Integral definido. Exercicios resolvidos.

Matemática 1

Anatolie Sochirca ACM DEETC ISEL

Integral definido. Exercícios resolvidos. a) Calcular os integrais definidos utilizando a fórmula de Barrow.

1

∫

Exercício 1.

1 + x dx .

0

1

1   1 1  (1 + x) 2 +1   1 + x d (1 + x) = ∫ (1 + x) 2 d (1 + x) =  1  0 + 1    2 

1

∫

1 + x dx = ∫

0

0

3 2 = ⋅ (1 + x ) 2 3

1

0

(

1

∫ (x

−1

∫ (x

−1

x dx 2

)

+1

x dx 2

)

+1

2

1

= 0

.

1 d (x2 ) 1 1 1 d (x2 ) d ( x 2 + 1) 1 1 1 =∫ 2 = ⋅ = ⋅ = ⋅ x2 +1 ∫ ∫ ∫ 2 2 2 2 2 2 2 −1 x + 1 2 −1 x + 1 2 −1 −1 x + 1

(

     

)

(

1

2

0

3   (1 + x) 2 =  3   2

3 3  2  32  2 2  2 2 = ⋅ (1 + 1) − (1 + 0)  = ⋅  2 − 1 = ⋅ 2 2 − 1 . 3   3   3

Exercício 2.

1

1

(

)

)

1

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

−1

(

)

((

)

1 = − ⋅ x2 +1 2

−1

(

)

1 −1

)

1  1  = − ⋅ 2  2  x + 1

1

−1

)

−2

d ( x 2 + 1) =

 1  1 1 =0 = − ⋅  2 − 2 2  1 + 1 (−1) + 1 

(1 + l nx) dx . x 1

e

Exercício 3.

∫

(1 + l nx) dx dx =∫ (1 + l nx) ⋅ = ∫ (1 + l nx) ⋅ d (l nx) = ∫ d (l nx) + ∫ l nx d (l nx) = ∫1 x x 1 1 1 1 e

e

= (l nx ) 1

e

e

e

e

e

 ln2 x   l n 2e l n 21  1 3  = (l n e − l n1) +   = 1 + = . +  − 2  2 2  2 1  2

1

Matemática 1


2

dx

∫

Exercício 4.

16 − x 2

0

2

∫ 0

2

dx 16 − x

2

dx

=∫

4 −x 2

0

2

.

 x = arcsen   4

2

0

2 0 = arcsen  − arcsen  = 4 4

π π 1 = arcsen  − arcsen(0 ) = − 0 = . 6 6 2

1

∫x

Exercício 5.

0

2

dx . + 4x + 5

O polinómio do denominador não tem raizes reais: − 4 ± 16 − 20 − 4 ± − 4 = . 2 2 Portanto é uma fracção elementar do terceiro tipo. x 2 + 4x + 5 = 0 ⇒ x =

dx dx dx d ( x + 2) ∫0 x 2 + 4 x + 5 = ∫0 x 2 + 4 x + 4 + 1 = ∫0 ( x + 2) 2 + 1 = ∫0 12 + ( x + 2) 2 = 1

1

1

1

= (arctg ( x + 2) ) 0 = arctg (1 + 2) − arctg (0 + 2) = arctg (3) − arctg (2) . 1

π 2

Exercício 6.

∫ sen(2 x) dx . 0

1o π

método : π

π

π 1 1 2 1 2 = ( ) sen ( 2 x ) dx = sen ( 2 x ) ⋅ ⋅ d ( 2 x ) = ⋅ sen ( 2 x ) ⋅ d ( 2 x ) = − ⋅ c os ( 2 x ) ∫0 ∫0 0 2 2 ∫0 2 2

2

 1   π 1 1 = − ⋅  cos 2 ⋅  − c os (2 ⋅ 0)  = − ⋅ (cos (π ) − c os (0) ) = − ⋅ (− 1 − 1) = 1 . 2   2 2 2 

2

Matemática 1

2o


método :

π

π

π

π

 sen x  ∫0 sen(2 x) dx = ∫0 2 ⋅ senx ⋅ cos x ⋅ dx = 2 ⋅ ∫0 senx ⋅ d ( senx) = 2 ⋅  2  2

2

2

2

  π   sen 2    2  12 0 2 2  sen 0    = 2⋅ − = 2 ⋅  −  2 2  2 2    

2

=

0

  = 1 . 

π 2

cosx dx . 3 x

∫ π sen

Exercício 7.

6

π

π

π − 3+1

 ( sen x) c osx d ( senx) −3 ∫π sen 3 x dx = π∫ sen 3 x = π∫ (sen x) ⋅ d ( sen x) =  − 3 + 1 2

2

6

2

6

π

  

2

π

π

1  1  = − ⋅  2  sen 2 x 

π

=

6

6

6

2

        1  1 1 1 1 1  1 3  =− ⋅ − = − ⋅ 2 − = − ⋅ (1 − 4 ) = . 2    2 2 1 1 2 2 2π  2π     sen   sen        2  6   2  

∫ (x − 2e )⋅ dx . 1

x

Exercício 8.

0

∫( 1

0

)

 x2 x − 2e ⋅ dx = ∫ x ⋅ dx − ∫ 2e ⋅ dx = ∫ x ⋅ dx − 2 ⋅ ∫ e ⋅ dx =   2 0 0 0 0 x

 12 0 2 =  − 2 2

1

1

1

1

x

(

x

1

( )

  − 2 ⋅ e x 0

1 0

=

)

 1 5  − 2 ⋅ e1 − e 0 = − 2e + 2 = − 2e . 2 2 

∫(

)

8

Exercício 9.

2 x + 3 x ⋅ dx .

0

∫( 8

0

)

8

8

8

1 2

8

1 3

2 x + x ⋅ dx = ∫ 2 x ⋅ dx + ∫ x ⋅ dx = 2 ⋅ ∫ x ⋅ dx + ∫ x ⋅ dx = 3

3

0

0

0

0

3

Matemática 1


8

 1 +1   1 +1   x3   x2     = 2⋅ + 1   1 +1    +1 2  0 3 

=

2 2 3

8 8

3 4 2  2 3  3    = 2 ⋅ ⋅ x  + ⋅ x  3   4   0

8

= 0

0

3 4  2 2  3  3  4 3 2 2 3 100 . ⋅  8 2 − 0 2  + ⋅  8 3 − 0 3  = ⋅ 83 + ⋅ 3 84 = ⋅ 16 2 + ⋅ 16 = 4 3 4 3 4 3    

2

1

A função

2x − 1 ⋅ dx . 3 +x

∫x

Exercício 10.

2x −1 é racional. x3 + x

2x − 1 2x − 1 A Bx + C = = + 2 3 2 x + x x( x + 1) x x +1

⇒ 2 x − 1 = Ax 2 + A + Bx 2 + Cx = ( A + B ) x 2 + Cx + A

Obtemos o sistema: A + B = 0 ,  ⇔  C = 2,  A = −1, 

 A = −1,   B = 1,  C = 2. 

Portanto 2x − 1 1 x+2  −1 x + 2  ∫1 x 3 + x ⋅ dx = ∫1  x + x 2 + 1  ⋅ dx = − ∫1 x ⋅ dx + ∫1 x 2 + 1 ⋅ dx = 2

2

2

2

2

2

2

2

2

1 2  1 x 2  x = − ∫ ⋅ dx + ∫  2 + 2 ⋅ dx + ∫ 2 ⋅ dx =  ⋅ dx = − ∫ ⋅ dx + ∫ 2 x x + + x + 1 x + 1 x 1 x 1   1 1 1 1 1 2

2

2

1 1 1 1 = − ∫ ⋅ dx + ⋅ ∫ 2 ⋅ d ( x 2 + 1) + 2 ⋅ ∫ 2 ⋅ dx = x 2 1 x +1 1 1 x +1 = − (l n x

)

2 1

+

(

1 ⋅ l n x2 +1 2

+ 2 ⋅ (arctg 2 − arctg 1) =

)

2 1

+ 2 ⋅ (arctg x ) 1 = − (l n 2 − l n 1) + 2

1 ⋅ (l n 5 − l n 2 ) + 2

1 3 π 1 π  5  ⋅ l n 5 − ⋅ l n 2 + 2 ⋅  arctg 2 −  = ⋅ l n   + 2 ⋅  arctg 2 −  . 2 2 4 2 4  8 

4

Matemática 1


b) Calcular os integrais efectuando a substituição de variável. π 2

∫ senx ⋅ cos

Exercício 11.

2

x ⋅ dx .

0

Fazemos a substituição senx = t . Então temos: dt senx = t ⇒ x = arcsen t ⇒ dx = e cos 2 x = 1 − sen 2 x = 1 − t 2 . 2 1− t Determinamos os limites de integração para a variável t : xi nf = 0 ⇒ t i nf = sen(0) = 0 ,

x s up =

π

π  ⇒ t s up = sen  = 1 2

2

Portanto π 1

2

1

1

1

1 2 2 ∫0 senx ⋅ cos x ⋅ dx = ∫0 t ⋅ (1 − t ) ⋅ 1 − t 2 ⋅ dt = ∫0 t ⋅ 1 − t ⋅ dt = 2 ⋅ ∫0 1 − t ⋅ d (t ) = 2

2

1

1

2

(

1 1 = − ⋅ ∫ 1 − t 2 ⋅ d (1 − t 2 ) = − ⋅ ∫ 1 − t 2 2 0 2 0  1  1 − t 2 =− ⋅ 3 2    2

(

)

3 2

     

(

1  = − ⋅  1 − t 2 3 

(

)

= 0

)

3 2

  

1

0

(

1  = − ⋅  1 − 12 3 

) − (1 − 0 ) 3 2

3 2 2

 1 =3. 

0

∫e 0

1 2

1

1

1

Exercício 12.

)

  1  2 2 +1  1 1− t  ⋅ d (1 − t 2 ) = − ⋅  2  1 + 1    2 

x

1 ⋅ dx . + e−x

Fazemos a substituição e x = t . Então temos: dt e x = t ⇒ x = l n t ⇒ dx = . t Determinamos os limites de integração para a variável t :

xi nf = 0 ⇒ t i nf = e 0 = 1 , Portanto 1 1 1 ∫0 e x + e − x ⋅ dx = ∫0

x s up = 1 ⇒ t s up = e1 = e .

e

e

1 1 e ⋅ dx = ∫ ⋅ ⋅ dt = ∫ 2 ⋅ dt = (arctg t ) 1 = 1 1 t 1 t + 1 t +1 ex + x t e 1

= arctg e − arctg 1 = arctg e −

π

4

1

.

5

Matemática 1


4

Exercício 13.

∫x⋅

x 2 + 9 ⋅ dx .

0

Fazemos a substituição

x 2 + 9 = t . Então temos:

x2 + 9 = t ⇒ x2 + 9 =t 2

t ⋅ dt

⇒ x = t 2 − 9 ⇒ dx =

t2 −9

.

Determinamos os limites de integração para a variável t : xi nf = 0 ⇒ t i nf = 9 = 3 ,

x s up = 4 ⇒ t s up = 4 2 + 9 = 5 .

Portanto 4

∫x⋅

5

x + 9 ⋅ dx = ∫ 2

0

3

2

Exercício 14.

∫ 0

5

 t3  53 33 98 t − 9 ⋅t⋅ ⋅ dt = ∫ t ⋅ dt =   = − = . 3 3 t2 −9 33 3 3 5

t

2

dx x + 1 + ( x + 1) 3

2

.

Fazemos a substituição x + 1 = t . Então temos: x + 1 = t ⇒ x + 1 = t 2 ⇒ x = t 2 − 1 ⇒ dx = 2t ⋅ dt . Determinamos os limites de integração para a variável t : xi nf = 0 ⇒ t i nf = 1 = 1 ,

x s up = 2 ⇒ t s up = 2 + 1 = 3 .

Portanto 2

∫ 0

2

dx x + 1 + ( x + 1) 3 3

= 2⋅

dt

∫1+ t 1

2

=∫ 0

2t ⋅ dt t ⋅ dt = ∫ = 2⋅ ∫ = 3 2 3 x + 1 + ( x + 1) 1 t +t 1 t (1 + t ) dx

3

(

3

)

3 π π  π = 2 ⋅ (arctg t ) 1 = 2 ⋅ arctg 3 − arctg 1 = 2 ⋅  −  = . 3 4 6

π 2

Exercício 15.

dx

∫ 1 + sen x + cos x . 0

 x Fazemos a substituição tg   = t . Então temos:  2 senx =

2t 1− t2 , c os x = . 1+ t2 1+ t2 6

Matemática 1


 x tg   = t ⇒ 2

x = arctg t ⇒ 2

x = 2 ⋅ arctg t ⇒ dx =

2 ⋅ dt . 1+ t2

Determinamos os limites de integração para a variável t : xi nf = 0 ⇒ t i nf = tg (0) = 0 ,

x s up =

π 2

π  ⇒ t s up = tg   = 1 . 4

Portanto π 1

2

dx ∫0 1 + sen x + cos x = ∫0

2t 1− t2 1+ + 1+ t2 1+ t2

1

1 ⋅ d (1 + t ) = (l n 1 + t 1 + t 0

=∫

4

dx

∫1+

Exercício 16.

x

0

)

1 0

2 ⋅ dt 2 ⋅ dt dt =∫ =∫ = 2 2 2 1 t + 1+ t 1 t 2 t 1 t + + + − 0 0 1

1

⋅

1

= (l n 1 + 1 − l n 1 + 0 ) = l n 2 .

.

Fazemos a substituição x = t . Então temos: x = t ⇒ x = t 2 ⇒ dx = 2t ⋅ dt . Determinamos os limites de integração para a variável t : xi nf = 0 ⇒ t i nf = 0 = 0 ,

x s up = 4 ⇒ t s up = 4 = 2 .

Portanto

2t ⋅ dt t ⋅ dt 1+ t −1 1  1+ t ∫0 1 + x = ∫0 1 + t = 2 ⋅ ∫0 1 + t = 2 ⋅ ∫0 1 + t ⋅ dt = 2 ⋅ ∫0  1 + t − 1 + t  ⋅ dt = 4

dx

2

2

2

2

2

1 2 ⋅ d (1 + t ) = 2 ⋅ (t ) 0 − 2 ⋅ (l n 1 + t 1+ t 0

= 2 ⋅ ∫ dt − 2 ⋅ ∫ 0

2

)

2 0

=

= 2 ⋅ (2 − 0) − 2 ⋅ (l n 1 + 2 − l n 1 + 0 ) = 4 − 2 ⋅ l n 3 . 0

Exercício 17.

∫1+

−1

dx 3

x +1

.

Fazemos a substituição 3 x + 1 = t . Então temos: 3 x + 1 = t ⇒ x + 1 = t 3 ⇒ x = t 3 − 1 ⇒ dx = 3t 2 ⋅ dt .

7

Matemática 1


Determinamos os limites de integração para a variável t : xi nf = −1 ⇒ t i nf = 3 − 1 + 1 = 0 ,

x s up = 0 ⇒ t s up = 3 0 + 1 = 1 .

Portanto 0

∫1+

−1

dx 3

x +1

1 2 1 1  1 3t 2 dt t dt 1+ t 2 −1 t 2 −1  ⋅ dt = = 3⋅ ∫ = 3⋅ ∫ ⋅ dt = 3 ⋅ ∫  + 1 + t 1 + t 1 + t 1 + t 1 + t   0 0 0 0

1

=∫

(t − 1)(t + 1)   1  1  = 3⋅ ∫ + + t − 1 ⋅ dt =  ⋅ dt = 3 ⋅ ∫  1+ t 1+ t 1+ t   0 0 1

1

1

1

1

1 = 3⋅ ∫ ⋅ dt + 3 ⋅ ∫ t ⋅ dt − 3 ⋅ ∫ dt = 3 ⋅ (l n 1 + t 1+ t 0 0 0

)

1 0

t2 + 3 ⋅  2

1

 1  − 3 ⋅ (t ) 0 = 0

 12 0 2  3 = 3 ⋅ (l n 1 + 1 − l n 1 + 0 ) + 3 ⋅  −  − 3 ⋅ (1 − 0) = 3 ⋅ l n 2 − . 2  2 2 3

x ⋅ dx . 6−x

∫

Exercício 18.

0

x = t . Então temos: 6− x


x =t ⇒ 6− x  6t 2 dx =  2 1+ t

x =t2 6− x

⇒ x = 6t 2 − x ⋅ t 2

⇒

′  12t (1 + t 2 ) − 6t 2 ⋅ 2t 12t  ⋅ dt = ⋅ dt = 2 2 1+ t 1+ t2 

(

)

(

)

2

x=

6t 2 1+ t2

⇒

⋅ dt .

Determinamos os limites de integração para a variável t : xi nf = 0 ⇒ t i nf = 0 ,

x s up = 3 ⇒ t s up = 1 .

Portanto 3

∫ 0

1

x 12t ⋅ dx = ∫ t ⋅ 6− x 1+ t2 0

(

1

)

2

⋅ dt = 12 ⋅ ∫ 0

1

t2

(1 + t )

2 2

⋅ dt = 6 ⋅ ∫ 0

t ⋅ 2t ⋅ dt

(1 + t )

2 2

1

= 6⋅ ∫t ⋅ 0

2t ⋅ dt

(1 + t )

2 2

=

Na continuação integramos por partes: U = t , dU = dt ;

8

Matemática 1

dV =

2t ⋅ dt

(1 + t )

2 2


, V =∫

2t ⋅ dt

(1 + t )

2 2

=∫

d (t 2 )

(1 + t )

2 2

=∫

d (1 + t 2 )

(1 + t )

2 2

(1 + t ) =

2 − 2 +1

− 2 +1

=−

1 . 1+ t2

Obtemos

  t  1 1 dt    1 0  1 = 6 ⋅ −  +∫ − + (arctg t ) 0  =  = 6 ⋅ −  2  2 2 2   1+1 1+ 0     1 + t  0 0 1 + t   1 π  3 ⋅ (π − 2) . = 6⋅− +  = 2  2 4

1

Exercício 19.

∫e 0

dx . +1

x

Fazemos a substituição e x = t . Então temos: dt e x = t ⇒ x = l n t ⇒ dx = . t Determinamos os limites de integração para a variável t :

xi nf = 0 ⇒ t i nf = e 0 = 1 ,

x s up = 1 ⇒ t s up = e1 = e .

Portanto e e e 1  1+ t  dx dt 1+ t − t t ∫0 e x + 1 = ∫1 (t + 1) ⋅ t = ∫1 (t + 1) ⋅ t ⋅ dt = ∫1  (t + 1) ⋅ t − (t + 1) ⋅ t  ⋅ dt = e e e 1 1  1 1  ⋅ dt = ∫ ⋅ dt − ∫ = ∫  − ⋅ d (t + 1) = (l n t t ( t + 1 ) t t + 1   1 1 1

) − (l n t + 1 ) e

e

1

1

=

= (l ne − l n1) − (l n(e + 1) − l n 2 ) = l ne + l n 2 − l n(e + 1) = l n (2e) − l n(e + 1) = l n

3

Exercício 20.

dx

∫

(1 + x 2 ) 3

1

2e . e +1

.

Fazemos a substituição x = tg t . Então temos: dt dx = , c os 2 t

1 + x = 1 + (tg t ) 2

2

2

 sen t  sen 2 t cos 2 t + sen 2 t 1  = 1 + = 1 +  = = . 2 2 cos t cos t cos 2 t  cos t 

Porque x = tg t ⇒ t = arctg x , determinamos os limites de integração para a variável t :

9

Matemática 1


xi nf = 1 ⇒ t i nf = arctg (1) =

π 4

x s up = 3 ⇒ t s up = arctg ( 3 ) =

,

π 3

.

Portanto π 3

3

dx

∫

=

(1 + x 2 ) 3

1

∫ 1

dx

( 1+ x )

3

2

π

3

∫

=

dt cos 2 t

π 3

1  4   cos 2 t   

π

=∫

3

dt cos 2 t

 1  4   c os t   

π

π

3

=

3

∫

π

4

dt π 2 c os t 3 c os 3t ⋅ dt =∫ = 1 cos 2 t π cos 3 t 4

π

3 2 3− 2 π  π  = ∫ c os t ⋅ dt = (sent ) π3 = sen  − sen  = − = . 2 2 2 3 4 π 3 3

4

c) Calcular os integrais aplicando o método de integração por partes. 1

∫ xe

Exercício 21.

−x

dx .

0

Fazemos: U = x ⇒ dU = dx ,

dV = e − x dx ⇒ V = ∫ e − x dx = −e − x .

Portanto 1

∫ xe

−x

dx = (x ⋅ (− e )) −x

1 0

0

− ∫ (− e )dx = −(x ⋅ (e )) 1

−x

−x

0

+ ∫ e − x dx = − x ⋅ e − x

0

( ( )) − (e )

= − x ⋅ e−x

1

−x

0

1 0

( ( )) + (− e )

1

1

1

0

−x

1 0

=

0

(

) (

)

1 1 2 = − 1 ⋅ e −1 − 0 ⋅ e 0 − e −1 − e 0 = − − + 1 = 1 − . e e e

1 2

∫ arcsen x dx .

Exercício 22.

0

Fazemos:

′ U = arcsen x ⇒ dU = (arcsen x ) dx =

1 1− x2

dx ,

dV = dx ⇒ V = ∫ dx = x .

Portanto 1 2

∫ arcsen x dx = (x ⋅ arcsen x )

1 2 0

1 2

−∫

0

0

1 2 0

= ( x ⋅ arcsen x ) +

1 2

1

x 1− x2

dx = (x ⋅ arcsen x )

1 2 0

1 2 d (x2 ) − ⋅∫ = 2 0 1− x2

1 2

1 d (− x ) 1 d (1 − x 2 ) ⋅∫ = ( x ⋅ arcsen x ) + ⋅ ∫ = 2 0 1− x2 2 0 1− x2 2

1 2 0

10

Matemática 1


1

= (x ⋅ arcsen x ) 02 +

1 2

(

1 ⋅ ∫ 1− x2 2 0

)

−

1 2

  1  2 − 2 +1  1 1− x  + ⋅ 1 2    − +1   2 

(

1

⋅ d (1 − x 2 ) = ( x ⋅ arcsen x ) 02

1  2 2 1    1   1 =  ⋅ arcsen  − 0 ⋅ arcsen(0 ) +  1 −    − 1 − 0 2 2 2     2   

(

)

1 2

)

1 2

= 0

  1 π π 3 3 −1 = + − 1. = ⋅ + 4 12 2  2 6 

π 3

x ⋅ dx . 2 x

∫ π sen

Exercício 23.

4

Fazemos: U = x ⇒ dU = dx ,

dV =

dx sen 2 x

⇒ V =∫

dx = − ctg x . sen 2 x

Portanto π

π

π

π

π

3 cos x x ⋅ dx 3 3 ( ) ( ) = − x ⋅ ctg x − ( − ctg x ) ⋅ dx = − x ⋅ ctg x + ⋅ dx = π π ∫π sen 2 x ∫ ∫ π sen x π 4 4 3

3

4

4

π

π 3

= − ( x ⋅ ctg x ) π3 + ∫ 4

4

π

π

d ( sen x) = − ( x ⋅ ctg x ) π3 + (l n senx sen x 4

π

)π

3

=

4

4

π π π  π   π  π  = −  ⋅ ctg   − ⋅ ctg    +  l n sen  − l n sen   3 4  4   3 4  3

π

Exercício 24.

∫x

3

 π (9 − 4 3 ) 1 3 = + ⋅ ln  .  36 2 2 

⋅ senx ⋅ dx .

0

Fazemos: U = x3

dV = sen x dx ⇒ V = ∫ sen x dx = − c os x .

⇒ dU = 3 x 2 dx ,

Portanto π

∫x

3

π

⋅ senx ⋅ dx = (− x ⋅ cos x ) − ∫ 3x 2 ⋅ (−c os x) ⋅ dx = (− π 3 ⋅ cos π + −0 3 ⋅ cos 0 ) + π

3

0

0

0

π

π

+ 3 ⋅ ∫ x ⋅ cos x ⋅ dx = π + 3 ⋅ ∫ x 2 ⋅ cos x ⋅ dx = 2

0

3

0

11

Matemática 1


Integramos por partes: U = x2

dV = c os x dx ⇒ V = ∫ cos x dx = sen x .

⇒ dU = 2 xdx ,

Portanto na continuação temos:  = π + 3 ⋅  x 2 ⋅ senx 

(

3

)

π 0

π

 − ∫ 2 x ⋅ sen x ⋅ dx  = 0  π

π

0

0

= π 3 + 3 ⋅ (π 2 ⋅ sen(π ) − 0 2 ⋅ sen(0) ) − 6 ⋅ ∫ x ⋅ sen x ⋅ dx = π 3 − 6 ⋅ ∫ x ⋅ sen x ⋅ dx = Mais uma vez integramos por partes: U = x ⇒ dU = dx ,


Portanto obtemos: π   π ( ) = π − 6 ⋅  − x ⋅ c os x 0 − ∫ (−c os x) ⋅ dx  = 0   3

π

= π + 6 ⋅ (π ⋅ c os (π ) − 0 ⋅ c os (0) ) − 6 ⋅ ∫ cos x ⋅ dx = 3

0

= π − 6π − 6 ⋅ (sen x ) 0 = π − 6π − 6 ⋅ (sen (π ) − sen (0) ) = π 3 − 6π . π

3

3

π 2

Exercício 25.

∫e

2x

⋅ cos x ⋅ dx .

0

Fazemos: U = e2x

dV = cos x dx ⇒ V = ∫ c os x dx = sen x .

⇒ dU = 2 ⋅ e 2 x ⋅ dx ,

Portanto π

(

2

2x 2x ∫ e ⋅ cos x ⋅ dx = e ⋅ sen x

)

π 2 0

π 2

− ∫ 2 ⋅ e 2 x ⋅ sen x ⋅ dx =

0

0

π

π

π

2  2⋅  π  =  e 2 ⋅ sen   − e 2⋅0 ⋅ sen (0)  − 2 ⋅ ∫ e 2 x ⋅ sen x ⋅ dx = e π − 2 ⋅ ∫ e 2 x ⋅ sen x ⋅ dx = 2 0 0   2

Mais uma vez integramos por partes: U = e2x

⇒ dU = 2 ⋅ e 2 x ⋅ dx ,


12

Matemática 1


Na continuação temos:   π = e − 2 ⋅  − e 2 x ⋅ c os x  

(

  π = e + 2 ⋅  e 2 x ⋅ cos x   Portanto obtemos:

(

)

)

π 2 0

π 2 0

π

  − ∫ 2 ⋅ e ⋅ (− c os x) ⋅ dx  = 0   2

2x

π

π  2  π 2x − ∫ 2 ⋅ e ⋅ c os x ⋅ dx  = e − 2 − 4 ⋅ ∫ e 2 x ⋅ cos x ⋅ dx . 0 0   2

π

π

2

2

0

0

π 2x 2x ∫ e ⋅ cos x ⋅ dx = e − 2 − 4 ⋅ ∫ e ⋅ cos x ⋅ dx .

Resolvemos a equação obtida em relação ao integral: π

π

2

2

5 ⋅ ∫ e 2 x ⋅ cos x ⋅ dx = e π − 2 e 0

2x ∫ e ⋅ cos x ⋅ dx = 0

eπ − 2 . 5

e

Exercício 26.

∫ sen (l nx) dx . 1

Fazemos: 1 ′ U = sen (l nx) ⇒ dU = (sen (l nx) ) ⋅ dx = c os (l nx) ⋅ ⋅ dx , x dV = dx ⇒ V = ∫ dx = x . Portanto e

e

1 e ∫1 sen (l nx) dx = (x ⋅ sen (l nx) ) 1 − ∫1 x ⋅ cos (l nx) ⋅ x ⋅ dx = e

e

1

1

= (e ⋅ sen (l n(e)) − 1 ⋅ sen (l n(1)) ) − ∫ c os (l nx) ⋅ dx = e ⋅ sen(1) − ∫ cos (l nx) ⋅ dx = Mais uma vez integramos por partes: 1 ′ U = c os (l nx) ⇒ dU = (cos (l nx) ) ⋅ dx = − sen (l nx) ⋅ ⋅ dx , x dV = dx ⇒ V = ∫ dx = x .

13

Matemática 1


Na continuação temos: e   e = e ⋅ sen(1) −  ( x ⋅ cos (l nx) ) 1 − ∫ (− sen (l nx)) ⋅ dx  = 1  

e

= e ⋅ sen(1) − (e ⋅ c os (l n(e)) − 1 ⋅ cos (l n(1)) ) − ∫ sen (l nx) ⋅ dx = 1 e

= e ⋅ sen(1) − e ⋅ cos (1) + 1 − ∫ sen (l nx) ⋅ dx . 1

Portanto obtemos: e

∫

e

sen (l nx) ⋅ dx = e ⋅ sen(1) − e ⋅ cos (1) + 1 − ∫ sen (l nx) ⋅ dx .

1

1

Resolvemos a equação obtida em relação ao integral: e

e

2 ⋅ ∫ sen (l nx) ⋅ dx = e ⋅ sen(1) − e ⋅ cos (1) + 1 e

∫ sen (l nx) ⋅ dx =

1

1

1

∫

Exercício 27.

arcsen x 1+ x

0

e ⋅ sen(1) − e ⋅ c os (1) + 1 . 2

⋅ dx .

Fazemos:

′ U = arcsen x ⇒ dU = (arcsen x ) ⋅ dx =

1 1− x2

⋅ dx ,

− +1 ( 1 + x) 2 ⋅ dx = ∫ (1 + x ) ⋅ d (1 + x) = 1

dV =

1 1+ x

⋅ dx ⇒ V = ∫

1

−

1+ x

1 2

1 − +1 2

= 2 1+ x .

Portanto 1

∫

arcsen x 1+ x

0

(

⋅ dx = 2 1 + x ⋅ arcsen x

) − ∫2 1

0

1

1+ x ⋅

0

(

)

1

1 1− x2

= 2 1 + 1 ⋅ arcsen (1) − 2 1 + 0 ⋅ arcsen (0) − ∫ 2 1 + x ⋅ 0

=2 2⋅

π 2

1

− 2⋅∫ 0

1

1 1− x

 = 2π + 4 ⋅  (1 − x) 

1 2

   

⋅ dx = 2π + 2 ⋅ ∫ 0

1

0

⋅ dx =

1 1− x ⋅ 1+ x

⋅ dx =

1   − +1   2 1 (1 − x)   ⋅ d (1 − x) = 2π + 2 ⋅  − 1 +1  1− x    2 

 = 2π + 4 ⋅  (1 − 1) 

1 2

1

= 0

1  − (1 − 0) 2  = 2π − 4 . 

14

Matemática 1


1

Exercício 28.

∫ arctg

x ⋅ dx .

0

Fazemos: U = arctg x

(

)

′ ⇒ dU = arctg x ⋅ dx =

1

1+

dV = dx ⇒ V = x . Portanto

(

1

∫ arctg

x ⋅ dx = x ⋅ arctg x

0

(

) − ∫x⋅ 2 1

1

0

)

0

( x)

2

1 x ⋅ (1 + x)

⋅

( x )′ ⋅ dx =

dx 2 x ⋅ (1 + x)

,

⋅ dx =

1 x π 1 x = 1 ⋅ arctg 1 − 0 ⋅ arctg 0 − ⋅ ∫ ⋅ dx = − ⋅ ∫ ⋅ dx = 2 0 1+ x 4 2 0 1+ x Fazemos a substituição x = t ⇒ x = t2

1

1

⇒ dx = 2t dt ;

Determinamos os limites de integração para a variável t : xi nf = 0 ⇒ t i nf = 0 = 0 ,

x s up = 1 ⇒ t s up = 1 = 1 .

Na continuação temos:

=

=

=

π 4

π π 1 t t2 1+ t 2 −1 2 dt ⋅∫ ⋅ t ⋅ dt = − ⋅ = − ⋅ dt = 2 0 1+ t2 4 ∫0 1 + t 2 4 ∫0 1 + t 2 1

−

1

π

1 1+ t 2 1 − ∫  − 2 4 0 1+ t 1+ t2

π 4

1  π 1 π 1 1  1  ⋅ dt = − ∫ 1 − ⋅ dt = − dt + ⋅ dt =  2 2 ∫ ∫ 4 0  1+ t  4 0  0 1+ t

− (t ) 0 + (arctg t ) 0 = 1

1

1

π 4

− (1 − 0) + (arctg (1) − arctg (0) ) =

π 2

−1.

1

Exercício 29.

∫ x ⋅ l n (1 + x

2

) ⋅ dx .

0

Fazemos: ′ U = l n (1 + x 2 ) ⇒ dU = l n (1 + x 2 ) ⋅ dx =

(

dV = x dx ⇒ V =

)

′ 1 2 x ⋅ dx ⋅ 1 + x 2 ⋅ dx = , 2 1+ x 1+ x2

(

)

x2 . 2

Portanto

15

Matemática 1


1

1  x2 x3 2    x ⋅ l n ( 1 + x ) ⋅ dx = ⋅ l n ( 1 + x ) − ⋅ dx = 2 ∫0 ∫  2  1 + x  0 0 1

2

 12  1 x + x3 − x 02 =  ⋅ l n (1 + 12 ) − ⋅ l n (1 + 0 2 )  − ∫ ⋅ dx = 2 2 2  0 1+ x 1

=

1

1

1  1 d (1 + x 2 ) 1 1 1  + ⋅ ∫ = ⋅ l n (2) − + ⋅ l n (1 + x 2 ) 2 2 2 2  0 2 0 1+ x

 x2 1 = ⋅ l n (2) −  2  2 =

1

1 x  1 x  ⋅ l n (2) − ∫  x − ⋅ dx = ⋅ l n (2) − ∫ x ⋅ dx + ∫ ⋅ dx = 2  2 2 2 1+ x  0 0 0 1+ x

(

(

)

1 0

=

)

1 1 1 1 ⋅ l n (2) − + ⋅ l n (1 + 12 ) − l n (1 + 0 2 ) = l n (2) − . 2 2 2 2

1

Exercício 30.

∫ l n (1 + x) ⋅ dx . 0

Fazemos: 1 ′ U = l n (1 + x) ⇒ dU = (l n (1 + x) ) ⋅ dx = ⋅ dx , 1+ x dV = dx ⇒ V = x . Portanto 1

∫ l n (1 + x) ⋅ dx = (x ⋅ l n (1 + x) ) 0

1

x 1+ x −1 −∫ ⋅ dx = (1 ⋅ l n (1 + 1) − 0 ⋅ l n (1 + 0) ) − ∫ ⋅ dx = 2 1+ x 0 0 1+ x 1

1 0

1

1

1

1  1  = l n (2) − ∫ 1 − ⋅ d (1 + x) =  ⋅ dx = l n (2) − ∫ dx + ∫ 1+ x  1+ x 0 0 0 1

1

1 1 1 ⋅ d (1 + x) = l n (2) − ( x ) 0 + (l n (1 + x) ) 0 = 1+ x 0

= l n (2) − ∫ dx + ∫ 0

= l n (2) − (1 − 0) + (l n (1 + 1) − l n (1 + 0) ) = 2 ⋅ l n (2) − 1 .

16

Matemática 1


d) Calcular os integrais. π 2

Exercício 31.

x + sen x

⋅ dx . ∫ π 1 + c os x 6

 x Fazemos a substituição tg   = t . Então temos:  2

senx =

2t 1− t2 , c os x = . 1+ t2 1+ t2

 x tg   = t ⇒ 2

x = arctg t ⇒ 2

x = 2 ⋅ arctg t ⇒ dx =

2 ⋅ dt . 1+ t2

Determinamos os limites de integração para a variável t : xi nf =

π 6

π  ⇒ t i nf = tg   ,  12 

x s up =

π 2

π  ⇒ t s up = tg   = 1 . 4

Portanto π 2

x + sen x

1

⋅ dx = ∫ ∫ π 1 + c os x π

  tg    12 

6

∫

2t 1 2t   1 + t 2 ⋅ 2 ⋅ dt =  2 ⋅ arctg t +  ⋅ dt = 2 ∫ 2 1+ t 1+ t2   π  tg    12  1+ t2

2 ⋅ arctg t +

1

=

2t 2t 2 ⋅ arctg t + 1 2 1 + t ⋅ 2 ⋅ dt = 1 + t 2 ⋅ 2 ⋅ dt = ∫ 1+ t2 +1− t2 1+ t2 1− t2 1+ t2 π  1+ tg    12  1+ t 2 1+ t2

2 ⋅ arctg t +

π  tg    12 

= 2⋅

1

1

  tg    12 

 tg    12 

∫πarctg t ⋅ dt + 2 ⋅ ∫π

t ⋅ dt = 1+ t2 

Integramos por partes o primeiro integral: U = arctg t ⇒ dU =

1 ⋅ dt ; 1+ t2

dV = dt ⇒ V = t .

Na continuação temos:

17

Matemática 1


  1 1 1   t t  = 2 ⋅  t ⋅ arctg (t )  − ∫ ⋅ dt  + 2 ⋅ ∫ ⋅ dt = 2 1+ t2  tg  π  tg  π  1 + t π    tg      12   12   12    1 1    π  t t π    = 2 ⋅ 1 ⋅ arctg (1) − tg   ⋅ arctg  tg     − 2 ⋅ ∫ ⋅ dt + 2 ⋅ ∫ ⋅ dt = 2 1+ t 1+ t2  12    12     π  π  tg   tg    12 

 12 

π π  π  π π π  = 2 ⋅  − ⋅ tg    = − ⋅ tg   .  12   2 6  12   4 12

π 2

∫ sen

Exercício 32.

3

x ⋅ cos x ⋅ dx .

0

1o

método :

π

π

π

2

2

3 2 ∫ sen x ⋅ cos x ⋅ dx = ∫ sen x ⋅ cos x ⋅ sen x ⋅ dx =

∫ (1 − cos x )⋅

0

0

0

π

2

2

π

= − ∫ (1 − cos x ) ⋅ 2

2

π

cos x ⋅ d (cos x) = − ∫ (c osx )

2

0

1 2

2

⋅ d (cos x) + ∫ (c osx ) 2 ⋅ d (cos x) =

0

π

   (c osx ) 12 +1   = −  1 +1     2 

2

0

cos x ⋅ d (−cos x) =

5

0

π

   (cosx ) 52 +1   +  5 +1     2 

2

π 3 2   = − ⋅  (c osx ) 2  3  

2 0

π 7 2   + ⋅  (cosx ) 2  7  

2

=

0

0

3 7     3  7  2     2  π 2 π 2 2 8     2 = − ⋅   cos    − (cos (0 )) 2  + ⋅   c os    − (c os (0 )) 2  = − = . 3   2   2   7   3 7 21    

2o

método :

Porque a função sen x tem exponente impar fazemos a seguinte substituição de variável: cos x = t ,

sen 2 x = 1 − c os 2 x = 1 − t 2 ,

senx dx = − d (c os x) .

18

Matemática 1


Determinamos os limites de integração para a variável t : xi nf = 0 ⇒ t i nf = c os (0) = 1 ,

x s up =

π 2

π  ⇒ t s up = c os  = 0 2

Portanto π

π

π

2

2

2

0

0

0

(

)

3 2 2 ∫ sen x ⋅ cos x ⋅ dx = ∫ sen x ⋅ cos x ⋅ sen x ⋅ dx = ∫ 1 − cos x ⋅ cos x ⋅ (−d (cos x)) =

5 0 1 1  1  = − ∫ 1 − t 2 ⋅ t ⋅ d t = ∫ 1 − t 2 ⋅ t ⋅ d t = ∫  t 2 − t 2  ⋅ d t = 1 0 0 

(

)

(

)

1

 1 +1   5 +1  1 1 1 5  t2   2   − t  = ∫t 2 ⋅ d t − ∫t 2 ⋅ d t =  1 +1  5 + 1 0 0     2 0 2 

1 1

3 7 2   2   = ⋅  t 2  − ⋅  t 2  3   7   0

1

= 0

2 2 8 − = . 3 7 21

0

2

Exercício 33.

∫ (x

3

− 2 x) ⋅ l n x ⋅ dx .

1

Integramos por partes. Fazemos: 1 ′ U = l n x ⇒ dU = (l n x ) ⋅ dx = ⋅ dx , x dV = ( x 3 − 2 x) ⋅ dx ⇒ V = ∫ ( x 3 − 2 x) ⋅ dx ⇒ V =

x4 − x2 . 4

Portanto 2

2  x4   x4 2 2 1   ∫1 ( x − 2 x) ⋅ l n x ⋅ dx =   4 − x  ⋅ l n x  − ∫1  4 − x  ⋅ x ⋅ dx =  1 2

3

2  24  2  x3  14   x3  2 2   =   − 2  ⋅ l n 2 −  − 1  ⋅ l n 1 − ∫  − x  ⋅ dx = − ∫  − x  ⋅ dx = 4  4    1  4  1 4

2

2

2  x4   x2   2 4 14   2 2 12  x3 15 3 9 = − ∫ ⋅ dx + ∫ x ⋅ dx = −  +   = − −  +  −  = − + = . 4 2 16 2 16  16  1  2  1  16 16   2 1 1 2

19

Matemática 1


4 2− x − 2+ x

2

Exercício 34.

∫(

)

2 + x + 4 2 − x ⋅ ( x + 2) 2

0

⋅ dx .

Transformamos a função integranda de modo a obter expressões de forma 2

∫( 0

2−x . 2+ x

2− x −1 4 2−x − 2+ x 2+ x ⋅ dx = ∫ ⋅ dx = (∗) 2 + x + 4 2 − x ⋅ ( x + 2) 2 2−x  0  2 1 + 4 ⋅  ⋅ ( x + 2)   2 x +   4⋅

2

)


2−x = t2. 2+ x

Então

2−x = t2 2+ x x=

⇒ 2 − x = t 2 ⋅ (2 + x) ⇒ 2 − x = 2 ⋅ t 2 + t 2 ⋅ x ⇒ 2 − 2 ⋅ t 2 = x ⋅ (t 2 + 1) ⇒

2 − 2 ⋅ t 2 4 − 2 − 2 ⋅ t 2 4 − (2 + 2 ⋅ t 2 ) 4 (2 + 2 ⋅ t 2 ) 4 = = = − = 2 −2. 2 2 2 2 2 t +1 t +1 t +1 t +1 t +1 t +1

Portanto 8t  4   4  dx = d  2 − 2 = d  2 =− 2  t +1   t + 1 t +1

(

)

2

⋅ dt .

Determinamos os limites de integração para a variável t :

xi nf = 0 ⇒ t i nf =

2−0 = 1, 2+0

x s up = 2 ⇒ t s up =

2−2 = 0. 2+2

Na continuação temos:  8t − ⋅ 2  2 1 (1 + 4 ⋅ t ) ⋅  2 4 − 2 + 2   t + 1  t +1  0

(∗) = ∫

4 ⋅ t −1

(

)

2

 0 (4 ⋅ t − 1) ⋅ 8t ⋅ dt  = ∫ ⋅ dt =  1 16 ⋅ (1 + 4 ⋅ t ) 

(4 ⋅ t − 1) ⋅ 8t 1 4⋅t2 − t ⋅ dt = − ⋅ ∫ ⋅ dt = 16 ⋅ (1 + 4 ⋅ t ) 2 0 4⋅t +1 1 0

=∫

1

A função integranda é racional irregular. Dividimos o polinómio do numerador pelo polinómio do denominador e obtemos: 1 1 1 1 1  1 1 1  1 1  1 1 1 1  = − ⋅ ∫ t − + ⋅  ⋅ dt = − ⋅ ∫ t ⋅ dt − ⋅ ∫  −  ⋅ dt + ⋅ ∫  ⋅  ⋅ dt = 2 0  2 2 4 ⋅ t + 1 2 0 2 0 2 2 0  2 4 ⋅ t +1

20

Matemática 1


1

1

1

1

1

1

1 1 1 1 1 1 1 1 = − ⋅ ∫ t ⋅ dt + ⋅ ∫ dt + ⋅ ∫ ⋅ dt = − ⋅ ∫ t ⋅ dt + ⋅ ∫ dt + ⋅ ∫ ⋅ d (4t + 1) = 2 0 4 0 4 0 4⋅t +1 2 0 4 0 16 0 4 ⋅ t + 1 1

1 1 t2  1   1  = − ⋅   + ⋅  t  + ⋅  l n 4 ⋅ t + 1 2  2  0 4   0 16 

   

1

= 0

1  12 0 2 = − ⋅  − 2 2 2

 1 1 1 1 ln5 ln5  + ⋅ (1 − 0) + ⋅ l n 4 ⋅ 1 + 1 − l n 4 ⋅ 0 + 1 = − + + = . 16 4 4 16 16  4

Exercício 35.

Calcular a área da região plana

{

}

A = ( x, y ) ∈ R 2 : y ≤ 2 − x 2 ∧ y ≥ x . A região é limitada pelo gráfico da parábola y = 2 − x 2 orientada em baixo e pela recta y = x . O esboço da região é apresentado na figura 1. Os limites de variação (integração) da variável x é o segmento que é a projecção da região sobre o eixo O x . Portanto para determinar os limites de integração determinamos as abcissas dos pontos de intersecção da parábola y = 2 − x 2 com a recta y = x: y = x ∧ y = 2 − x2 x=

⇒ x = 2 − x2

⇒ x2 + x − 2 = 0 ⇒

−1± 9 −1± 3 = ⇒ x = −2 ∨ x = 1 . 2 2

21

Matemática 1


Porque a linha que delimita a região na parte de baixo é dada analiticamente só por uma função e a linha que delimita a região na parte de cima também é dada analiticamente só por uma função temos:

∫ (2 − x 1

SA =

2

−2

)

1

1

− x dx = ∫ 2 dx − ∫ x dx − ∫ x dx = (2 x) −2

−2

 1 (−2) = (2 ⋅ 1 − 2 ⋅ (−2)) −  − 3 3 3

Exercício 36.

1

2

3

−2

 1 (−2)  −  − 2  2 2

2

1

−2

 x3  −    3

1

−2

 x2 −   2

  

1

= −2

 9 3 9  = 6 − + = . 3 2 2 


{

}

A = ( x, y ) ∈ R 2 : y 2 ≥ 2 x + 4 ∧ y ≥ x − 2 . A região é limitada pelo gráfico da parábola y 2 = 2 x + 4 orientada no sentido positivo do eixo O x e pela recta y = x − 2 . O esboço da região é apresentado na figura 2.

Porque y 2 = 2x + 4 ⇔

x=

1 2 y − 2 concluímos que o vértice B da parábola tem as 2 e a parábola intersecta o eixo O y nos pontos C = ( 0 , − 2 ) e

coordenadas ( − 2 , 0 ) D = ( 0 , 2 ). Determinamos as coordenadas dos pontos de intersecção da parábola 2 y = 2 x + 4 com a recta y = x − 2 : y 2 = 2 x + 4 ∧ y = x − 2 ⇒ ( x − 2) 2 = 2 x + 4 ⇒ x 2 − 4 x + 4 = 2 x + 4 ⇒

22

Matemática 1


x 2 − 6x = 0 ⇒ x = 0 ∨ x = 6 . Calculemos a área da região. 1o

método

A projecção da região sobre o eixo O x é o segmento [ − 2 , 6 ] . Porque a linha que delimita a região na parte de baixo é dada analiticamente por duas funções a área da região representa a soma das áreas das regiões (BCD) e (CDE ) . A região (BCD) é limitada na parte de baixo pelo ramo y = − 2 x + 4 , na parte de cima pelo ramo y = 2 x + 4 da parábola e a sua projecção sobre o eixo O x é o segmento [ − 2 , 0 ] . Portanto

∫( 0

S ( BCD ) =

−2

(

))

0

2 x + 4 − − 2 x + 4 dx = ∫ 2 ⋅ 2 x + 4 dx = −2

1   1 0  (2 x + 4) 2 +1   = ∫ ( 2 x + 4) 2 d ( 2 x + 4) =  1  −2 + 1    2 

0

∫

2 x + 4 d ( 2 x + 4) =

−2

0 3  2   = ⋅  (2 x + 4) 2  3  

0

3

−2

2 2 16 = ⋅ 42 = ⋅ 4 4 = . 3 3 3

−2

A região (CDE ) é limitada na parte de baixo pela recta y = x − 2 , na parte de cima pelo ramo y = 2 x + 4 da parábola e a sua projecção sobre o eixo O x é o segmento [ 0 , 6 ] . Portanto 6

S ( CDE ) = ∫

(

)

6

2 x + 4 − ( x − 2) dx = ∫

0

(

)

6

6

6

0

0

2 x + 4 − x + 2 dx = ∫ 2 x + 4 dx − ∫ x dx + 2 ∫ dx =

0

0 6

1   6 +1  2  x2  1 1  (2 x + 4)  6 = ⋅ ∫ (2 x + 4) d (2 x + 4) − ∫ x dx + 2 ∫ dx = ⋅ −   + 2 ⋅ ( x ) 0 = 2 0 2  1  2 0 0 0 + 1    2 0 6

1 2

6

6

6

6

3 3 3    62 02   x2  1  1  6 = ⋅  (2 x + 4) 2  −   + 2 ⋅ ( x ) 0 = ⋅  (2 ⋅ 6 + 4) 2 − (2 ⋅ 0 + 4) 2  −  −  + 2 ⋅ (6 − 0) = 3  3  2  0  2 0   2

1 56 38 ⋅ (64 − 8) − 18 + 12 = −6 = . 3 3 3 Portanto 16 38 54 S A = S ( BCD ) + S (CDE ) = + = = 18 . 3 3 3 =

2o

método

Observamos que em relação ao eixo O y a região é limitada à esquerda 1 pela parábola x = y 2 − 2 , à direita pela recta x = y + 2 e a sua projecção sobre o eixo 2 O y é o segmento [ − 2 , 4 ] . Portanto 23

Matemática 1


4

4

4

4

1 1  1    S A = ∫  y + 2 − y 2 + 2  dy = ∫  4 + y − y 2  dy = ∫ (4 + y ) d (4 + y ) − ⋅ ∫ y 2 dy = 2 2  2 −2  − 2 − 2 −2

 (4 + y) 2 =  2 

4

  

1  y3  − ⋅   2  3 

−2

Exercício 37.

4

−2

 82 2 2 =  − 2  2

 1  4 3 (−2) 3   − ⋅  −  = 30 − 12 = 18 . 3   2  3


{

}

A = ( x, y ) ∈ R 2 : x 2 + ( y − 1) 2 ≥ 1 ∧ x 2 + ( y − 2) 2 ≤ 4 ∧ y ≥ x 2 .

A região é situada fora da circunferência

x 2 + ( y − 1) 2 = 1 , dentro da

circunferência x 2 + ( y − 2) 2 = 4 e entre os ramos da parábola y = x 2 . O esboço da região é dado na figura 3 e porque as funções que delimitam a região são pares concluímos que a região é simétrica em relação ao eixo O y . Portanto para determinar a área S A da região calculemos a área S da metade da região situada á direita do eixo O y e multiplicamos o resultado obtido por dois. A metade da região situada á direita do eixo O y é limitada na parte de baixo pela circunferência x 2 + ( y − 1) 2 = 1 (mais precisamente pela semicircunferência y = 1+ 1− x2 )

e pela parábola y = x 2 . Na parte de cima é limitada pela

semicircunferência y = 2 + 4 − x 2 . Determinamos os pontos de intersecção da parábola com as circunferências:  y = x2  y = x2 y = x2  y = x2     ⇔  ⇔  ⇔  ⇔  2 2 2 2  x + ( y − 1) = 1 y + y − 2y +1 = 1  y −y=0 y = 0 ∨ y = 1     y = x2   y =0 

y = x2  ∨   y =1 

⇔ ( x, y ) = (0, 0) ∨ ( x, y ) = (−1, 1) ∨ ( x, y ) = (1, 1) .

24

Matemática 1


Para metade da região situada á direita do eixo O y temos os pontos ( x, y ) = (0, 0) e ( x, y ) = (1, 1) de intersecção da parábola com a circunferência x 2 + ( y − 1) 2 = 1 .  y = x2  y = x2 y = x2  y = x2     ⇔  ⇔  ⇔  ⇔  2 2 2 2  x + ( y − 2) = 4 y + y − 4y + 4 = 4  y − 3y = 0 y = 0 ∨ y = 3     y = x2   y =0 

y = x2  ∨   y=3 

⇔ ( x, y ) = (0, 0) ∨ ( x, y ) = (− 3 , 3) ∨ ( x, y ) = ( 3 , 3) .

Para metade da região situada

á direita do eixo O y temos os pontos

( x, y ) = (0, 0) e ( x, y ) = ( 3 , 3) de intersecção da parábola

com a circunferência

x + ( y − 2) = 4 . Portanto a projecção da parte direita da região sobre o eixo O x é o segmento 2

2

[ 0 , 3 ] . Calculemos a área dela S : 1

(

)

S = ∫ 2 + 4 − x 2 − 1 − 1 − x 2 dx + 0

1

∫ (2 +

4 − x 2 − x 2 dx =

1

(

)

= ∫ 1 + 4 − x − 1 − x dx + 2

2

0

∫ (2 +

)

3

4 − x 2 − x 2 dx =

1

1

1

1

3

= ∫ d x + ∫ 4 − x dx − ∫ 1 − x dx + 2 ∫ d x + 2

0

2

0

1

= ∫d x+ 0

0

3

∫

= ( x) 0 +

∫

1

∫

4 − x dx − ∫ x 2 dx = 1

3

1

1

1

4 − x dx − ∫ 1 − x dx + 2 ⋅ ( x) 1 2

0

1

3

 x3  −    3 

4 − x 2 dx − ∫ 1 − x 2 dx + 2 3 − 2 − 3 +

0

2 = 3− + 3

3

2

1

3

2

0

3

∫

1

0

3

3

4 − x 2 dx − ∫ 1 − x 2 dx + 2 ∫ d x − ∫ x 2 dx =

0

1

= 1+

)

3

0

3

1

∫

4 − x dx − ∫

0

0

2

= 1

1 = 3

2 1 − x dx = 3 − + 2 ⋅ 3 2

3

3

∫ 0

2

1

 x 1 −   dx − ∫ 1 − x 2 dx = (∗) 2 0

25

Matemática 1


No primeiro integral fazemos a substituição x = 2 sen t . Então 2

 x 1 −   = 1 − sen 2 t = cos 2 t = cos t 2 Determinamos os limites de integração para a variável t : d x = 2 ⋅ c os t ⋅ d t e

 xi nf   = arcsen (0) = 0 , xi nf = 0 ⇒ t i nf = arcsen  2    3 π  x s up    = arcsen  x s up = 3 ⇒ t s up = arcsen    2 = 3. 2     No segundo integral fazemos a substituição x = sen u . Então d x = c os u ⋅ d u e

1 − x 2 = 1 − sen 2 u = cos 2 u = c os u

Determinamos os limites de integração para a variável u :

xi nf = 0 ⇒ u i nf = arcsen (xi nf ) = arcsen (0) = 0 , x s up = 1 ⇒ u s up = arcsen (x s up ) = arcsen (1) =

π 2

.

Na continuação temos: π

π

3

2 2 2 (∗) = 3 − + 2 ⋅ ∫ 2 ⋅ cos t ⋅ dt − ∫ cos 2 u du = 3 0 0 1 + c os ( 2 α ) 1 cos (2α ) Como cos 2α = = + temos: 2 2 2

π

= 3−

π

2 2  1 c os(2 u )  + 2 ⋅ ∫ (1 + c os(2t ) ) ⋅ dt − ∫  +  du = 3 2 2  0 0 3

π

π

π

π

3 3 2 1 2 1 2 = 3 − + 2 ⋅ ∫ d t + ∫ cos(2t ) ⋅ d (2 t ) − ⋅ ∫ du − ⋅ ∫ c os(2 u ) ⋅ d (2 u ) = 3 2 0 4 0 0 0

π

2 = 3 − + 2 ⋅  t    3 = 3−

3 0

π

  +  sen (2t )   

3

0

π

1 − ⋅  u  2  

2 0

π

1   − ⋅  sen (2 u )  4  

2

=

0

 1 π 2 π    2π   1 + 2 ⋅  − 0  +  sen   − sen (0)  − ⋅  − 0  − ⋅ (sen (π ) − sen (0) ) = 3 3    3   4  2 2

26

Matemática 1

= 3−


2 π 3 π 9 3 − 4 5π + 2⋅ + − = + . 3 3 2 4 6 12

Portanto

 9 3 − 4 5π  9 3 − 4 5π = . S A = 2 S = 2 ⋅  + +  6 12 3 6  

Exercício 38.

[

]

Calcular o comprimento da linha dada pela função f ( x) = l n x , com

x∈ 2 2, 2 6 . b

Para calcular o comprimento da linha utilizamos a fórmula l = ∫ 1 + [ f ′( x)] ⋅ dx . 2

a

Temos: a = 2 2 , b = 2 6 , f ′( x) =

1 . x

Portanto

x2 +1 x2 +1 1 1 +   ⋅ dx = ∫ ⋅ dx = ∫ ⋅ dx = (∗) x x2  x 2 2 2 2 2

2 6

l=

∫

2 2

2 6

2 6

Integramos por substituição.

x2 +1 = t ⇒ x2 +1 = t 2 ′ t t 2 − 1 ⋅ dt = ⋅ dt . 2 t −1

Fazemos

dx=

(

)

⇒ x2 = t 2 −1 ⇒

x = t 2 −1 .


xi nf = 2 2 ⇒ t i nf =

(x )

x s up = 2 6 ⇒ t s up =

(x )

2

i nf

2

s up

+1 =

(2 2 )

+1 =

(2 6 )

2

2

+1 = 9 = 3, + 1 = 25 = 5 .

Na continuação temos: 5

(∗) = ∫ 3

5

t t 2 −1

⋅

5 2 5  t 2 −1 t2 t −1+1 1   2  ⋅ dt = ⋅ dt = ⋅ dt = + 2 2 2 ∫ ∫ t −1 t − 1  3 t −1 3 3  t −1 5

t t 2 −1 5

⋅ dt = ∫ 5

1  1  = ∫ 1 + 2 ⋅ dt =  ⋅ dt = ∫ dt + ∫ 2 t −1 3 3 3 t −1 Representamos a fracção racional como soma de fracções elementares:

27

Matemática 1


1  A = ,  A + B = 0 A B 1 2 = + ⇒ 1 = ( A + B) ⋅ t + ( A − B) ⇒  ⇒  1 t 2 −1 t −1 t +1 A− B =1 B = − . 2  Então: 1   1 5 5 5   1 1 1 1 = ∫ dt + ∫  2 − 2  ⋅ dt = ∫ dt + ⋅ ∫ ⋅ dt − ⋅ ∫ ⋅ dt = t −1 t −1 2 3 t −1 2 3 t +1 3 3 3     5

5

5

5

5

1  1      =  t  + ⋅  l n (t − 1)  − ⋅  l n (t + 1)  =  3 2  3 2  3 = (5 − 3) +

= 2+

1   1   ⋅ l n (5 − 1) − l n (3 − 1) − ⋅ l n (5 + 1) − l n (3 + 1) = 2   2  

    1  1 1 2  16  ⋅ l n (4) − l n (2) − l n (6)+ l n (4) = 2 + ⋅ l n (16)− l n (12) = 2 + ⋅ l n   = 2 + l n .   2  2  2 3  12     

Exercício 39.

Calcular o comprimento da linha dada pela função f ( x) = l n (cos x) ,

 π  com x ∈  0 ,  .  6 b


a

Temos: a = 0 , b =

′ sen x   , f ′( x) =  l n (cos x)  = − . 6 c os x  

π

Portanto π

l=

6

∫ 0

2

π

π

π

6  sen x  sen x cos x + sen x 1  ⋅ dx = ∫ 1 + 1 +  − ⋅ dx = ⋅ dx = ⋅ dx = (∗) 2 ∫ 2 ∫ c os x c os x c os x  cos x  0 0 0 6

2

6

2

2

Integramos por substituição.

 x Fazemos tg   = t ⇒ 2

x = 2 ⋅ arctg (t ) ⇒ dx =

1− t2 2 . c os x = . ⋅ d t 1+ t2 1+ t2


28

Matemática 1


 xi nf   = tg (0) = 0 , xi nf = ⇒ t i nf = tg  2    x s up  π π  = tg   . x s up = ⇒ t s up = tg  6  12   2  Na continuação temos: π  tg    12 

π  tg    12 

1 2 2 ⋅ ⋅ dt = ∫ ⋅ dt = 2 2 2 1 − t 1 + t 1 − t 0 0 1+ t2 Representamos a fracção racional como soma de fracções elementares:  A = 1, A − B = 0 2 A B = + ⇒ 2 = ( A − B) ⋅ t + ( A + B ) ⇒  ⇒  2 1− t 1+ t 1− t  B = 1. A + B = 2

∫

(∗) =

Então: π  tg    12 

=

∫ 0

1   1 +   ⋅ dt = 1− t 1+ t 

π  tg    12 

=−

∫ 0

1 ⋅ d (1 − t ) + 1− t

π  tg    12 

π  tg    12 

∫ 0

∫ 0

1 ⋅ dt + 1− t

π  tg    12 

∫ 0

1 ⋅ dt = 1+ t

 1 ⋅ d (1 + )t = −  l n 1 − t 1+ t 

 π  = −  l n 1 − tg   − l n 1 − 0  12  

  

π  tg    12 

0

  π  +  l n 1 + tg   − l n 1 + 0    12   

 +  l n 1 + t 

  

π  tg    12 

=

0

 =  

   π  π   π   1 + tg     cos   + sen   π  π   12   = l n   12   12   = = l n 1 + tg   − l n 1 − tg   = l n     π  π   π   12   12   1 − tg     c os   − sen    12   12    12         c os  π  + sen π   ⋅  cos  π  − sen π         12   12     12   12    = ln  = 2  π   π    cos  12  − sen 12               3  π  π  π    c os 2   − sen 2     cos      12  12  6        2   = ln 3 . = ln = ln = ln       1 π π π π π           2 2  c os   − 2 ⋅ c os   ⋅ sen  + sen     1 − sen   1−  2  12   12   12   12    6    

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Matemática 1


Exercício 40. Calcular o comprimento da linha dada pela função f ( x) = x 3 , com  4 x ∈  0,  .  3 b


a

′

′   1  3  3 4 3 Temos: a = 0 , b = , f ′( x) =  x  =  x 2  = ⋅ x 2 .   3 2       Portanto 4 3

l=∫ 0

4

2

4

1

3  3 12  9 4 3 9  2  9    1 +  ⋅ x  ⋅ dx = ∫ 1 + x ⋅ dx = ⋅ ∫ 1 + x  ⋅ d 1 + x  = 4 9 0 4   4  0 2 

  9  1 + x  4  4  = ⋅ 1 9  +1  2 

1 +1 2

      

4 3

 4 2  9  = ⋅ ⋅  1 + x  9 3  4  

3 2

    

4 3

3 3   8   9 4  2  9  2  56 = ⋅  1 + ⋅  − 1 + ⋅ 0   = . 27   4 3   4   27  

0

0

Exercício 41.

Calcular o comprimento da linha dada pela função  1 f ( x) = 1 − x 2 + arcsen x , com x ∈  0 ,  .  2 b


a

1 Temos: a = 0 , b = , 2

′ ′ ′      −x 1  2 2     ′ f ( x) = 1 − x + arcsen x = 1 − x +  arcsen x  = + = 2       1− x 1− x2     −x 1 1− x 1− x 1− x 1− x 1− x = + = = = = = . 2 2 2 1+ x (1 − x) ⋅ (1 + x) 1− x ⋅ 1+ x 1+ x 1− x 1− x 1− x Portanto 1 2

1 2

2

1 2

1 2

 1− x  1− x 1+ x +1− x 2  ⋅ dx = ∫ 1 + l = ∫ 1 +  ⋅ dx = ∫ ⋅ dx = ∫ ⋅ dx =  1+ x 1+ x 1+ x 0 0 0 0  1+ x  1 2

= 2⋅∫ 0

1 1+ x

1 2

⋅ dx = 2 ⋅ ∫ (1 + x) 0

−

1 2

1   (1 + x) 2 ⋅ d (1 + x) = 2 ⋅  1    2

     

1 2

 3  = 2 2 ⋅  − 1 = 2( 3 − 2 ) .  2  0

30