# Respostas livro calculo a 6°

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• Páginas : 14 (3288 palavras )
• Publicado : 21 de março de 2013

Amostra do texto
8.4 – EXERCÍCIOS – pg. 344
Nos exercícios de 1 a 14, encontrar o comprimento de arco da curva dada. 1. y = 5 x − 2 , − 2 ≤ x ≤ 2

s=

a

b

1 + f ′( x ) dx
2 2

= =

−2

2

1 + 5 dx =

−2

2

2

26 dx = 26 x
−2

26 (2 + 2) = 4 26 u. c.

2. y = x

2

3

−1 , 1 ≤ x ≤ 2

y′ =

2 −13 x 3
2

s=

1 2

1+

4 9x
2 3

dx = ∫
1 1 2

29x

2

3

+4
2 3

dx

9x

2 = ∫ 9x 3 + 4     1

dx .
1

1 − 13 . x dx 3

2 −1 2 1 1 = . ∫  9 x 3 + 4  . 6 . x 3 dx    3 6 1

2

 9x 23 + 4 32    1   = . 3 18 2

2

1 3 3   3  2 2 2 1 2   23   9 x + 4  − 13 2  = 1   9 . 2 3 + 4  − 13 3  = .     27      18 3      

3. y =

3 1 (2 + x 2 ) 2 , 0 ≤ x ≤ 3 3

y′ =

1 1 3 .(2 + x 2 ) 2 . 2 x 3 2

s= = = =

0

3

1 + x 2 2 + x 2 dx 1 + x 2 + x 4 dx
2

(

)

0 3 0

3

∫ (x ∫(
0 2 3 2

+ 1 dx
3

)

x3 33 x + 1 dx = +x = + 3 = 12 3 3 0

)

4. x

3

+y

2

3

=2

2

3

 x = 2 cos 3 t    y = 2 sen 3t 
π

s = 4 ∫ 36 cos 4 t sen 2 t + 36 sen 4 t cos 2 t dt
0

2

π

= 4 ∫ 6. sen 2 t cos 2 t cos 2 t + sen 2 tdt
0

2

(

)

sen 2 t = 24 ∫ sen t . cos t dt = 24 . 2 0
2

π

π

2

0

= 12 u. c.

5. y =

1 4 1 x + 2 , 1≤ x ≤ 2 4 8x

y′ =

1 3 1 x + (− 2 ) x −3 4 8

s= = = = =
2

1

2

1   1 +  x 3 − 3  dx 4x   1 + x 6 − 2x3. 1 + x 6 − 2x3. 1 1 + dx 3 16 x6 4x 1 dx 16 x 6

2

1

2

1

2

1 2 1

2

8 x 6 + 16 x12 + 1 dx 16 x 6
6 3

1∫ 4 x (4 x
6

+1

)

2

dx

= =

1 ∫ 4 x (4 x
3 1 2

+ 1 dx

)

2

1 3 −3 ∫ 4 x + x dx 41

(

)

1  x 4 x −2   = 4 . + 4 4 2 1   = =

2

1 4 1 1 2 − −1 +  2  4 2.2 2  123 32

6. x =

1 3 1 y + , 1≤ y ≤ 3 3 4y

x′ =

1 1 . 3 y 2 + (− 1) . y − 2 3 4 1 = y2 − 2 4y = 4y4 −1 4y2

 4 y 4 −1 16 y 8 − 8 y 4 + 1 16 y 4 + 16 y 8 − 8 y 4 + 1  =1+ 1+  =  4y2  16 y 4 16 y 4  
16 y 8 + 8 y 4 + 1 4 y 4 − 1 = = 16 y 4 16 y 4

2

(

)

2

s= =

1 3

3

(4 y

+1 16 y 4

4

)

2

dy
3

4y4 +1  2 1 −2  ∫ 4 y 2 dy = ∫  y + 4 y  dy  1 1
3

 y 3 1 y −1  =  3 + 4 . −1    1
=

33 1 1 1 − − + 3 4.3 3 4 = 53 6

 e + e −1  1 x −x  7. y = e + e de (0,1) a 1,  2 2   

(

)

y′ =

1x e − e −x 2

(

)

s= = = = =

0

1

1+ 1+ 1+ 1+

1 x e − e−x 4

( ( (

)

2

dx

0

1

1 2x e − 2 . e x . e − x + e − 2 x dx 4

)

0

1

1 2x e − 2 + e −2 x 4

)

dx

0

1

1 2 x 1 1 −2 x e − + e dx 4 2 4

0

1

4 + e 2 x − 2 + e −2 x dx 4

= =

0 1

1

2 + e2 x +

1 e
−2 x

dx dx

1 1 2 ∫ ex 0
1

(e (2x

+1

)

2

1 1 = ∫ x e 2 x + 1 dx 20e

)

=

1 x −x ∫ e + e dx 20

1

(

)

1 = e x + e−x 2

( (

)

1

0

1 e − e −1 − 1 + 1 2 = sen h 1 =

)

8. y = ln x , y′ = 1 x

3≤x≤ 8

s=

8

1+

3

1 dx = x2

8

3

x2 +1 dx x

x2 +1 = t 2 x2 = t 2 −1 x = t 2 −1 dx =

(

)

1

2

1 2 t −1 2

(

)

1

2

. 2t dt

I= =∫

x2 +1 t dx = ∫ x t 2 −1

(

) (t
2

1

.

t dt
2

−1

)

1

2

t 2 dt ∫ t 2 −1 1   = ∫ 1 + 2  dt  t −1 dt = ∫ dt + ∫ (t − 1) (t + 1) 1 1 = t + ∫ 2 dt − ∫ 2 dt t −1 t +1 1 1 = t + ln t − 1 − ln t + 1 + C 2 2 1 t −1 +C = t + ln 2 t +1
= 1 x 2 + 1 + ln 2 x2 +1 −1 x2 +1 +1
2

+C

 1 s =  x 2 + 1 + ln  2 

x +1 −1   2 x +1 +1  

8

3

 1 2 11 =  3 + ln − 2 − ln   2 4 2 3   1 3 = 1 + ln 2 2

9. y = 1 − ln (sen x ) ,
cos x sen x

π
6

≤x≤

π
4

y′ = −

π

s=

π π

4

cos 2 x 1+ dx sen 2 x
4 sen 2 x + cos 2 x dx dx = ∫ 2 x sen x π sen 6

6 4

π

=

π π

6 4

=

∫ cos ec x dx
π
6

= ln cos ec − cot g x = ln = ln = ln

π π

4 6

2 3 − 1 − ln 2 − 3 2 2− 2 3 . 2 2 3 −1 2 3− 6 2...