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