C2 Assignment 24 5 - Diamond Bar High Schooldbhs.wvusd.k12.ca.us/ourpages/auto/2010/8/2... ·...

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CALCULUS II Assignment #24.5 Monday, 9/22 (due Tuesday, 9/23) 1. 3 0 sin lim x x x x = 2. ( ) 0 ln 1 lim ln x x e x + = 3. 0 3 1 1 lim sin x x x x + + = 4. ( ) 1 lim ln x x x →∞ = 5. 3 3 2 2 3 2 lim 3 3 5 x x x x x x →−∞ + = + 6. 2 2 2 2 1 x dx x x + = + 7. 1 9 4 x 2 dx = 8. 2 2 cos x π π 9. 2 cos3 x e x dx = 10. 2 cos 1 sin x x = + 11. 2 cos t t dt 12. 1 1 sin dx x = + 13. x 2 x 2 7 x + 12 dx = 14. 2 lim x x xe →∞ = 15. dy Find if ln dx 1 x x e y e = 16. 2 2 x dx = 17. dx x ln x dx 2 8 = Answer must be in SIMPLEST TERMS !!!!! 18. cos 2 x x dx = 19. /2 3 0 cos x dx π = 20. Find the equation of y if ( ) 1/3 2 4 8 dy xx dx = + and () 0 0 y = . Answers: 1. – 1/6 2. 1 3. 3 4. 1 5. – 2/3 6. 2ln x 1 4 x 1 ( ) 1 + C 7. 1 2 sin 1 2x 3 + C or 1 2 cos 1 2x 3 + C 8. π/4 9. 1 13 e 2 x 2cos3x + 3sin3x ( ) + C 10. tan 1 sin x ( ) + C 11. t 2 sin t + 2t cos t 2sin t + C 12. tan x sec x + C 13. ln x 3 ( ) 2 x 4 ( ) + C 14. 0 15. 1 1 e x 16. 4 17. ln 3 18. 2 x sin x 2 + 4cos x 2 + C 19. 2/3 20. y = 3 2 x 2 + 8 ( ) 4/3 24

Transcript of C2 Assignment 24 5 - Diamond Bar High Schooldbhs.wvusd.k12.ca.us/ourpages/auto/2010/8/2... ·...

CALCULUS II – Assignment #24.5 Monday, 9/22 (due Tuesday, 9/23)

1. 30

sinlimx

x xx→

− = 2. ( )

0

ln 1lim

ln

x

x

ex+→

−= 3.

0

3 1 1limsinx

xx x+→

+ − = 4. ( )1

lim ln xx

x→∞

=

5. 3

3 2

2 3 2lim3 3 5x

x xx x x→−∞

− + − =+ −

6. 2

2 22 1x dx

x x+ =

− +∫ 7.

−1

9− 4x2∫ dx = 8. 2

2

cos xπ

π∫

9. 2 cos3 xe x dx =∫ 10. 2

cos1 sin

xx=

+∫ 11. 2 cos t t dt∫ 12. 1 1 sin

dxx

=+∫

13.

x − 2x2 − 7x +12

dx∫ = 14. 2lim x

xx e−

→∞= 15.

dyFind if lndx 1

x

x

eye

⎛ ⎞= ⎜ ⎟−⎝ ⎠

16.

2

2

x dx−

=∫

17.

dxx ln x

dx2

8

∫ =

Answer must be in SIMPLEST TERMS !!!!!⎡

⎣⎢

⎦⎥

18. cos 2xx dx =∫ 19.

/ 23

0

cos x dxπ

=∫ 20. Find the equation of

y if ( )1/324 8dy x xdx

= +

and ( )0 0y = .

Answers: 1. – 1/6 2. 1 3. 3 4. 1 5. – 2/3 6.

2ln x −1 − 4 x −1( )−1+C

7.

−12

sin−1 2x3

⎛⎝⎜

⎞⎠⎟+C or

12

cos−1 2x3

⎛⎝⎜

⎞⎠⎟+C

8. π/4

9.

113

e2x 2cos3x + 3sin3x( ) +C

10. tan−1 sin x( ) +C 11. t

2 sin t + 2t cos t − 2sin t +C 12. tan x − sec x +C

13.

lnx − 3( )2

x − 4( ) +C 14. 0

15.

11− ex

16. 4

17. ln 3 18.

2xsin

x2+ 4cos

x2+C

19. 2/3 20.

y = 3

2x2 +8( )4/3

− 24

CALCULUS II – Assignment #24.5 Monday, 9/22 (due Tuesday, 9/23) SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS 1

30

sinlimx

x xx→

− =

limx→0

sin x − xx3 = 0

0 ⇒ L'H lim

x→0

cos x −13x2 = 0

0 ⇒ L'H lim

x→0

−sin x6x

= 00

⇒ L'H limx→0

−cos x6

= −16

2 ( )

0

ln 1lim

ln

x

x

ex+→

−=

( )0 0 0

0 0

ln 1 0 01lim L'H lim L'H lim 1ln 0 1 0

L'H lim lim

xx xx

xx x x

xx x

xx x

ee xeex e

xee xe

e

+ + +

+ +

→ → →

→ →

− −= ⇒ ⇒ = ⇒−

+= ⇒ = ( )1x

xe+

1=

3

0

3 1 1limsinx

xx x+→

+ − =

0 0 0

0 0 0

3 1 1 3 1 1 sinlim lim lim 3 sin sin sin

cos 1 sin 0L'H lim 3 lim L'H 3 lim 3 3 sin cos cos cos sin 1 1 0

x x x

x x x

x x x xx x x x x x x

x xx x x x x x x

+ + +

+ + +

→ → →

→ → →

+ −− ⇒ + − ⇒ + ⇒

− −+ ⇒ + = + =+ + − + −

4 ( )

1

lim ln xx

x→∞

=

Let y = ln x( )1x ⇒ ln y = 1

xln ln x( ) ⇒ lim

x→∞ln y = lim

x→∞

1x

ln ln x( ) ⇒ limx→∞

ln y = limx→∞

ln ln x( )x

= ∞∞

L'H limx→∞

ln y = limx→∞

1ln x

⎛⎝⎜

⎞⎠⎟

1x

⎛⎝⎜

⎞⎠⎟

1= 0

1= 0 ⇒ Thus, lim

x→∞ln x( )

1x = lim

x→∞y = lim

x→∞eln y = e 0 = 1

5 3

3 2

2 3 2lim3 3 5x

x xx x x→−∞

− + − =+ −

3 3 2 3

3 2 3

2

2 3 22 3 2 1 2lim lim 3 53 3 5 1 33x x

x x x x xx x x x

x x→−∞ →−∞

⎛ ⎞− + −⎜ ⎟⎛ ⎞ ⎡ ⎤− + − ⇒ = −⎜ ⎟⎜ ⎟ ⎢ ⎥+ −⎝ ⎠ ⎣ ⎦ ⎜ ⎟+ −⎝ ⎠

CALCULUS II – Assignment #24.5 Monday, 9/22 (due Tuesday, 9/23) SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS

6. 2

2 22 1x dx

x x+ =

− +∫

2x + 2x2 − 2x +1

dx∫ = 2x + 2

x −1( )2 = Ax −1( ) +

B

x −1( )2 ⇒ A x −1( ) + B = 2x + 2

Let x = 1 ⇒ B = 4 AND Let x = 2 ⇒ A+ 4 = 6 ⇒ A = 2 ⇒

2x −1( ) +

4

x −1( )2 dx∫ = 2ln x −1 − 4 x −1( )−1+C

7. 2

19 4

dxx

− =−∫

−1

9− 4x2∫ dx ⇒ −1

9 1− 49

x2⎛⎝⎜

⎞⎠⎟

∫ dx ⇒ −13

1

1− 23

x⎛⎝⎜

⎞⎠⎟

2⎛

⎝⎜

⎠⎟

∫ dx ⇒

G: sin−1 2x3

⎛⎝⎜

⎞⎠⎟

CK: 1

1− 23

x⎛⎝⎜

⎞⎠⎟

2

23

⎣⎢⎤

⎦⎥⇒ ANS: −1

2sin−1 2x

3⎛⎝⎜

⎞⎠⎟+C OR 1

2cos−1 2x

3⎛⎝⎜

⎞⎠⎟+C

8. 2

2

cos xπ

π∫ = ( )2 2

2 2 22

1 1 1 1cos cos cos2 1 sin 2 0 02 2 2 2 2 4

x x x x xππ π π

π π ππ

π ππ⎡ ⎤⎡ ⎤ ⎛ ⎞= = + = + = + − + =⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦∫ ∫ ∫

9. 2 cos3 xe x dx =∫

e2x cos3x dx∫ = 12

e2x cos3x + 32

e2x sin3x dx ⇒∫

1 e2x cos3x dx∫ = 12

e2x cos3x + 34

e2x sin3x − 94

e2x cos3x dx∫1+ 9

4⎡

⎣⎢⎤

⎦⎥e2x cos3x∫ dx = 1

4e2x 2cos3x + 3sin3x( )

e2x cos3x∫ dx = 113

e2x 2cos3x + 3sin3x( ) +C

21 1

21 1

22 2

22 2

1cos3 2

3sin 3 1sin 3 2

3cos3

x

x

x

x

u x v e

du x dv e dx

u x v e

du xdx dv e dx

= =

= − =

= =

= =

10.

cos x1+ sin2 x∫ = G: tan−1 sin x( ) CK:

11+ sin2 x

⎛⎝⎜

⎞⎠⎟

cos x( ) ANS: tan−1 sin x( ) +C

CALCULUS II – Assignment #24.5 Monday, 9/22 (due Tuesday, 9/23) SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS 11. 2 cos t t dt∫ =

t2 cos t∫ dt = t2 sin t − 2t sin t∫ dt = t2 sin t − −2t cos t − −2cos t dt∫⎡⎣

⎤⎦ =

t2 sin t + 2t cos t − 2sin t +C

21 1

1 1

sin2 cos

u t v tdu tdt dv tdt= == =

2 2

2 2

2 cos2 sin

u t v tdu dt dv tdt

= = −= =

12. 1 1 sin

dxx

=+∫

11+ sin x∫ dx = 1− sin x

1− sin2 x∫ dx = 1− sin xcos2 x∫ dx = sec2 x − sec x tan x⎡⎣ ⎤⎦∫ dx = tan x − sec x +C

13. 2

27 12x dx

x x− =

− +∫

x − 2x2 − 7x +12

dx = x − 2x − 4( ) x − 3( ) dx∫∫ ⇒ x − 2

x − 4( ) x − 3( ) =A

x − 4( ) +B

x − 3( ) ⇒

A x − 3( ) + B x − 4( ) = x − 2 ⇒ Ax − 3A+ Bx − 4B = x − 2 ⇒

A+ B = 1 A = 1− B⎡⎣ ⎤⎦ AND − 3A− 4B = −2 ⇒ − 3 1− B( )− 4B = −2 ⇒ − 3+ 3B − 4B = −2 ⇒ B = −1

A = 1− −1( ) = 2 ⇒ A = 2 ⇒

−1x − 4

⎛⎝⎜

⎞⎠⎟+ 2

x − 3⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟∫ dx = −1ln x − 4 + 2ln x − 3 +C = ln

x − 3( )2

x − 4( ) +C

14. 2lim x

xx e−

→∞=

2 2 2 2lim L'H lim L'H lim 0x x xx x x

x xe e e→∞ →∞ →∞

∞ ∞= ⇒ ⇒ = ⇒ ⇒ = =∞ ∞ ∞

15. dyFind if lndx 1

x

x

eye

⎛ ⎞= ⎜ ⎟−⎝ ⎠

y = ln ex

ex −1⎛⎝⎜

⎞⎠⎟= ln ex

ex −1⎛⎝⎜

⎞⎠⎟= lnex − ln ex −1( ) = x − ln ex −1( )⇒ y ' = 1− ex

ex −1⎡

⎣⎢

⎦⎥ =

ex −1− ex

ex −1= −1

ex −1= 1

1− ex

CALCULUS II – Assignment #24.5 Monday, 9/22 (due Tuesday, 9/23) SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS SOLUTIONS

16. 2

2

x dx−

=∫

( ) [ ]

( )( ) ( )( )

0 22 0 2 2 2

2 2 0 2 0

0 2 2 0 2 2 42 2

1 1OR: GRAPH IT! AREA 2 2 2 2 42 2

x xx dx x dx x dx− − −

⎤ ⎤−= − + = + = − − + − = + =⎡ ⎤⎥ ⎥ ⎣ ⎦⎦ ⎦

= + =

∫ ∫ ∫

17. [ ]8

2

Answer must be in SIMPLEST TERMS !!!!!lndx dxx x

=∫

88

2 2

ln8 3 ln 2ln ln ln ln8 ln ln 2 ln lnln ln 2dx dx xx x

⎛ ⎞= ⎤ = − = =⎜ ⎟⎦ ⎝ ⎠∫ ln 2ln3

⎛ ⎞=⎜ ⎟

⎝ ⎠

18. cos 2xx dx =∫

xcos

x2∫ dx = 2xsin

x2− 2sin

x2∫ dx = 2xsin

x2+ 4cos

x2+C 2sin

2

cos2

xu x v

xdu dx dv

= =

= =

19. / 2

3

0

cos x dxπ

=∫

( )

( ) ( ) ( )

/ 2/ 2 / 2 / 23 2 2 3

0 0 0 0

33

1cos cos 1 sin cos cos sin sin sin3

1 1 1 2sin sin sin 0 sin 0 1 02 3 2 3 3 3

x dx x x dx x x x dx x xππ π π

π π

⎤= − = − = − ⎥

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤ ⎛ ⎞= − − − = − − =⎡ ⎤⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎣ ⎦⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎝ ⎠⎢ ⎥⎣ ⎦

∫ ∫ ∫

20. Find the equation of y if ( )1/324 8dy x xdx

= + and ( )0 0y = .

dydx

= 4x x2 +8( )1/3 ⇒ dy = 4x x2 +8( )∫

1/3dx ∫ ⇒ y = 3

2x2 +8( )4/3

+C ⇒

y 0( ) = 0 SO, 0 = 32

0+8( )4/3+C ⇒ 0 = 3

224( ) +C ⇒ C = −24 ∴ y = 3

2x2 +8( )4/3

− 24

−2 −1 1 2

−1

1

2

x

y