangle

4

2

42

y27. Trapezoid

A 20

2x 5dx 14

A b1 b

2

2 h 5 92 2

x

9

6

321

3

Trapezoid

y

29. Triangle

A 11

1 xdx 1

A 1

2bh

1

221

1 Triangle

11x

y 31. Semicircle

A 33

9 x2dx 9

2

A 1

2r2

1

232

x

4 Semicircle

4224

y

33.24

xdx 42

xdx 6 35.42

4xdx 442

xdx 46 24

In Exercises 3339,42

x3dx 60, 42

xdx 6, 42

dx 2

37.42

x 8dx 42

xdx 842

dx 6 82 10 39.

1

260 36 22 16

42

12x3 3x 2dx 1

24

2

x3 dx 342

xdx 242

dx

41. (a)

(b)

(c)

(d)50

3fxdx 350

fxdx 310 30

55

fxdx 0

0

5

fxdx

5

0

fxdx 10

70

fxdx 50

fxdx 75

fxdx 10 3 13 43. (a)

(b)

(c)

(d)62

3fxdx 362

fxdx 310 30

62

2gxdx 262

gxdx 22 4

2 10 12

6

2

g x fxdx 6

2

gxdx 6

2

fxdx

10 2 8

62

fx gxdx 62

fxdx 62

gxdx

45. (a) Quarter circle belowx-axis:

(b) Triangle:

(c) Triangle Semicircle belowx-axis:

(d) Sum of parts (b) and (c):

(e) Sum of absolute values of (b) and (c):

(f) Answer to (d) plus 3 2 20 23 2210 20:

4 1 2 5 2

4 1 2 3 2

1

2

21 1

222 1 2

12bh

12 42 4

14r

2 142

2

47. The left endpoint approximation will be greater than the

actual area: >

49. Because the curve is concave upward, the midpoint

approximation will be less than the actual area:


16/33

Section 4.3 Riemann Sums and Definite Integrals 191

51.

is not integrable on the interval andfhas a

discontinuity atx 4.

3, 5

fx 1

x 4

53.

a. square unitsA 5

x1 2 3 4

1

2

3

4

y

55.

d.1

0

2 sin x dx1

212 1

y

x1

1

1 2

2

2

12

32

32

57.30

x3 xdx

4 8 12 16 20

3.6830 3.9956 4.0707 4.1016 4.1177

4.3082 4.2076 4.1838 4.1740 4.1690

3.6830 3.9956 4.0707 4.1016 4.1177Rn

Mn

Ln

n

59.2

0

sin2xdx

4 8 12 16 20

0.5890 0.6872 0.7199 0.7363 0.7461

0.7854 0.7854 0.7854 0.7854 0.7854

0.9817 0.8836 0.8508 0.8345 0.8247Rn

Mn

Ln

n

61. True 63. True 65. False

2

0

xdx 2

67.

41 102 404 881 272

4

i1

fc

ix f1x

1f2x

2f5x

3f8x

4

c4 8c

3 5,c

2 2,c

1 1,

x4 1x

3 4,x

2 2,x

1 1,

x4 8x

3 7,x

2 3,x

1 1,x

0 0,

0, 8fx x2 3x,


17/33

192 Chapter 4 Differentiation

Section 4.4 The Fundamental Theorem of Calculus

69.

is not integrable on the interval As or in each subinterval since there

are an infinite number of both rational and irrational numbers in any interval, no matter how small.

fci 0fc

i 10,0, 1.

fx 1,0,xis rational

xis irrational

71. Let and The appropriate Riemann Sum is

limn

2n2 3n 1

6n2 lim

n

13 1

2n

1

6n2 1

3

limn

1

n312 22 32 . . . n2 lim

n

1

n3

n2n 1n 1

6

n

i1

fcix

i

n

i1

in2 1

n

1

n3n

i1

i 2.

xi 1n.fx x2, 0 x 1,

1.

is positive.

0

4

x2 1dx

2

5 5

fx 4

x2 1 3.

2

2

xx2 1 dx 0

5

5 5

fx xx2 1

5.10

2xdx x21

0

1 0 1 7.01

x 2dx x2

2 2x

0

1

0 12 2 5

2

9.11

t2 2dt t3

3 2t

1

1

13 2 1

3 2 103

11.

1

0

2t 12dt

1

0

4t2 4t 1dt 43

t3 2t2 t1

0

4

3

2 1 1

3

13.21

3x2 1dx 3

xx

2

1

32 2 3 1 1

2

15.41

u 2

udu 4

1

u12 2u12du 23u32 4u124

1

2343 44 23 4

2

3

17.11

3t 2dt 34t43 2t1

1

34 2 3

4 2 4

19.

1

0

x x

3

dx 1

3

1

0

x x12dx 1

3

x2

2

2

3

x32

1

0

1

3

1

2

2

3

1

18

21.01

t13 t23dt 34t43 3

5t53

0

1

0 34 3

5 27

20

23. split up the integral at the zero

3x x232

0

x2 3x3

32 92

9

4 0 9 9 9

4

9

2 29

2

9

4 9

2

x 3

23

0

2x 3dx 320

3 2xdx 332

2x 3dx


18/33

Section 4.4 The Fundamental Theorem of Calculus 193

25.

23

3

8 83 9 12 8

3 8

4x x3

3 2

0

x3

3 4x

3

2

30

x2 4dx 20

4 x2dx 32

x2 4dx

27.0

1 sinxdx x cosx

0

1 0 1 2

29.66

sec2xdx tanx6

63

3 33

23

3

31.33

4 sec tan d 4 sec 3

3 42 42 0

33.30

10,000t 6dt 10,000t2

2 6t

3

0

$135,000 35. A 10

x x2dx x2

2

x3

3 1

0

1

6

37. A 30

3 xxdx 30

3x12 x32dx 2x32 25x523

0

xx5 10 2x3

0

123

5

39. A 20

cosxdx sinx2

0

1 41. Since on

A 20

3x2 1dx x3 x2

0

8 2 10.

0, 2,y 0

43. Since on

A 20

x3 xdx x4

4

x2

2 2

0

4 2 6.

0, 2,y 0 45.

c 0.4380 or c 1.7908

c 1 6 423 2

c 1 6 423

c 12 6 42

3

c 2c 1 3 42

3 1

c 2c 3 42

3

fc2 0 6 82

3

20

x 2xdx x2

2

4x32

3 2

0

2 82

3


19/33


47.

arccos

2 0.4817

c arcsec 2

sec c 2

sec2c 4

2 sec2c 8

fc4

4 4

44

2 sec2xdx 2 tanx4

4 21 21 4

49.

Average value

when orx 23

3 1.155.x2 4

8

34 x2

8

3

8

3

5

1

3 3

2 3

3 3

8,( ( 2 33 38,( (1

2 2

2

2

4 x2dx 1

44x 1

3x3

2

2

1

48 8

3 8 8

3 8

3

51.

x 0.690, 2.451

sinx 2

1

2

2

0.690,

2.451,

((

((

2

2

3

2

Average value 2

1

0

0

sinxdx 1cosx

0

2

53. If is continuous on and

then ba

fxdx Fb Fa.

Fx fxon a, b,a, bf

55.20

fxdx area of regionA 1.5 57.60

fxdx 20

fxdx 62

fxdx 1.5 5.0 6.5

59.

12 3.5 15.5

60

2 fxdx 60

2 dx 60

fxdx

61. (a)

Fx 500 sec2x

F0 k 500

Fx ksec2x (b)

newtons

newtons 827

826.99

1500

3 0

1

3 03

0

500 sec2xdx 1500

tanx3

0

63.1

5 05

0

0.1729t 0.1552t2 0.0374t3dt1

50.08645t2 0.05073t3 0.00935t45

0

0.5318 liter


20/33

Section 4.4 The Fundamental Theorem of Calculus 195

65. (a)

The area above the -axis equals the area below the

-axis. Thus, the average value is zero.x

x

24

1

0

1

67. (a)

(b)

(c) meters600

vtdt 8.61 104t4

4

0.0782t3

3

0.208t2

2 0.0952t

60

0

2476

70

10

10

90

v 8.61 104t3 0.0782t2 0.208t 0.0952

(b)

The average value of appears to be g.S

24

0

0

10

69.

F8 64

2 58 8

F5 25

2 55

25

2

F2 4

2 52 8

Fx x0

t 5dt t2

2 5t

x

0

x2

2 5x 71.

F8 1078 35

4

F5 1045 8

F2 1012 5

10

x 10 101 1x

Fx x1

10

v2dv x

1

10v2 dv 10

v x

1

73.

F8 sin 8 sin 1 0.1479

F5 sin 5 sin 1 1.8004

F2 sin 2 sin 1 0.0678

Fx x1

cos d sin x

1

sinx sin 1 75. (a)

(b) d

dx1

2x2 2x x 2

x0

t 2dt t2

2 2t

x

0

1

2x2 2x

77. (a)

(b) d

dx3

4x43 12 x13 3x

x8

3tdt 34t43x

8

3

4x43 16

3

4x43 12 79. (a)

(b) d

dxtanx 1 sec2x

xx4

sec2tdt tan tx

x4 tanx 1

81.

Fx x2 2x

Fx x2

t2 2tdt 83.

Fx x4 1

Fx x1

t4 1 dt 85.

Fx xcosx

Fx x0

tcos tdt


21/33


87.

Fx 8

8x 10

2x 22 x 2 2x2 x

2t2 tx2

x

Fx x2x

4t 1dt Alternate solution:

Fx 4x 1 4x 2 1 8

x

0

4t 1dt x2

0

4t 1dt

0x

4t 1dt x20

4t 1dt

Fx x2x

4t 1dt

89.

Alternate solution

Fx sinx ddx sinx sinxcosx

Fx sin x0

tdt

Fx sinx12 cosx cosxsinx

Fx sin x0

tdt 23t32

sin x

0

2

3sinx32 91.

Fx sinx32 3x2 3x2sinx 6

Fx x30

sin t2dt

93.

has a relative maximum at x 2.g

x

y

1

1 2 3 4

2

2

1

f g

g0 0, g11

2, g2 1, g3

1

2, g4 0

gx x0

ftdt 95. (a)

(b)

C10 1000125 121054 $338,394

C5 1000125 12554 $214,721

C1 1000125 121 $137,000

500025 125x54 1000125 12x54

500025 345t54x

0

Cx 500025 3x0

t14dt

97. True 99. False;

Each of these integrals is infinite.

has a nonremovable discontinuity atx 0.

fx x2

11

x2dx 01

x2dx 10

x2dx

101.

By the Second Fundamental Theorem of Calculus, we have

Since must be constant.fxfx 0,

1

1 x2

1

x2 1 0.

fx 1

1x2 11

x2 1

x2 1

fx

1x

0

1

t2

1

dt

x

0

1

t2

1

dt


22/33

Section 4.5 Integration by Substitution 197

103.

28 units

4 4 20

310

t2 4t 3dt 331

t2 4t 3dt 353

t2 4t 3dt

50

3t 3t 1dt

Total distance

5

0

xt

dt

3t 3t 1

3t2 4t 3

xt 3t2 12t 9

xt t3 6t2 9t 2

105.

22 1 2 units

2t124

1

41

1

tdt

4

1

vtdt

Total distance 41

xtdt

Section 4.5 Integration by Substitution

du gxdxu gxfgxgxdx1. 10x dx

3. 2x dx

5. tanx sec2xdxtan2xsec2xdxx2 1 x

x2 1dx

5x2 15x2 1210xdx

7.

Check: d

dx1 2x5

5 C 21 2x4

1 2x42 dx 1 2x5

5 C

9.

Check: d

dx2

39 x232 C 23

3

29 x2122x 9 x22x

9 x2122xdx 9 x23232

C2

39 x232 C


23/33


11.

Check: d

dxx4 33

12 C 3x

4 32

12 4x3 x4 32x3

x3x4 32dx 14x4 324x3dx 1

4x4 33

3 C

x4 33

12 C

13.

Check: d

dxx3 15

15 C 5x

3 143x2

15 x2x3 14

x2x3 14dx

1

3x3 143x2dx

1

3x3 15

5 Cx3 15

15 C

15.

Check: d

dtt2 232

3 C 32t

2 2122t

3 t2 212t

tt2 2 dt 12t2 2122tdt 1

2t2 232

32 C

t2 232

3 C

17.

Check: d

dx15

81 x243 C 158

4

31 x2132x 5x1 x213 5x 31 x2

5x1 x213dx 521 x2132xdx 5

2

1 x243

43 C

15

81 x243 C

19.

Check: d

dx1

41 x22 C 1421 x232x

x

1 x23

x1 x23dx 121 x232xdx 121 x22

2 C

1

41 x22 C

21.

Check: ddx 131 x3 C 1311 x323x2 x

2

1 x32

x21 x32dx 131 x323x2dx 131 x31

1 C 1

31 x3 C

23.

Check: d

dx1 x212 C

1

21 x2122x

x

1 x2

x1 x2

dx 1

21 x2122xdx 1

21 x212

12 C 1 x2 C

25.

Check:

d

dt

1 1t4

4

C

1

441

1

t3

1

t2

1

t21

1

t3

1 1t3

1t2dt 1 1t3

1t2dt 1 1t4

4 C

27.

Check: d

dx2x C 1

22x122

1

2x

12x

dx 1

22x122 dx 1

22x12

12 C2x C


24/33


29.

Check: d

dx2

5x52 2x32 14x12 C x

2 3x 7

x

x2 3x 7x

dx x32 3x12 7x12dx 25x52 2x32 14x12 C

2

5xx2 5x 35 C

31.

Check: d

dt1

4t4 t2 C t3 2t t2t 2t

t2t

2

tdt

t3 2tdt

1

4t4 t2 C

33.

Check: d

dy2

5y3215 y C ddy6y32

2

5y52 C 9y12 y32 9 yy

9 yydy 9y12 y32dy 923y32 25y52 C 25y3215 y C

35.

2x2 416 x2 C

4x2

2 216 x212

12 C

4xdx 216 x2122xdxy

4x

4x

16 x2

dx 37.

1

2x2 2x 3 C

1

2x2 2x 31

1 C

1

2x2 2x 322x 2dx

y

x 1

x2

2x 32dx

39. (a)

x

y

2 2

1

3

(b)

2 2

1

2

y 1

34 x232 2

2 1

34 2232 C C 22, 2:

1

2

2

34 x232 C

1

34 x232 C

y

x4

x

2

dx

1

2

4

x2

12

2xdx

dy

dxx4 x2, 2, 2

41. sin xdx cos x C 43. sin 2xdx 12sin 2x2xdx 12cos 2x C

45. 12

cos1

d cos 1

1

2d sin1

C


25/33


47. OR

OR

sin 2xcos 2xdx

1

2

2 sin 2xcos 2xdx

1

2

sin 4xdx

1

8cos 4x C

2

sin 2xcos 2xdx 12cos 2x2 sin 2xdx 1

2cos 2x2

2 C

1

1

4cos22x C

1

sin 2xcos 2xdx 12sin 2x2 cos 2xdx 1

2sin 2x2

2 C

1

4sin22x C

49. tan4xsec2xdx tan5x5 C

1

5tan5x C

51.

cotx2

2 C

1

2 cot2x C

1

2tan2x C

1

2sec2x 1 C

1

2sec2x C

1

csc2xcot3x

dx cotx3csc2xdx

53. cot2xdx csc2x 1dx cotx x C 55.Since Thus,

fx 2 sinx

2 3.

C 3.f0 3 2 sin 0 C,

fx cosx2

dx 2 sinx

2 C

57.

2

15x 2323x 4 C

2

15x 2323x 2 10 C

2u32

15 3u 10 C

2

5u52

4

3u32 C

u32 2u12duxx 2 dx u 2udu

dx dux u 2,u x 2,

59.

2

1051 x3215x2 12x 8 C

2

1051 x3235 421 x 151 x2 C

2u32

10535 42u 15u2 C

23u32 4

5u52

2

7u72 C

u12 2u32 u52dux21 xdx 1 u2udu

dx dux 1 u,u 1 x,


26/33


61.

1

152x 13x2 2x 13 C

1

602x 112x2 8x 52 C

2x 1

60 32x 12 102x 1 45 C

u12

603u2 10u 45 C

1

82

5u52

4

3u32 6u12 C

1

8u32 2u12 3u12du

1

8u12u2 2u 1 4du

x2 12x 1

dx 12u 12 1u

1

2du

dx 1

2dux

1

2u 1,u 2x 1,

63.

where C1 1 C.

x 2x 1 C1

x 2x 1 1 C

x 1 2x 1 C

u 2u C

u

2u12

C

1 u12du u 1u 1

uu 1du

xx 1 x 1

dx u 1u u

du

dx dux u 1,u x 1,

65. Let

11

xx2 13dx 1

211

x2 132xdx 18x2 141

1

0

du 2xdx.u x2 1,

67. Let

4

927 22 12 8

92

4

9x3 1322

1

23x3 132

32 2

1

21

2x2x3 1 dx 2 1

321

x3 1123x2dx

u x3 1, du 3x2dx


27/33


69. Let

40

1

2x 1dx

1

240

2x 1122dx 2x 14

0

9 1 2

du 2 dx.u 2x 1,

71. Let

91

1

x1 x2dx 29

1

1 x2 12x

dx 21 x

9

1

1

2 1

1

2

du 1

2x

dx.u 1 x,

73.

When When

01

u32 u12du 25u52 2

3u32

0

1

25 2

3 4

1521

x 12 xdx 01

2 u 1udu

u 0.x 2,u 1.x 1,

dx dux 2 u,u 2 x,

75. 20

cos23xdx 3

2sin23x

2

0

3

23

2 33

4

77.

When When

81

u43 u13du 37u73 3

4u43

8

1

3847 12 3

7

3

4 1209

28

Area 70

x3x 1 dx 81

u 13udu

u 8.x 7,u 1.x 0,

dx dux u 1,u x 1,

79. A 0

2 sinx sin 2xdx 2 cosx 12cos 2x

0

4

81. Area 232

sec2x2dx 223

2sec2x2

1

2dx 2 tanx

223

2 23 1

83.

1

1 5

3

40

x

2x 1dx 3.333

10

3 85.

0

0 8

15

73

xx 3 dx 28.8 144

5 87.

1

1 4

30

cos 6d 7.377

89.

They differ by a constant: C2 C

1

1

6.

2x 12dx 4x2 4x 1dx 43x3 2x2 x C

2

2x 12dx 122x 122 dx 1

62x 13 C

1

4

3x3 2x2 x

1

6 C

1


28/33


91. is even.

2325 8

3 272

15

22

x2x2 1dx 220

x4 x2dx 2x5

5

x3

3 2

0

fx x2x2 1 93. is odd.

22

xx2 13dx 0

fx xx2 13

95. the function is an even function.

(a)

(c)20

x2dx 20

x2dx 8

3

02

x2dx 20

x2dx 8

3

x220

x2dx x3

3 2

0

8

3;

(b)

(d) 02

3x2dx 320

x2dx 8

22

x2dx 220

x2dx 16

3

97. 0 240

6x2 3dx 22x3 3x4

0

23244

x3 6x2 2x 3dx 44

x3 2xdx 44

6x2 3dx

99. Answers will vary. See Guidelines for Making a Change of Variables on page 292.

101. is odd. Hence, 22

xx2 12dx 0.fx xx2 12 103.

Solving this system yields and

. Thus,

When V4 $340,000.t 4,

Vt 200,000

t 1 300,000.

C 300,000

k 200,000

V1 1

2k C 400,000

V0 k C 500,000

Vt kt 12dt kt 1 CdV

dt

k

t 12

105.

(a) thousand units

(b) thousand units

(c) thousand units11274.50t 262.5 cos t6

12

0

112894 262.5 262.5 74.5

1

374.50t262.5

cos

t

66

3

1

3447 262.5

223.5 102.352

1

374.50t262.5

cos

t

63

0

1

3223.5 262.5

102.352

1

b aba

74.50 43.75 sin t6dt1

b a74.50t262.5

cos

t

6b

a


29/33


109. False

2x 12dx 122x 122 dx 1

62x 13 C

111. True

Odd Even10

10

ax3 bx2 cx ddx

10

10

ax3 cxdx

10

10

bx2 ddx 0 2

10

0

bx2 ddx

113. True

4sinxcosxdx 2sin 2xdx cos 2x C

115. Let then When When Thus,

ba

fx hdx bhah

fudu bhah

fxdx.

u b h.x b,u a h.x a,du dx.u x h,

107.

(a) amps

(b)

amps

(c) amps1

130 01

30cos60t

1

120sin120t

130

0

30 130 1

30 0

2

5 22 1.382

1

1240 01

30cos60t

1

120sin120t

1240

0

240 1302

1

120 1

30

1

160 01

30cos60t

1

120sin120t

160

0

60 130 0 1

30 4

1.273

1

b aba

2 sin60t cos120tdt1

b a1

30cos60t

1

120 sin120t

b

a

Section 4.6 Numerical Integration

1. Exact:

Trapezoidal:

Simpsons: 20

x2dx1

60 41

22

212 4322

22 83 2.6667

20

x2dx1

40 21

22

212 2322

22 114 2.7500

20

x2dx 13x32

0

8

3 2.6667

3. Exact:

Trapezoidal:

Simpsons: 20

x3dx1

60 41

23

213 4323

23 246 4.0000

20

x3dx1

40 21

23

213 2323

23 174 4.2500

2

0

x3dx x4

4 2

0

4.000


30/33

Section 4.6 Numerical Integration 205

5. Exact:

Trapezoidal:

Simpsons: 20

x3dx1

120 41

43

2243

4343

213 4543

2643

4743

8 4.0000

20

x3dx1

80 21

43

2243

2343

213 2543

2643

2743

8 4.0625

20

x3dx 14x42

0

4.0000

7. Exact:

Trapezoidal:

Simpsons: 94

xdx5

242 4378 21 4478 26 4578 31 4678 3 12.6667 12.6640

94

xdx5

162 2378 2214 2478 2264 2578 2314 2678 3

94

xdx 23x329

4

18 16

3

38

3 12.6667

9. Exact:

Trapezoidal:

Simpsons:

1

121

4

64

81

8

25

64

121

1

9 0.1667

21

1

x 12dx

1

121

4 4 154 12 2

1

32 12 41

74 12 1

9

1

81

4

32

81

8

25

32

121

1

9 0.1676

21

1

x 12dx

1

81

4 2 154 12 2

1

32 12 21

74 12 1

9

2

1

1x 12

dx 1x 12

1

13 1

2 1

6 0.1667

11. Trapezoidal:

Simpsons:

Graphing utility: 3.241

20

1 x3dx1

61 41 18 22 41 278 3 3.240

2

0

1 x3dx1

41 21 18 22 21 278 3 3.283

13.

Trapezoidal:

Simpsons:


10

x1 xdx1

120 4141 14 2121 12 4341 34 0.372

10

x1 xdx1

8

0 2141

1

4 21

21 1

2 23

41 3

4

0.342

10

x1 xdx 10

x1 xdx


31/33


15. Trapezoidal:

Simpsons:


0.978

2

0

cosx2dx2

12 cos 0 4 cos24 2

2 cos22 2

4 cos24 2

cos22

0.957

2

0

cosx2dx2

8 cos 0 2 cos24 2

2 cos22 2

2 cos24 2

cos22

17. Trapezoidal:

Simpsons:


1.11

sinx2dx1

120sin1 4 sin1.0252 2 sin1.052 4 sin1.0752 sin1.12 0.089

1.11

sinx2dx1

80sin1 2 sin1.0252 2 sin1.052 2 sin1.0752 sin1.12 0.089

19. Trapezoidal:

Simpsons:


40

xtanxdx

480 4

16tan

16 22

16tan2

16 43

16tan3

16

4 0.186

4

0xtanxdx

320 2

16tan

16 2

2

16tan2

16 2

3

16tan3

16

4 0.194

21. (a)

The Trapezoidal Rule overestimates the area if the graph

of the integrand is concave up.

xa b

y f x= ( )

y 23.

(a) Trapezoidal: since

is maximum in when

(b) Simpsons: since

f4x 0.

Error 2 05

180440 0

x 2.0, 2fx

Error 2 03

1242 12 0.5

f4

x

0

fx 6

fx 6x

fx 3x2

fx x3

25. in .

(a) is maximum when and

Trapezoidal: let

in

(b) is maximum when and when

Simpsons: let n 26.n > 25.56;n4 > 426,666.67,Error 25

180n424 < 0.00001,

f41 24.x 1f4x

1, 3f4x 24

x5

n 366.n > 365.15;n2 > 133,333.33,Error 23

12n22 < 0.00001,

f1

2.x 1

fx

1, 3fx 2

x3


32/33

Section 4.6 Numerical Integration 207

27.

(a) in .

is maximum when and

Trapezoidal: Error < 0.00001, > 16,666.67,n > 129.10; let

(b) in

is maximum when and

Simpsons: Error < 0.00001, > 16,666.67, n > 11.36; let n 12.n432

180n415

16f40

15

16.x 0f4x

0, 2f4x 15

161 x72

n 130.n2

8

12n21

4

f0 1

4.x 0fx

0, 2fx 1

41 x32

fx 1 x

29.

(a) in .

is maximum when and

Trapezoidal: Error < 0.00001, > 412,754.17, n > 642.46; let

(b) in

is maximum when and

Simpsons: Error < 0.00001, > 5,102,485.22,n > 47.53; let n 48.n41 05

180n4 9184.4734

f41 9184.4734.x 1f4x

0, 1f4x 8 sec2x212x2 3 32x4tanx2 36x2tan2x2 48x4tan3x2

n 643.n21 03

12n2 49.5305

f1

49.5305.x 1

fx

0, 1fx 2 sec2x21 4x2tanx2

fx tanx2

31. Let Then

Simpsons: Error

Therefore, Simpsons Rule is exact when approximating the integral of a cubic polynomial.

Example:

This is the exact value of the integral.

10

x3dx 1

60 41

23

1 14

b a5

180n4 0 0

f4x 0.fx Ax3 Bx2 Cx D.

33. on .0, 4fx 2 3x2

n

4 12.7771 15.3965 18.4340 15.6055 15.4845

8 14.0868 15.4480 16.9152 15.5010 15.4662

10 14.3569 15.4544 16.6197 15.4883 15.4658

12 14.5386 15.4578 16.4242 15.4814 15.4657

16 14.7674 15.4613 16.1816 15.4745 15.4657

20 14.9056 15.4628 16.0370 15.4713 15.4657

SnTnRnMnLn


33/33


35. on .0, 4fx sinx

n

4 2.8163 3.5456 3.7256 3.2709 3.3996

8 3.1809 3.5053 3.6356 3.4083 3.4541

10 3.2478 3.4990 3.6115 3.4296 3.4624

12 3.2909 3.4952 3.5940 3.4425 3.4674

16 3.3431 3.4910 3.5704 3.4568 3.4730

20 3.3734 3.4888 3.5552 3.4643 3.4759

SnTnRnMnLn

37.

Simpsons Rule:

x

4 2

1

1

2

y

0.701

20

xcosxdx

840 cos 0 4

28cos

28 214cos

14 4328cos

3

28 . . . 2 cos

2

n 14

A 20

xcosxdx

39.

Simpsons Rule:

4001512125 15

123

. . . 0 10,233.58 ft lb

50

100x125 x3dx5

3120 400 512125 5123

2001012125 10

123

n 12

W

5

0 100x125

x

3

dx

41. Simpsons Rule,

1

36113.098 3.1416

1

2

0

36 6 46.0209 26.0851 46.1968 26.3640 46.6002 6.9282

n 6120

6

1 x2dx

43. Area1000

210125 2125 2120 2112 290 290 295 288 275 235 89,250 sq m

45 10t i d 2

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