01_Avaliação N3 - Cálculo I - Gabarito Comentado

Prévia do material em texto

• A resolução da prova deve ser a caneta azul ou preta; 
• Só será aceita como resposta correta a resolução completa correta de cada item; 
• Em todas as questões é necessário apresentar cálculos como justificativas; 
• Proibido durante a resolução da prova o porte de celular, mesmo desligado. 
 
QUESTÃO 1. (Valor 5) Se W dt
t
−
= ∫
1
2
3
1
2
, então W2 é 
A) – 5 
B) – 4 
C) – 3 
D) 9 
E) 16 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
FAPAC - Faculdade Presidente Antônio Carlos 
INSTITUTO TOCANTINENSE PRESIDENTE ANTÔNIO 
CARLOS LTDA - ITPAC-PORTO – TO 
 
Avaliações 
N1 ( ) N2 ( ) N3 ( x ) 
1ª Chamada AE ( ) 
Curso: Engenharia Civil Disciplina: Cálculo I Data: 07/06/2014 
Código: 54 Valor da Avaliação: 40,0 
Professor: Dr. Antônio Rafael Bôsso Duração da Avaliação: 2 Aulas 
Nota: Aluno (a): Material de Estudo 
 
 
( )
( )
−
−
−
−
−
−
=
=
= ⋅
−
= − ⋅
 
 = − − −
 
−   
 
 
= − − −
= − ⇒ =
∫
∫
1
2
3
1
1
2
3
1
1
2 2
1
1
2
2
1
2 2
2
2W dt
t
W 2 t dt
tW 2
2
1W
t
1 1W
1 1
2
W 4 1
W 3 W 9
 
QUESTÃO 2. (Valor 5) O resultado da Integral x x dx
x
+ −
⋅∫
2 4 4
 é 
A) x x x C⋅ + ⋅ − ⋅ +
5 3 1
2 2 22 8 8
5 3
 
B) x x x C⋅ + ⋅ − ⋅ +
5 3 1
2 2 25 4 2
2 3
 
C) x x x C+ ⋅ − ⋅ +
5 3 1
2 2 24 4 
D) x x⋅ + ⋅ −
5 3
2 22 8 8
5 3
 
E) x x C⋅ + ⋅ − +22 8 8
5 3
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
QUESTÃO 3. (Valor 5) Calcular a integral indefinida ( )3 4x x 1 dx+ ⋅∫ 
( )
( ) ( )
( )
( )
( )
( )
( ) ( )
3 4
4 3 3
3 4 4 3
3 4
1
3 4 2
3
2
3 4
3
3 4 2
3
3 4 4 2
x x 1 dx
du
u x 1 du 4x dx x dx
4
x x 1 dx x 1 x dx
du
x x 1 dx u
4
1
x x 1 dx u du
4
1 u
x x 1 dx C34
2
2 1
x x 1 dx u C
12 4
1
x x 1 dx x 1 A
6
+ ⋅
= + ⇒ = ⇒ =
+ ⋅ = + ⋅
+ ⋅ = ⋅
+ ⋅ = ⋅
 
 
+ ⋅ = + 
 
 
+ ⋅ = +
+ ⋅ = + +
∫
∫ ∫
∫ ∫
∫ ∫
∫
∫
∫ 
 
 
 
−
 + −
⋅ = + − ⋅ 
 
+ −
⋅ = ⋅ + ⋅ − ⋅
+ −
⋅ = ⋅ + ⋅ − ⋅
+ −
⋅ = + ⋅ − ⋅ +
+ −
⋅ = + ⋅ − ⋅ +
∫ ∫
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
∫
∫
2 2
2 2
1 1 1
2 2 2
3 1 12
2 2 2
5 3 1
2 2 2 2
5 3 12
2 2 2
x 4x 4 x 4x 4dx dx
x x x x
x 4x 4 x x 1dx dx 4 dx 4 dx
x
x x x
x 4x 4 dx x dx 4 x dx 4 x dx
x
x 4x 4 x x xdx 4 4 C5 3 1x
2 2 2
x 4x 4 2 8dx x x 8 x C
5 3x
 
QUESTÃO 4. (Valor 5) Calcular o valor da Integral ( )+ ⋅∫ 3x x 1 dx , então 
 
( )
( ) ( )
( ) ( )
( )
( )
( )
+ ⋅
= + ⇒ =
= + =
+ ⋅ = ⋅
+ ⋅ = − + − ⋅
 
+ ⋅ = − + − ⋅ 
 
+ ⋅ = − + − +
+ ⋅ = − + − +
∫
∫ ∫
∫ ∫
∫ ∫
∫
∫
3
33
1
3 3 2 2
7 5 3 1
3 2 2 2 2
9 7 5 3
2 2 2 2
3
9 7 5 3
3 2 2 2 2
3
x x 1 dx
u x 1 du dx
se u x 1, logo x u -1
x x 1 dx u -1 u du
x x 1 dx u 3u 3u 1 u du
x x 1 dx u 3u 3u 1u du
u u u u
x x 1 dx 3 3 C9 7 5 3
2 2 2 2
2 6 6 2
x x 1 dx u u u u C
9 7 5 3
x x( ) ( ) ( ) ( ) ( )+ ⋅ = + − + + + − + +∫
9 7 5 3
2 2 2 2
2 6 6 21 dx x 1 x 1 x 1 x 1 C
9 7 5 3
 
 
QUESTÃO 5 . (Valor 5) Seja ( )f x x= −2 6 . Calcular a área delimitada pela função f, pelo eixo das 
abscissas (x) e pelas retas x = -2 e x = 5. 
Orientação. Utilizar integral definida. 
 
( ) ( ) ( )
( )
( ) ( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) [ ] ( )
( )
x dx x dx x dx
x dx x x x x
x dx
x dx
x dx
x dx
− −
−
−
−
−
−
−
− ⋅ = − − ⋅ + − ⋅
   − ⋅ = − − + −   
   
− ⋅ = − − ⋅ − − − ⋅ − + − ⋅ − − ⋅  
− ⋅ = −  − − +  +  − − −    
− ⋅ = − − − + − − −  
− ⋅
∫ ∫ ∫
∫
∫
∫
∫
5 3 5
2 2 3
5 3 52 2
2 3
2
5
22 2 2
2
5
2
5
2
2
2 6 2 6 2 6
2 6 6 6
2 6 3 6 3 2 6 2 5 6 5 3 6 3
2 6 9 18 4 12 25 30 9 18
2 6 9 16 5 9
2 6 [ ] u.a= − − + =∫
5
25 4 29
 
 
 
 
 
 
 
 
QUESTÃO 6. (Valor 5) Seja f(x) = x2 e g(x) = - x2 + 4 x. Calcular usando integral definida a área 
delimitada pelas funções f e g. Construir um esboço do gráfico. 
 
Ponto de Intersecção: 
( ) ( )
( ) ( )
( ) ( )
f x x g x x x
Fazendo : f x g x
x x x
x
x x
x
Temos : f ;f
= = − +
=
= − +
=
− = ⇒ 
=
= =
2 2
2 2
2
4
4
0
2 4 0
2
0 0 2 4
 
 
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
g x f x dx x x x dx
g x f x dx x x dx
g x f x dx x dx x dx
g x f x dx x x
g x f x dx
g x f x dx
  −  ⋅ = − + − ⋅   
  −  ⋅ = − + ⋅   
 −  ⋅ = − ⋅ + ⋅ 
 
 −  ⋅ = − +    
 
 −  ⋅ = − ⋅ + ⋅ − − ⋅ + ⋅    
 −  ⋅ = − 
∫ ∫
∫ ∫
∫ ∫ ∫
∫
∫
∫
2 2
2 2
0 0
2 2
2
0 0
2 2 2
2
0 0 0
22
3 2
0 0
2
3 2 3 2
0
2
0
4
2 4
2 4
2 2
3
2 22 2 2 0 2 0
3 3
16
( ) ( )g x f x dx u.a
+
 −  ⋅ = ∫
2
0
8
3
8
3
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
QUESTÃO 7. (Valor 5) Calcular xx e dx∫
23
 utilizando a Técnica de Integral por Partes: udv uv vdu= −∫ ∫ . 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
QUESTÃO 8. (Valor 5) Calcular x sec x tgx dx⋅ ⋅ ⋅∫ utilizando a Técnica de Integral por Partes: 
udv uv vdu= −∫ ∫ . 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
x x x
x x
x x
x x
x
x
udv uv vdu
x e dx x e e xdx
x e dx x e
x e dx x e C
e xdx
e
x e dx e x C
= −
= ⋅ −
= ⋅ −
= ⋅ − +
 = − + 
∫
∫ ∫
∫ ∫
∫
∫
∫
2 2
2
2
2 2
2 2
2 2
2
3 2
3 2
3 2
3 2
1 1 2
2 2
1
2
1
2
1 1
2
1
2
 
x x
x
w
x
x
w
w x
x
x e dx x xe dx
Fazendo :
u x du xdx
dv xe dx
Cálculo de v:
v dv
v
Fazendo :
dw
w x dw xdx xdx
dw
v e
v e dw
v e v
e xdx
e x
e
dx e
=
= =
=
=
=
= = ⇔ =
=
=
= ⇒ =
=
∫ ∫
∫
∫
∫
∫
∫
2 2
2
2
2
2 2
3 2
2
2
2
2
2
2
1
2
1 1
2 2
1
2
 
Sabemos : secxdx
udv u
x secx tgx dx x sec x ln sec x tg
ln sec
v vdu
x secx tgx dx x sec x sec xdx
x tgx
x C
= −
⋅
⋅ ⋅ ⋅ = ⋅ − +
= +
= ⋅ −
+
⋅ ⋅
∫
∫
∫ ∫
∫ ∫
 
x secx tgx dx
Fazendo :
u x du dx
dv secx tgx dx
Cálculo de v:
v dv
v secx tgx dx
v secx
⋅ ⋅ ⋅
= =
= ⋅ ⋅
=
= ⋅ ⋅
=
∫
∫
∫