Exerc´ıcio 1

(1)

Gabarito Lista 1

Higor Mendes Garcia 07/09/2021

Exerc´ıcio 1

a)

Figure 1: Tomandoξ_i a extremidade esquerda do intervalo.

i 1 2 3 4

ξ_i 0 1

2 1 3

2 f(ξ_i) -1 −3

4 0 5 4 f(ξ_i)×∆x −1

2 −3

8 0 5 8

4

X

i=1

f(ξ_i)×∆x=

−1 2

+

−3 8

+ 0 +

5 8

=−1 4

(2)

Figure 2: Tomando ξ_i a extremidade direita do intervalo.

i 1 2 3 4

ξi

1

2 1 3

2 2 f(ξ_i) −3

4 0 5 4 3 f(ξ_i)×∆x −3

8 0 5 8

3 2

4

X

i=1

f(ξi)×∆x=

−3 8

+ 0 +

5 8

+

12 8

= 14 8

(3)

Figure 3: Tomando ξ_i o ponto m´edio do intervalo.

i 1 2 3 4

ξi

1 4

3 4

5 4

7 4 f(ξ_i) −15

16 −14 32

18 32

33 16 f(ξ_i)×∆x −15

32 − 7 32

9 32

33 32

4

X

i=1

f(ξi)×∆x=

−15 32

+

− 7 32

+

9 32

+

33 32

= 20 32

(4)

b)

i 1 2 3 4

ξ_i 0 1

4

2 4

3 4 f(ξ_i) 0 − 1

16 −1 4

9 16 f(ξi)×∆x 0 − 1

64 − 4

64 − 9 64

4

X

i=1

f(ξ_i)×∆x= 0 +

− 1 64

+

− 4 64

+

− 9 64

=−14 64

(5)

i 1 2 3 4

ξi

1 4

1 2

3

4 1

f(ξ_i) − 1

16 −1

4 − 9 16 -1 f(ξ_i)×∆x − 1

64 − 1

16 − 9 64 −1

4

Xf(ξ_i)×∆x=

− 1 64

+

− 4 64

+

− 9 64

+

−16 64

=−30 64

(6)

i 1 2 3 4

ξi

1 8

3 8

5 8

7 8 f(ξ_i) − 1

64 − 9

64 −25

64 −49 64 f(ξ_i)×∆x − 1

256 − 9

256 − 25

256 − 49 256

4

X

i=1

f(ξi)×∆x=

− 1 256

+

− 9 256

+

− 25 256

+

− 49 256

=−21 64

(7)

c)

i 1 2 3 4

ξ_i −π −π

2 0 π

2 f(ξ_i) 0 −1 0 1 f(ξi)×∆x 0 −π

2 0 π

2

4

Xf(ξ_i)×∆x= 0 +

−π 2

+ 0π 2

= 0

(8)

i 1 2 3 4

ξi −π

2 0 π

2 0 f(ξ_i) −1 0 1 0 f(ξ_i)×∆x −π

2 0 π

2 0

4

X

i=1

f(ξ_i)×∆x=

−π 2

+ 0 +π 2

+ 0 = 0

(9)

i 1 2 3 4

ξi −3π

4 −π

4

π 4

3π 4 f(ξ_i) −

√2

2 −

√2 2

√2 2 f(ξ_i)×∆x −

√2π

4 −

√2π 4

√2π

4 −

√2π 4

4

Xf(ξi)×∆x= −

√2π 4

!

+ −

√2π 4

! +

√2π 4

! +

√2π 4

!

= 0

(10)

d)

Figure 10: Tomando ξ_i a extremidade esquerda do intervalo.

i 1 2 3 4

ξ_i −π −π

2 0 π

2 f(ξ_i) 1 0 1 2 f(ξi)×∆x π

2 0 π

2 π

4

X

i=1

f(ξ_i)×∆x=π 2

+ 0 +π 2

+π = 2π

(11)

i 1 2 3 4

ξi −π

2 0 π

2 π f(ξ_i) 0 1 2 1 f(ξ_i)×∆x 0 π

2 π π

2

4

Xf(ξi)×∆x= 0 + π

2

+π+ π

2

= 2π

(12)

i 1 2 3 4

ξ_i −3π

4 −π

4

π 4

3π 4 f(ξ_i) −

√2

2 + 1 −

√2 2 + 1

√2 2 + 1 f(ξ_i)×∆x −π√

2 4 + π

2 −π√ 2 4 + π

2 π√

2 4 +π

2 π√

2 4 + π

2

4

X

i=1

f(ξ_i)×∆x= −π√ 2 4 +π

2

!

+ −π√ 2 4 + π

2

!

+ π√ 2 4 + π

2

!

+ π√ 2 4 +π

2

!

= 2π

Exerc´ıcio 2

a)

Z 2 0

x²dx

b)

Z 0

−1

2x³dx

c)

Z 5

−7

(x²−3x)dx

(13)

d)

Z 3 2

1 1−x

dx

e)

Z 0

−^π₄

sec(x)dx

Exerc´ıcio 3

a)

Figure 13: Regi˜ao (R) delimitada porf(x) e g(x).

Para encontrarmos os limites de integra¸c˜ao basta resolvermos a equa¸c˜ao:

2x=x²+x 2x−x²−x= 0

x−x² = 0 x(1−x) = 0

⇒ x = 0e x = 1

(14)

Logo, os limites de integra¸c˜ao s˜ao [0, 1], logo:

A(R) = Z 1

0

f(x)−g(x)dx

= Z 1

0

(2x)−(x²+x)dx

= Z 1

0

2x−x² −x dx

= Z 1

0

x−x²dx

= x²

2 − x³ 3

1 0

= 1 6

Portanto, a área da região delimitada for f(x) = 2x e g(x) =x²+xé 1

6 unidades de

´ area.

b)

(15)

x² =−2x³ x²+ 2x³ = 0 x²(1 + 2x) = 0

⇒ x₁ = 0e x₂ =−1 2 Assim, os limites de integra¸c˜ao s˜ao [−¹₂, 0], logo:

A(R) = Z 0

−¹

2

f(x) − g(x)dx

= Z 0

−¹₂

(x²) − (−2x³)dx

= Z 0

−¹₂

x²+ 2x³dx

= x³

3 +x⁴ 2

0

−¹₂

= 1 96

Portanto, a área da região delimitada pory=x² ey=−2x³ é 1

96 de unidade de ´area.

c)

(16)

−2x⁴ = 2x²−4x 2x²+ 2x⁴−4x= 0

2x(x³+x−2) = 0 2x(x³−x+ 2x−2) = 0 2x(x(x²−1) + 2(x−1)) = 0

⇒ x₁ = 0e x₂ = 1 Logo, os limites de integra¸cão são [0, 1], então:

A(R) =

Z 1 0

g(x)dx

−

Z 1 0

f(x)dx

=

Z 1 0

(2x² −4x)dx

−

Z 1 0

(−2x⁴)dx

=

2x³ 3 −2x²

1 0

−

−2x⁵ 5

1 0

=|2

3−2| −

−2 5

= 14 15

Portanto, a regi˜ao limitada porg(x) e f(x) possui 14

15 de unidade de ´area.

(17)

d)

x+ 1 = 4 + 3x−x² 3 + 2x−x² = 0

x²−3−2x= 0 x²+x−3x−3 = 0 x(x+ 1)−3(x+ 1) = 0

⇒ x₁ =−1e x₂ = 3

(18)

Logo, os extremos de integra¸cão são [−1, 3], então:

A(R) = Z 3

−1

g(x)−f(x)dx

= Z 3

−1

(4 + 3x−x²)−(x+ 1)dx

= Z 3

−1

(3 + 2x−x²)dx

=

3x+x²− x³ 3

3

−1

=

3×3 + 3³ −3³ 3

−

3×(−1) + (−1)²− −1³ 3

= 32 3

Portanto, a regi˜ao limitada por f(x) = x+ 1 e g(x) = 4 + 3x−x² possui 32

3 unidades de ´area.

Exerc´ıcio 4

a)

Comecemos calculandof⁰(x):

f(x) = e^x+e^−x

2 ⇒f⁰(x) = d dx

e^x+e^−x 2

= 1 2 × d

dx(e^x+e^−x)

= 1

2 ×(e^x−e^−x)

= e^2x−1 2e^x

(19)

Assim, temos que:

l(f(x)) = Z 1

0

p1 + (f⁰(x))²dx

= Z 1

0

s 1 +

e^2x−1 2e^x

2

dx

= Z 1

0

s 1 +

(e^2x−1)² 4e^2x

dx

= Z 1

0

s

4e^2x+ (e^2x−1)² 4e^2x

dx

= Z 1

0

s

e^4x+ 2e^2x+ 1 4e^2x

dx

= Z 1

0

p(e^2x+ 1)² 2e^x

! dx

= Z 1

0

e^2x+ 1 2e^x dx

= 1 2×

Z 1 0

e^2x+ 1 e^x dx

= 1 2×

Z 1 0

e^x+ 1 e^x dx

= 1 2×

e^x− 1 e^x

1 0

= e 2− 1

2e

b)

f(x) = 1−ln(sin(x))⇒f⁰(x) = d

dt(1−ln(sin(x)))

= d

dx(1)− d

dx(ln(sin(x)))

= 0− d dg

g=sin(x)

(ln(g))× d

dx(sin(x))

!

−

(20)

Assim temos que:

l(f(x)) = Z ^π₄

π 6

p1 + (f⁰(x))²dx

= Z ^π₄

π 6

q

1 + (cot²(x))dx

= Z ^π₄

π 6

s

1 + cos²(x) sin²(x) dx

= Z ^π₄

π 6

s

1 + 1−sin²(x) sin²(x) dx

= Z ^π₄

π 6

s

1 + 1

sin²(x) −1dx

= Z ^π₄

π 6

s 1 sin²(x)dx

= Z ^π₄

π 6

1

|sin(x)|dx

= Z ^π₄

π 6

1 sin(x)dx

= Z ^π₄

π 6

1

sin(x)× sin(x) sin(x)dx

= Z ^π₄

π 6

sin(x) sin²(x)dx

= Z ^π₄

π 6

sin(x) 1−cos²(x)dx

=|_t=cos(x) Z

− t 1−t² dt

=|t=cos(x)

1 2 ×ln

t−1 t+ 1

=|t=cos(x)

1 2 ×ln

cos(x)−1 cos(x) + 1

= 1 2 ×ln

−tan x

2 2

= 1

2 ×ln

tan²x 2

=h ln

tanx 2

i^π₄

π 6

= ln

2 tan π

8

+ tan π

8

×√ 3

(21)

c)

f(x) = ln(x)⇒f⁰(x) = d

dx(ln(x)) = 1 x Assim temos:

l(f(x)) = Z

√8

√3

p1 + (f⁰(x))²dx

= Z

√8

√3

s 1 +

1 x

2

dx

= Z

√8

√ 3

r 1 + 1

x² dx

= Z

√ 8

√3

1 x

√

x²+ 1dx

=|_u=^√_x2+1

Z

1 + 1 u² −1

du

=|_u=^√_x2+1 [u] + Z

1 u² −1

du

=|_u=^√_x2+1

h√

x²+ 1 i

√ 8

√3+ Z

1 u²−1

du

=|_u=^√_x2+1 1 + Z

1

2(u−1)− 1 2(u+ 1)

du

=|_u=^√_x2+1 1 + Z

1 2(u−1)

du−

Z 1 2(u+ 1)

du

=|_u=^√_x2+1 1 + 1

2 ×ln (|u−1|)−1

2 ×ln (|u+ 1|)

= 1 + 1

2×ln(|√

x²+ 1−1|)−1

2 ×ln(|√

x²+ 1 + 1|)

√ 8

√ 3

= 1 + ln(√ 9−1)

−ln(√ 4−1)

− ln(√ 9 + 1)

+ln(√ 4 + 1)

(22)

d)

f(x) =√

x³ ⇒f⁰(x) = d dx(x√

x)

= d dt(x)√

x+x d dx(√

x)

= 1√

x+x 1 2√ x

= 3√ x 2 Assim:

l(f(x)) = Z 4

0

s 1 +

3√ x 2

2

dx

= Z 4

0

r

1 + 9x 4 dx

= Z 4

0

r4 + 9x 4 dx

= Z 4

0

√4 + 9x

2 dx

= 1 2×

Z 4 0

√

4 + 9x dx

=|_t²_=4+9x 1 2×

Z

√ 40

√ 4

2t² 9 dt

= 1 18×

Z

√40

√ 4

2t²dt

= 1 18×

2t³ 3

√40

√ 4

= 80√ 10−8 27

(23)

Exerc´ıcio 5

a)

γ(t) = (t³, t²),1≤t ≤3

⇒γ⁰(t) = (3t²,2t)

⇒ ||γ⁰(t)||=p

(3t²)²+ (2t)²

=t√

9t²+ 4 Desta forma:

Z 3 1

||γ⁰(t)||dt= Z 3

1

t√

9t²+ 4dt

=|_u=9t²₊₄ Z 1

18

√u du

=|_u=9t²₊₄ 1

18 ×2u√ u 3

=

"

(9t²+ 4)√

9t²+ 4 27

#3

1

= 85√

85−13√ 13 27

b)

γ(t) = (2(t−sin(t),2(1−cos(t)),0≤t≤π

⇒γ⁰(t) = (2−2 cos(t),2 sin(t))

⇒ ||γ⁰(t)||= 2p

2−2 cos(t)

(24)

Assim

Z π 0

||γ⁰(t)||dt = Z π

0

2p

2−2 cos(t)dt

= 2× Z π

0

p2−2 cos(t)dt

=|_u=tan(^t

2) 2× Z ∞

0

4u (u²+ 1)³² du

= 8× Z ∞

0

u

(u²+ 1)³² du

=|_v=^√_u2+1 8× Z ∞

1

1 v² dv

= 8

c)

γ(t) = (−sin(t),cos(t)),0≤t ≤π

⇒γ⁰(t) = (−cos(t),−sin(t))

⇒ ||γ⁰(t)||= 1 Assim

Z 2π 0

||γ⁰(t)||dt= Z 2π

0

1dt = 2π

d)

γ(t) = (e^tcos(t), e^tsin(t)),1≤t≤2

⇒γ⁰(t) = (e^tcos(t)−e^tsin(t), e^tcos(t) +e^tsin(t))

⇒ ||γ⁰(t)||=√ 2e^t Logo

Z 2 1

||γ⁰(t)||dt= Z 2

1

√2e^tdt=√ 2

Z 2 1

e^tdt =√

2×[e^x]²₁ =√

2e²−√ 2e

(25)

Exerc´ıcio 6

a)

Do enunciado temos f(x) = x+ 1 e os limites de integra¸c˜ao [0,2], ent˜ao:

V(S) =π× Z 2

0

(x+ 1)²dx=π x³

3 +x²+x 2

0

= 26π 3

b)

Do enunciado temos f(x) = x²−x³ e os limites de integra¸c˜ao [0,1], ent˜ao:

V(S) =π× Z 1

0

(x²−x³)²dx=π x⁷

7 −x⁶ 3 + x⁵

5 1

0

= π 105

c)

Do enunciado temos f(x) = x³ e os limites de integra¸c˜ao [−1,1], ent˜ao:

V(S) =π× Z 1

−1

(x³)²dx=π× Z 1

−1

x⁶dx=π× x⁷

7 1

−1

= 2π 7

d)

Do enunciado temos f(x) = cos(x) e os limites de integra¸c˜ao [0,^π₄], ent˜ao:

V(S) =π× Z ^π₄

0

(cos(x))²dx=π× x

2 +sin(2x) 4

^π₄

0

= π² 8 +π

4

Exerc´ıcio 7

a)

Do enunciado temos f(x) = sin(x) e os limites de integra¸c˜ao [0, π], assim:

A(S) = Z π

0

2πsin(x)p

1 + cos²(x)dx

=|_u=cos(x) 2π×

− Z −1

1

√u²+ 1du

=|_u=tan(θ) 2π× Z ^π₄

−^π₄

sec³(θ)dθ

!

× √ √

(26)

Observe que podemos calcular R

sec³(θ)dθ por partes como se segue:

Z

u dv=uv− Z

v du







u= sec(θ)⇒du= sec(θ) tan(θ)dθ v =R

sec²(θ)dθ⇒dv = sec²(θ)dθ

Z

sec³(θ)dθ = sec(θ) tan(θ)− Z

sec(θ) tan²(θ)dθ tan²(θ) = sec²(θ)−1⇒sec(θ) tan(θ)−

Z

sec(θ)(sec²(θ)−1)dθ

= sec(θ) tan(θ)− Z

sec³(θ)−sec(θ)dθ

sec³(θ)dθ− Z

sec(θ)dθ

sec³(θ)dθ−ln|sec(θ) + tan(θ)|

⇒2× Z

sec³(θ) = sec(θ) tan(θ)−ln|sec(θ) + tan(θ)|

⇒ Z

sec³(θ) = sec(θ) tan(θ)

2 −ln|sec(θ) + tan(θ)|

2 + C,C∈R.

(27)

b)

Do enunciado temos f(x) = x² e os limites de integra¸c˜ao [0,¹₂], assim:

A(S) = Z ¹₂

0

2πx²p

1 + (2x)²dx

= 2π× Z ¹₂

0

x²√

1 + 4x²dx

=|_u=^√_4x2+1−2x 2π× Z

√2−1

1

− u³

128 +2u⁴−1 128u⁵

du

= 2π× Z

√2−1

1

− u³

128 + 2u⁴

128u⁵ − 1 128u⁵

du

= 2π× − 1 128

Z

√2−1

1

u³− 2

u+ 1 u⁵

du

!

= 2π× − 1 128

u⁴

4 + 2 ln(|u|)− 1 4u⁴

√2−1

1

!

= 2π× 3√ 2

64 − ln(1 +√ 2 64

!

= 3π√ 2

32 −πln(1 +√ 2) 32

c)

Do enunciado temos f(x) = √

xe os limites de integra¸c˜ao [1,4], logo:

A(S) = Z 4

1

2π√ x

s 1 +

1 2√

x 2

dx

= 2π× Z 4

1

√4x+ 1

2 dx

= 2π×1 2 ×

Z 4 1

√4x+ 1dx

=|_u2=√

4x+1 π× Z

√ 17

√ 5

u²du

= π 2 ×

u³ 3

√17

√ = π 6 ×

17√

17−5√ 5

(28)

d)

Do enunciado temos f(x) = e^x+e^−x

2 e os limites de integra¸c˜ao [−1,1], ent˜ao:

A(S) = Z 1

−1

2π

e^x+e^−x 2

s 1 +

e^2x−1 2e^x

2

dx

= 2π× Z 1

−1

e^4x+ 2e^2x+ 1 4e^2x

dx

= 2π× 1 4 ×

Z 1

−1

e^4x

e^2x +2e^2x e^2x + 1

e^2x

dx

= π 2 ×

Z 1

−1

e^2x+ 2 + 1 e^2x

dx

= π 2 ×

e^2x

2 + 2x− 1 2e^2x

1

−1

= π

2 ×(e²+ 4−e⁻²)