2 . question details scalcet8 7.1.005. [3805335] cha… · 3/7/2019 assignment previewer 2/21 3 ....

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Question

Chapter 7 Review (14266894)

1234567891011121314151617181920212223242526272829303132333435Due: Tue, Feb 5, 2019 12:00 AM PST

1. -Question Details SCalcET8 7.1.003.MI. [3805119]

Evaluate the integral. (Use C for the constant of integration.)

Solution or Explanation

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3x cos(8x) dx

2. -Question Details SCalcET8 7.1.005. [3805335]



Let

Then by the equation,

te dt−5t

u = t, dv = e dt du = dt, v = − e .−5t 15

−5t

= − te − = − te + = − te − e + C.te dt−5t 15

−5t − e dt15

−5t 15

−5t 15 e dt−5t 1

5−5t 1

25−5t

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Let Then by the equation,

(x − 3)sin(πx) dx

u = x − 3, dv = sin(πx) dx du = dx, v = − cos(πx).1π

(x − 3)sin(πx) dx = − (x − 3)cos(πx) − − cos(πx) dx = − (x − 3)cos(πx) + cos(πx) dx

= − (x − 3)cos(πx) + sin(πx) + C.

1π

1π

1π

1π

1π

1

π2





Note: We could start by using

ln( ) dxx

u = ln( ),x dv = dx du = · dx = dx,1

x

1

2 x12x

v = x.

ln( ) dx = x ln( ) − x · dx = x ln( ) − dx = x ln( ) − x + C.x x 12x

x 12

x 12

ln( ) = ln(x).x 12













t ln(t) dt4

u = ln(t), dv = t dt du = dt,4 1t

v = t .15

5

t ln(t) dt = t ln(t) − t · dt = t ln(t) − t dt = t ln(t) − t + C.4 15

5 15

5 1t

15

5 15

4 15

5 125

5




First let Then by the equation,

Next let to get

Thus,

where

(ln(x)) dx2

u = (ln(x)) , dv = dx du = 2 ln(x) · dx, v = x.2 1x

I = (ln(x)) dx = x(ln(x)) − 2 x ln(x) · dx = x(ln(x)) − 2 ln(x) dx.2 2 1x

2

U = ln(x), dV = dx dU = dx, V = x1x

ln(x) dx = x ln(x) − x · dx = x ln(x) − dx = x ln(x) − x + C .1x 1

I = x(ln(x)) − 2(x ln(x) − x + C ) = x(ln(x)) − 2x ln(x) + 2x + C,21

2 C = −2C .1














e sin(3θ) dθ2θ





9 sin (x) cos (x) dx2 3




The symbols and indicate the use of the substitutions and respectively.

sin (θ) cos (θ) dθ3 6

{u = sin(θ), du = cos(θ) dθ}{u = cos(θ), du = −sin(θ) dθ},

sin (θ) cos (θ) dθ = sin (θ) cos (θ) sin(θ) dθ = (1 − cos (θ))cos (θ) sin(θ) dθ

(1 − u )u (−du) = (u − u ) du = u − u + C

= cos (θ) − cos (θ) + C

3 6 2 6 2 6

2 6 8 6 19

9 17

7

19

9 17

7


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Evaluate the integral.



sin (θ) cos (θ) dθπ/2

9 5

0

{u = sin(θ), du = cos(θ) dθ}{u = cos(θ), du = −sin(θ) dθ},

= =

= =

= u − u + u = − + − 0 = =

sin (θ) cos (θ) dθπ/2

9 5

0sin (θ) cos (θ) cos(θ) dθ

π/29 4

0sin (θ)(1 − sin (θ)) cos(θ) dθ

π/29 2 2

0

u (1 − u ) du1

9 2 2

0u (1 − 2u + u ) du

19 2 4

0(u − 2u + u ) du

19 11 13

0

110

10 16

12 114

141

0

110

16

114

21 − 35 + 15210

1210





5 cos (θ) dθπ/2

2

0





5 sin (t) cos (t) dtπ

2 4

0





https://www.webassign.net/latex2pdf/44879978a1a271bcb30163a4f4c5e888.pdf




https://www.webassign.net/latex2pdf/e59a39009500e5dfe68c0262496ada0b.pdf








9 sin (x) cos (x) dxπ/2

2 2

0





tan (x) cos (x) dx2 3

{u = sin(x), du = cos(x) dx}{u = cos(x), du = −sin(x) dx},

tan (x) cos (x) dx = cos (x) dx = sin (x) cos(x) dx u du = u + C = sin (x) + C2 3 sin (x)

cos (x)

2

23 2 2 1

33 1

33




4 tan(x) sec (x) dx3

= = [u = sec(x), du = sec(x) tan(x) dx]

= u + C = sec (x) + C

4 tan(x) sec (x) dx3 4 tan(x) sec(x) sec (x) dx2 4u du2

43

3 43

3


https://www.webassign.net/latex2pdf/b40da11f609eadb2b5341531141efdbe.pdf














tan (x) dx2





4 tan (x) + tan (x) dx2 4




Let Then so

4 tan (x) sec (x) dx4 6

u = tan x. du = sec (x) dx,2

= 4 = 4

= 4 = 4

= 4 u + u + u + C = tan (x) + tan (x) + tan (x) + C.

4 tan (x) sec (x) dx4 6 tan (x) sec (x)(sec (x) dx)4 4 2 tan (x)(1 + tan (x)) (sec (x) dx)4 2 2 2

u (1 + u ) du4 2 2 (u + 2u + u ) du8 6 4

19

9 27

7 15

5 49

9 87

7 45

5


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8 tan (x) sec(x) dx3


Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)



5 tan (x) sec(x) dx2





dφ5 sin(φ)

cos (φ)3


https://www.webassign.net/latex2pdf/0117fda072d7fc35ee88e0586013c90b.pdf




https://www.webassign.net/latex2pdf/794eb2f9ea0beb331655084ee0593e23.pdf?epsf=1



https://www.webassign.net/latex2pdf/39a2f30da08b54e81c51d31299861974.pdf







Let where Then and

dxx2

9 − x2

x = 3 sin(θ), −π/2 ≤ θ ≤ π/2. dx = 3 cos(θ) dθ

= = = 3|cos(θ)| = 3 cos(θ).9 − x2 9 − 9 sin (θ)2 9 cos (θ)2

dx = 3 cos(θ) dθ = 9 sin (θ) dθ

= 9 (1 − cos(2θ)) dθ = θ − sin(2θ) + C = θ − (2 sin(θ) cos(θ)) + C

= sin − · · + C = sin − x + C

x2

9 − x29 sin (θ)3 cos(θ)

22

12

92

12

92

94

92

−1 x3

92

x3 3

9 − x2 92

−1 x3

12

9 − x2








The symbol indicates the use of the substitution

Let so and Then

4

2

dx

(x − 1)2 3/2

{u = sin(θ), du = cos(θ) dθ}.

x = sec(θ), dx = sec(θ) tan(θ) dθ, x = 2 θ = ,π3

x = 4 θ = sec (4).−1

= = dθ

du = − = + = − + .

4

2

dx

(x − 1)2 3/2

sec (4)−1

π/3

sec(θ) tan(θ) dθtan (θ)3

sec (4)−1

π/3

cos(θ)

sin (θ)2

√15/4

√3/2

1

u21u

√15/4

√3/2

−4√15

2√3

415

15 23

3




Let where a > 0 and Then and

Thus,

2 a > 0,a dx

(a + x )2 2 3/20

x = a tan θ, − < θ < .π2

π2

dx = a sec θ dθ, x = 02 θ = 0, x = a θ = .π4

2 = 2 = 2 = = sin θ

= − 0 = .

a dx

(a + x )2 2 3/20

π/4 a sec θ dθ[a (1 + tan θ)]

2

2 2 3/20

π/4 a sec θ dθa sec θ

2

3 30

2

a2 cos θ dθπ/4

0

2

a2

π/4

0

2

a22

2 2

a2 2











Let so and Thus,

6

0

dt

36 + t2

t = 6 tan(θ), dt = 6 sec (θ) dθ,2 t = 0 θ = 0, t = 6 θ = .π4

= = = sec(θ) dθ = ln(|sec(θ) + tan(θ)|)

= ln(| + 1|) − ln(|1 + 0|) = ln( + 1).

6

0

dt

36 + t2π/4

0

6 sec (θ) dθ2

36 + 36 tan (θ)2

π/4

0

6 sec (θ) dθ6 sec(θ)

2 π/4

0

π/4

0

2 2





dxx

x − 92

3





dx0.8 x2

16 − 25x20




https://www.webassign.net/latex2pdf/fdc125d2d80c840a57aadb88ca11aa89.pdf?epsf=1




https://www.webassign.net/latex2pdf/1a5d2ce09d7d09aaa8999c5a773b1b9a.pdf?epsf=1








dxxx − 1

4

dx = x + x + x + 1 + dx [by division] = x + x + x + x + ln(|x − 1|) + Cxx − 1

4 3 2 1x − 1

14

4 13

3 12

2




dt5t − 2t + 1

dt = 5 − dt = 5t − 7 ln(|t + 1|) + C5t − 2t + 1

7t + 1










Multiply both sides by to get

The coefficients of x must be equal and the constant terms are also equal, so and Adding theseequations gives us and hence, Thus,

Another method: Substituting 1 for x in the equation gives

Substituting for x gives

dx17x + 3(4x + 1)(x − 1)

= + .17x + 3(4x + 1)(x − 1)

A4x + 1

Bx − 1

(4x + 1)(x − 1)

17x + 3 = A(x − 1) + B(4x + 1) 17x + 3 = Ax − A + 4Bx + B 17x + 3 = (A + 4B)x + (−A + B).

A + 4B = 17 −A + B = 3.5B = 20 ⇔ B = 4, A = 1.

dx = + dx = ln(|4x + 1|) + 4 ln(|x − 1|) + C.17x + 3(4x + 1)(x − 1)

14x + 1

4x − 1

14

17x + 3 = A(x − 1) + B(4x + 1) 20 = 5B ⇔ B = 4.

− 14

− = − A ⇔ A = 1.54

54









We have the following.

Multiply both sides by to get the following.

The coefficients of y must be equal and the constant terms are also equal, so and Adding 4 times

the second equation and the first equation gives us and hence, Thus, we have the following.

Another method: Substituting for y in the equation gives Substituting for

y gives

dyy(y + 2)(4y − 1)

= + y(y + 2)(4y − 1)

Ay + 2

B4y − 1

(y + 2)(4y − 1)

y = A(4y − 1) + B(y + 2) y = 4Ay − A + By + 2B y = (4A + B)y + (−A + 2B)

4A + B = 1 −A + 2B = 0.

9B = 1 ⇔ B = 19

A = .29

= + dy = ln(|y + 2|) + · ln(|4y − 1|) + C

= ln(|y + 2|) + ln(|4y − 1|) + C

y dy(y + 2)(4y − 1) y + 2

29

4y − 1

19

29

19

14

29

136

14

y = A(4y − 1) + B(y + 2) = B ⇔ B = .14

94

19

−2

−2 = −9A ⇔ A = .29








Multiply both sides by to get

The coefficients of x must be equal and the constant terms are also equal, so and Subtracting the second equation from the first gives and hence, Thus,

Another method: Substituting −1 for x in the equation gives

Substituting for x gives

dx1

0

30

6x + 7x + 12

= = + .30

6x + 7x + 1230

(6x + 1)(x + 1)A

6x + 1B

x + 1(6x + 1)(x + 1)

30 = A(x + 1) + B(6x + 1). A + 6B = 0A + B = 30. B = −6, A = 36.

dx = − dx = 6 ln(|6x + 1|) − 6 ln(|x + 1|)

= (6 ln(7) − 6 ln(2)) − 0 = 6 ln .

1

0

30

6x + 7x + 12

1

0

366x + 1

6x + 1

1

0

72

30 = A(x + 1) + B(6x + 1) 30 = −5B ⇔ B = −6.

− 16

30 = A ⇔ A = 36.56




Multiplying both sides by to get

The coefficients of x must be equal and the constant terms are also equal, so and Adding twicethe first equation to the second gives us and hence, Thus,

Another Method: Substituting 4 for x in the equation gives Substituting 2 for x gives

dx1 x − 6

x − 6x + 820

= + .x − 6

x − 6x + 82A

x − 2B

x − 4(x − 2)(x − 4) x − 6 = A(x − 4) + B(x − 2)

x − 6 = Ax − 4A + Bx − 2B x − 6 = (A + B)x + (−4A − 2B).A + B = 1 −4A − 2B = −6.

−A = −2 ⇔ A = 2, B = −1.

= = 2 ln|x − 2| − ln|x − 4|

= (0 − ln 3) − (2 ln 2 − ln 4)

dx1 x − 6

x − 6x + 820

− dx1 2

x − 21

x − 40

1

0

x − 6 = A(x − 4) + B(x − 2) −1 = B.−2 = −A ⇔ A = 2.











Multiplying both sides by gives

Substituting −1 for x gives Substituting −2 for x gives Equating coefficients of gives so Thus,

dx1

0

x + x + 1

(x + 1) (x + 2)

2

2

= + + .x + x + 1

(x + 1) (x + 2)

2

2A

x + 1B

(x + 1)2C

x + 2(x + 1) (x + 2)2

x + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1) .2 2 1 = B.3 = C. x2 1 = A + C = A + 3, A = −2.

dx = + + dx = −2 ln(|x + 1|) − + 3 ln(|x + 2|)

= −2 ln(2) − + 3 ln(3) − (0 − 1 + 3 ln(2)) = + ln .

1

0

x + x + 1

(x + 1) (x + 2)

2

2

1

0

−2x + 1

1

(x + 1)23

x + 21

x + 1

1

0

12

12

2732





dx17

(x − 1)(x + 16)2




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Determine whether the integral is convergent or divergent.

If it is convergent, evaluate it. (If the quantity diverges, enter DIVERGES.)


dx∞

8

1

(x − 7)3/2

convergent

divergent

dx = (x − 7) dx = −2(x − 7) [u = x − 7, du = dx]

= + = 0 + 2 = 2. Convergent

∞

8

1

(x − 7)3/2 lim t→∞

t

8

−3/2 lim t→∞

−1/2 t

8

lim t→∞

−2

√t − 7

2

√1






dx0 1

5 − 8x−∞

convergent

divergent





https://www.webassign.net/latex2pdf/0e07bf262c5233377b935e7719a0a0cc.pdf









dx∞

0

x2

5 + x3

convergent

divergent

dx = dx = = − = ∞. Divergent∞

0

x2

5 + x3 lim t→∞

t

0

x2

5 + x3 lim t→∞

23

5 + x3t

0lim t→∞

23

5 + t323






9xe dx∞ −x

2

−∞

convergent

divergent





https://www.webassign.net/latex2pdf/646a71b2d03d3b6fcf788b43ff8e07da.pdf









dx∞

1

1

x + x2

convergent

divergent

dx = dx = − dx [partial fractions]

= ln(|x|) − ln(|x + 1|) = ln = ln − ln

= 0 − ln = ln(2). Convergent

∞

1

1

x + x2 lim t→∞

t

1

1x(x + 1)

lim t→∞

t

1

1x

1x + 1

lim t→∞

t

1lim t→∞

xx + 1

t

1lim t→∞

tt + 1

12

12





dx13

0

1

13 − x3

convergent

divergent

dx = (13 − x) dx = − (13 − x) = − (13 − t) − 13

= 13 . Convergent

13

0

1

13 − x3 lim t→13 −

t

0

−1/3 lim t→13 −

32

2/3 t

0lim

t→13 −

32

2/3 2/3

32

2/3








Name (AID): Chapter 7 Review (14266894)Submissions Allowed: 10Category: HomeworkCode:Locked: NoAuthor: Mkrtchyan, Tigran ( [email protected] )Last Saved: Mar 7, 2019 09:27 AM PSTPermission: ProtectedRandomization: PersonWhich graded: Last

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dx14 11

x + 24−2

convergent

divergent






dx3 44

x4−2

convergent

divergent

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