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Instituto Superior de Engenharia de Coimbra Licenciatura em Engenharia Mecˆanica An´alise Matem´atica I
Soluc¸˜oes do Cap´ıtulo 1 - Func¸˜oes reais de vari´avel real
Revis˜oes - Fun¸c˜oes trigonom´etricas
- (a) −2 cos(α) + cot(α) + tan(α) (b)
(c) cos(α) + tan(α) − cot(α)
- sin (x/2) =
5 /6, cos (x/2) =
1 /6, tan (x/2) =
(a) x = kπ, com k ∈ Z (b) x = π/4 + kπ/ 2 ∨ x = −π/2 + kπ, com k ∈ Z (c) x = −π/4 + kπ ∨ x = −π/8 + kπ/2, com k ∈ Z (d) x = −π/6 + 2kπ ∨ x = 7π/6 + 2kπ, com k ∈ Z (e) x = ± π/4 + 2kπ, com k ∈ Z (f) x = ± 2 π/3 + 4kπ, com k ∈ Z
x = 7π/ 6 ∨ x = 11π/ 6
Fun¸c˜oes trigonom´etricas inversas
- (a) arcsin(0) = 0 (b) arccos(1/2) = π/ 3 (c) arctan(
3 /3) = π/ 6 (d) tan(arccos(
(e) sin(arcsin(0.1)) = 0. 1 (f) sec(arctan(− 3 /5)) =
- (a) cos(y) =
8 /3, tan(y) =
2 /4, cot(y) =
8, sec(y) = 3
8 /8, cosec(y) = 3. (b) sin(y) = 2
5 /5, cos(y) = −
5 /5, tan(y) = −2, sec(y) = −
5, csc(y) =
- Df − 1 = [1, 3], D′ f − 1 =
[
5 − π 2 , 5 + π 2
]
e f −^1 (x) = arcsin(x − 2) + 5
- (a) f −^1 (x) = arcsin( 3 x)−^1 (b) f −^1 (x) = arctan( x− 2 3 ) (c) f −^1 (x) = 2 cot(x + 1)
- (a) sec(arctan(4x)) =
1 + 16x^2 (b) sin(arccos(x)) =
1 − x^2 (c) csc(arcsin(x − 1)) =
x − 1 (d) tan(arccos(x)) =
1 − x^2 x
Revis˜oes - Fun¸c˜oes exponenciais e logar´ıtmicas
- (^) (a) 6 (b)
5 (c) 32
- (^) (a) x = 0 (b) x = log 3 / 2
( (^) ln(3) ln(2)
(c) x = 4 (d) x = 3/ 4 (e) x = 1 (f) x = 0
- (^) (a) x > 9 / 4 (b) − 3 < x < 24
- (a) Df − 1 =] − 10 , +∞[, D′ f − 1 = IR, f −^1 (x) = log^6 (x 4 +10)
(b) Df − 1 = IR, D f′ − 1 =] − 5 , +∞[, f −^1 (x) = 3x−^3 − 5
Fun¸c˜oes hiperb´olicas
- (^) (a) e
2 e (b) 0 (c)
- Df − 1 = IR, D f′ − 1 = IR, f −^1 (x) = ln(x +
x^2 + 1)
Revis˜oes - Limites e continuidade
- − 25 / 3
- N˜ao existe (^) xlim→− 1 f (x) e lim x→ 2 f (x) = 5. A fun¸c˜ao n˜ao ´e cont´ınua.
- (a) Cont´ınua (b) Descont´ınua (c) Descont´ınua
Teoremas fundamentais do c´alculo diferencial em IR
- c = π
- Os resultados obtidos n˜ao est˜ao em contradi¸c˜ao com o teorema de Rolle. A fun¸c˜ao f n˜ao ´e diferenci´avel em x = 0, logo n˜ao est´a nas condi¸c˜oes do teorema para o podermos aplicar.
- c = 2
- c = 7 +
Indetermina¸c˜oes e regra de L’Hˆopital (regra de Cauchy)
- (a) (^) x→lim+∞ x^3 x + 1 = +∞^ (b)^ x→lim+∞
ex x + 1 = +∞^ (c)^ x→lim+∞
ex ln(x) = +∞
(d) lim x→ 1
x^3 − 1 x − 1 = 3 (e) (^) x→−∞lim^ e
−x x^2 + 1 = +∞ (f) lim x→ 0 +
e^1 /x ln(x)
(g) (^) x→lim+∞(x^3 − x + 1) = +∞ (h) (^) x→lim+∞(ex^ − x + 1) = +∞ (i) (^) x→−∞lim xex^ = 0
(j) lim x→ 0 1 − cos(3x) x^2
(k) lim x→ 0 1 − cos x x^3 − sin x = 0 (l) (^) xlim→ 0 ex^ − 1 + x x^2 + x
(m) lim x→ 0 sin^ x^ −^ x x^3
(n) (^) x→lim+∞ x
( arctan(3x) − π 2
) = − 13 (o) lim x→ 0 +
sin x
x
(p) (^) x→lim+∞ x (^1) x = 1 (q) (^) x→lim+∞(1 +
x )x^ = e^2 (r) lim x→ 0 +^ xx^ = 1
Acr´escimos e diferenciais
- (a) ∆f = 2x 0 ∆x + (∆x)^2 ; df = 2x 0 ∆x; l(x) = x^20 + 2x 0 ∆x
(b) ∆f = 1.25; df = 1; l(1.5) = 2
(c)
- l(x) = 1 + 4x; e^0.^08 = f (0.02) ≈ 1. 08
- ∆f ≈ df = 0. 5
- |∆V | ≈ |dV | ≤ 1. 25 π cm^3
Polin´omio de Taylor
p(x) = 3 − x + (x − 3)^2 + 16 (x − 3)^3 ; p′(2) = 116 ; p′(0) = − (^52)
0. 5 ≈ 0 .75; |R 1 (0.5)| ≤ 161 = 0. 0625
- (a) — (b) π 4 = arctan(1) ≈ (^23)
- ln(1.1) ≈ 0. 1 − 0. 201 + 0.^0013 − 0.^00014 = 0.095308(3); |R 4 (1.1)| ≤ 0. 2 × 10 −^5
- sin(89o) = sin
( (^) π 2 −^ π 180
≈ 1 − π 2 64800 ≈^0 .9998476912;^ |R^3
( (^) π 2 −^ π 180
| ≤ π 4 4!180^4 ≈^0.^4 ×^10 − 8
- Com f (x) = ex^ e x 0 = −1, e−^0.^9 ≈ e−^1 + e−^1 × 0 .1 + e − (^1) × 0. 12 2 +^ e−^1 × 0. 13 2 =^ e
− 1 × 1 .1051(6) ≈ 0. 4066
Primitiva¸c˜ao de fun¸c˜oes trigonom´etricas
- Com C ∈ IR:
(a) sin(x) − 1 3 sin^3 (x) + C (b) ln | cos(x)| + tan
(^2) (x) 2
+ C
(c) cos(x) +
cos(x)
sin(2x) +
sin(4x) + C
(e) x + cot(x) −
cot^3 (x) + C (f) x 8
sin(4x) + C
(g) − cot(x) − cot^3 (x) 3
- C (h) x − tan(x) + tan^3 (x) 3 − ln | cos(x)| + C
(i) tan
(^5) (x) 5
(^3) (x) 3
- tan(x) + C (j)^1 6 sin(3x) +^1 2 sin(x) + C
Primitiva¸c˜ao de fun¸c˜oes racionais
- Com C ∈ IR:
(a) 3 ln |x − 3 | − 2 ln |x − 2 | + C (b) x − 12 ln |x| +^23 ln |x − 1 | − 76 ln |x + 2| + C
(c) ln |x| + ln |x − 2 | − 2 ln |x + 1| + C (d) 4x + x
2 2 +^
x^3 3 + 2 ln^ |x|^ + 5 ln^ |x^ −^2 | −^ 3 ln^ |x^ + 2|^ +^ C (e) (^21) x +^54 ln x^ − x 2 + C (f) ln |x − 1 | − (^) 2(^4 xx −− 1)^32 + C
(g) 35 arctan(x) +^25 ln(x^2 + 1) − 15 7 ln | 3 x − 1 | + C (h) x + x
2 2 −^
1 2 ln(x
(^2) + 2) +
√ 2 2 arctan
( √ 2 2 x
)
(i) 32 1 ln x x^ −+ 2^2 − 16 1 arctan
( (^) x 2
)
- C (j) 16 1 ln x x^ + 1− 1 − (^) 8(x^1 − 1) + (^) 8(x + 1)^12 + C
(k) 3 ln |x − 1 | + 2 ln |x + 1| − ln |x| + C (l) ln |x − 1 | − 12 ln(x^2 + x + 1) −
√ 3 3 arctan
( 2 √ 3 x + √ 3 3
)
(m) ln (^) x + 2x + C (n) 2
√ 15 15 arctan
( 2 √ 15 x − √ 15 15
)
(o) ln(x^2 + 4x + 5) + 2 arctan(x + 2) + C (p) x^2 − 6 x + 18 ln |x + 3| + C
Primitiva¸c˜ao por partes
- Com C ∈ IR:
(a) x
(^3) ln(x) 3 −^
x^3 9 +^ C^ (b)^ sin(x)^ −^ (x^ + 1) cos(x) +^ C^ (c)^ e
x(x − 1) + C
(d) x ln^2 (x) − 2 x ln(x) + 2x + C (e) e
x 2 (cos(x) + sin(x)) +^ C^ (f)^
e−x 5 (2 sin(2x)^ −^ cos(2x)) +^ C (g) x 2 [sin(ln(x)) − cos(ln(x))] + C (h) x arcsin(x) +
√ 1 − x^2 + C (i) x
2 2 arctan(x)^ −^
x 2 +
1 2 arctan(x) +^ C (j) − e
− 2 x 5 [cos(x) + 2 sin(x)] +^ C^ (k)^
x 5 x ln(5) −^
5 x ln(5)^2 +^ C^ (l)^ cos(x)^ −^ cos(x) ln(1 + sin(x)) +^ x^ +^ C
(m) 12 [tan(x)sec(x)+ln |sec(x))+tan(x)|]+C (n) x
(^2) sin(x (^2) ) + cos(x (^2) ) 2 +^ C^ (o)^
ex^2 (x^2 − 1) 2 +^ C
Primitiva¸c˜ao por substitui¸c˜ao
- Com C ∈ IR:
(a)
arcsin
( (^) x 3
x
9 − x^2 + C (b)
arcsin(2x) +
x
1 − 4 x^2 + C
(c) a^2 2 b arcsin
bx a
x
a^2 − b^2 x^2 + C (d)
x^3 3
x + C
(e) − 2 ln
x + 1
x^3 3 − x + 2
x + C (f)
x^2 2
x^5 5
(g) − ln(ex^ + 1) + ex^ + C (h) 1 4 ln
4 x^2 + 1 + 2x
+^1
x
4 x^2 + 1 + C
(i) ln |
1 + x^2 + x| + C (j)
ln e^2 x^ − 1 e^2 x^ + 1
+ C
(k)
ln
3 x +
3 x^2 − 4 2
− arccos
3 x
(x + 1)^2 2
x + 1 + 3 ln
x + 1 + 1
+ C
(m) − 1 x
4 − x^2 − arcsin
( (^) x 2
x
18 x
x^2
+ C
(o) ln(
2 + x^2 + x) + x 2
x^2 + 2 + C (p) log 4 1 + 2x 1 − 2 x^ +^ C
(q)
3 ln(2) arctan
2 x
1 − tan(x/2)
+ C
(s) tan
( (^) x 2
− ln
tan^2
( (^) x 2
- C (t) sin(x) − sin^2 (x) 2
+ C
Exerc´ıcios de Revis˜ao - Primitiva¸c˜ao
- Com C ∈ IR:
(a) x ln(2x + 3) − x +
2 ln(2x^ + 3) +^ C^ (b)^
6 ln^
x − 3 x + 3 +^ C
(c) tan^4 (x) 4 +^ C^ (d) 2 ln^ |^ ln(x)|^ +
ln^3 (x) 3 +^ C (e) x^2 ln^2 (x) − x^2 ln(x) + x^2 2
(1 + x^3 )^3 + C
(g) 2
x +^3 2 ln
√ (^6) x − 1 √ (^6) x + 1 + 3 arctan( √ (^6) x) + C (h) 1 3 e^3 x^ − ex^ + arctan(ex) + C
(i) − x^2 cos(x) + 2x sin(x) + 2 cos(x) + C (j) x^2 2 + 2 ln^
x x − 1 −^
x +^ C (k) tan(1 + ln(x)) + C (l) log 5 | 5 x^ − 2 | + C
(m) ln |x − 1 | + arctan
( (^) x 2
tan(2x) ln(2)
+ C
(o) − ln | cos(x)| +
cos(x) +^ C^ (p) 2x^ sin(x) + 2 cos(x) +^ Cv (q) ln x 5
x 5
− 1 + C (r) 3 arcsin(ln(x)) + C
(s) arctan(ln(x)) + C (t)
ln |x^2 − 2 x + 2| + 4 arctan(x − 1) + C (u) 2
1 + sin(x) + C (v) − 2
1 − x + 2 arctan(
1 − x) + C
(w) 22 x−^1 ln(2)
23 x 3 ln(2)
x)^3 + C
(y) x^2 2 arcsin(x) − arcsin(x) 4
x
1 − x^2 + C (z)
ln(x) +
ln |x + 2| −
ln |x − 1 | + C
Soluc¸˜oes do Cap´ıtulo 4 - C´alculo Integral em R
Integral definido
- (a) 2/ 5 (b) 13/ 2 (c) 5
(d) 1/ 3 (e) (e − 1)/ 2 (f) ln(3) + 80/ 3
- 0
- (a) — (b) —
(a) π/ 4 − ln(2)/ 2 (b) ln(4) − 3 / 4
(c) ln(e + 1)/ 2 − ln(e − 1)/ 2 − ln(2)/ 2 (d) 4 − π
- Ver p´ag. 117 das NT; f (c) = 3/(2e^2 ) − 1 / 2
- (a) f (c) = 10, com c = 3
(b) f (c) = 38/15, com c = 2119/ 225
Aplica¸c˜oes do integral definido
- (a)
Area = 32^ ´ / 3 (b)
Area = 8^ ´ √ 3 (c)
Area = 9^ ´
(d)
Area = 49^ ´ / 6 (e)
Area = 2^ ´ √ 2 − 2 (f)
Area = 3^ ´ / 2
- (a) 4/ 3 (b) 4 (c) π/2 + 1/ 3 (d) π/4 + 5/ 6 (e) ln(2) + 1/ 3
- (a)
Volume = 2π (b)
Volume = π^2 / 2
Exerc´ıcios de Revis˜ao - Aplica¸c˜oes do integral definido
- (a) (b) i. ii. (c)
- (a) (b) (c)
- (a) (b) (c)
Integral indefinido
- F (x) =
x^3 / 3 se − 1 ≤ x ≤ 1 2 x − 5 / 3 se x > 1
- (a) F ′(x) = (2x^3 + 2x)e|x|
(b) F ′(x) = 3x^2
x
e −t 2 dt − x^3 e−x 2
(c) F ′(x) = 0
- F ′(1) = −π/4.
- —
- No intervalo [0, +∞[, o m´aximo relativo ´e igual a F (0) = 0 e o m´ınimo absoluto ´e igual a F (1) = 1 − e/2.
- (a) —
(b) (^) x→lim+∞ F (x) x
- (a) F ′(x) = x ln(x) − ln(x)/x^3 (b) — (c) y = 0
Integrais impr´oprios
- (a) 1 / 2 (b) divergente (c) π/ 2
(d) divergente (e) π (f) divergente
Area = 1^ ´ /e
- O integral ´e divergente quando 0 ≤ k ≤ 1 e ´e convergente para 1/(k − 1) quando k > 1.
