Lista de Exercícios de Cálculo I do Prof. Jair Salvador

Vista previa del material en texto

Prefa´cio
•
•
•
Suma´rio
Capı´tulo 0
Pre´-Ca´lculo
(2,−3) −4 (−4, 2) (3,−1)
(2,−4) x (1, 6) y
(4,−2) x+ 3y = 7 (5, 3) y + 7 = 2x
(−4, 3) (−2,−2) (1, 0)
2x+ 2y = 2 y = x+ 1 2x− 3y = 7 6y = −14 + 4x
f(x) = 2x2 − 3
f(5) f(0) f(a) f(−1/2) f(√3)
x ∈ R f(x) = −1 g(x) = |f(x)|
y =
1
x3 − 2 y =
3√
x− 1 y =
√
x− 3
x− 1
y =
√
1− x2
4 + x
y =
√
x+
√
x− 2 y = √3x− x2
y = f(x)
y = 3x2 − 6 y = x2 + 2x+ 3 y = x2 + 2x+ 1
y =
x2 − 4
(x− 1)3 y =
2x
x2 + 4
y =
4(x− 3)
3
√
(x− 2)5
f(x) = 2x2 + 3x g(x) =
√
3x2 + 4 h(x) =
1
x
m(x) = senx
f(x) + g(x) f(x) · g(x) f(5)− g(2) f(g(x))
h(g(0)) f(m(x)) f(1)g(0) −m(π/2)
A r
6 x
x
Capı´tulo 1
Limite
lim
x→+∞
(
√
x2 + 1− x) lim
x→+∞
(
3
√
x3 + 1− x) lim
x→+∞
√
x2 + 1−√x√
x
a b
lim
x→+∞
(
2x2 + bx+ 3
x+ 1
− ax
)
= 0.
Capı´tulo 2
Continuidade
c f [0, +∞)
f(x) =
⎧⎪⎨
⎪⎩
x+
√
x− 2
x− 1 , 0 ≤ x < 1,
cx+ 5
x2 + 3
, x ≥ 1.
a b f [−3, 3]
f(x) =
⎧⎪⎪⎪⎨
⎪⎪⎪⎩
a, x = −3,
9− x2
4−√x2 + 7 , −3 < x < 3,
b, x = 3.
f(x) =
2− 3√x2 + 7
x3 − 1 x ̸= 1 g(1) g
x = 1 g(x) = f(x) x ̸= 1
a b f R
f(x) =
⎧⎪⎨
⎪⎩
x4, x ≤ −1,
ax+ b, −1 < x < 2,
x2 + 3, x ≥ 2.
c f [2, +∞)
f(x) =
⎧⎨
⎩
x2 − 4
x2 − 3x+ 2 , x > 2,
c, x = 2.
c f [0, +∞)
f(x) =
⎧⎪⎨
⎪⎩
√
x2 + 16− 5
4−√x2 + 7 , x > 0 x ̸= 3,
c, x = 3.
c f x = 0
f(x) =
⎧⎨
⎩
1− cos x
sen2x
, x ̸= 0,
c, x = 0.
Capı´tulo 3
Derivada
a, b c f(x) = x2+ax+ b
g(x) = 3
√
(x3 − 2x)2 + c (1, 2)
a b f ′(1)
f(x) =
⎧⎨
⎩
ax2 + b, x ≤ 1,
1
x2
, x > 1.
a b f ′(2)
f(x) =
{
ax3 + bx2, x ≤ 2,√
x+ 2, x > 2.
a b f ′(π/2)
f(x) =
{
ax2 + b, 0 ≤ x ≤ π/2,
cos x, π/2 < x ≤ 2π.
f(x) = x3 sen2(5x) f(x) = tan3(senx) f(x) = esec(2x)
f(x) = log(log x) f(x) = 3 cos2(e−x)
f : R→ R x ∈ R x ̸= 1
x2 − 1
x− 1 ≤ f(x) ≤
x2 + 3
2
.
f x = 1 f(1) f ′(1)
x xarcsenx
f x = 0
g(x) = |x|f(x)
g x = 0
f(0) g x = 0
y = 2 sen2
(
x3 + π/2
x2 + 1
)
+ ax y =
√
ax2 + 6x+ b
(0, 2) a b
lim
h→0
3
√
tan2(a+ h)− 3
√
tan2 a
h
f ′ f(0) = 0 g(x) =
f(x)
x
(0, +∞)
f(x) = x5 + x3 + 3x
(f−1)′(5)
f(x) = (x5 + 7)1/3 x > 0 (f−1)′(2)
f−1 (2, 1)
f(x) = arctan
(
x3
3
− x
)
x > 1
f (f−1)′(0)
f−1 (0, f−1(0))
f(x) = arctan(x2 − 1) x > 0 (f−1)′(π/4)
f(x) = ex
5+x3+x f x ∈ R
e3
f(x) =
1
arctan(log x)
x > 0
f (f−1)′(π/2)
f−1 (π/2, f−1(π/2))
Capı´tulo 4
Aplicac¸o˜es da Derivada
f f(x) ≤ f(2) x ∈ [1, 3] f ′(2)
f g x = a f(a) > 0 g(a) > 0 f
g x = a h(x) = f(x)g(x)
x = a
lim
x→2
x3 + x2 − 11x + 10
x2 − x− 2 limx→−∞
x3 − 4x2 + x− 1
3x2 + 2x+ 7
lim
x→0
(
1
x
− 1
x2
)
lim
x→1
(
2
x2 − 1 −
1
x− 1
)
lim
x→1
(x− 1)2
1 + cos(πx)
lim
x→+∞
x sen
(
1
x
)
lim
x→
π
2
−
(sec x− tan x) lim
x→1
((1− x) tan(πx/2)) lim
x→0
sen(5x)
3x
lim
x→0
x
tan(2x)
lim
x→π/4
tan(x− π/4)
x− π/4 limx→0
sec(x)− 1
x
lim
x→0
x− arcsenx
sen3x
lim
x→0
(
1
x
− 1
tanx
)
lim
x→0
(
1
1− cos x −
2
sen2x
)
lim
x→1
x− 1
xn − 1 limx→0
tan(x)− x
x− senx limx→a
sen(x)− sen(a)
x− a limx→a
x3 − a3√
x−√a
lim
x→0
(sen x)tan x lim
x→0+
(tan x)1/log(sen x) lim
x→0
(cos(2x))1/x
2
lim
x→1
x1/(x−1) lim
x→0
(cos x)1/tan x lim
x→+∞
(
1 +
a
x
)x
lim
x→2
x2 + x− 6
x2 − 3x+ 2 limx→π/2
(
x tan x− π
2 cos x
)
lim
x→0
ex − e−x
senx
lim
x→1−
log(1− x) tan(1− x) lim
x→+∞
x2
ex
lim
x→+∞
x2e−x
lim
x→0+
xe1/x lim
x→+∞
x sen(π/x) lim
x→+∞
log x
x3
lim
x→0+
xsen x lim
x→0+
x2/(2+log x) lim
x→1−
(1− x)cos(πx/2)
y = f(x) R \ {0}
f(−5) = 2, f(−4) = 1, f(−3) = 3, f(−3/2) = 4 lim
x→+∞
f(x) = lim
x→−∞
f(x) = 0.
f ′(x)
x
y
−5 −4 −3 −32
1
2
−1
f ′(x)
f
f
f
f
f(x) =
1− x
x2
f(x) =
x3
(x− 1)2 f(x) =
2x
9− x2
f(x) = 1− x
(x+ 2)2
f(x) =
⎧⎨
⎩
x
(x− 1)2 , x < 1,
x2 − 1, x ≥ 1.
y = f(x) R
f(0) = 0, f(2) = 2 f(−1/2) = −2.
f ′
x
y
2−12
f ′(x)
f
f
f
f
f(x) = xe−x g(x) = x2e−x h(x) = x log x
PQR
Q QR
QR
P
Q
R
y
y
θ
P
P
θ
50cm2
1/2
y = −x2 + 1 t
P (x, y) P
x MQ PMQ
P0(1/
√
2, 1/2)
x
y
y = −x2 + 1
t
M
Q
P
P (x0, y0) t y =
1
x
x > 0
P
OPQ
OPQ P
x
y
y = 1/x
t
O
P
Q
x2 + 2xy + y3 = 1 y2 + 9x − (x − 2)3 − 9 = 0
(0, 1)
a b h(x) = a+ b sen2(x/2)
y cos x+ xy = 5πx
(π/2, 5π)
y = y(x) y4 − y2 + 4 sen(xy) = 0
y(0) = 1 y′(0)
2x2y2 + y3 cos(πx) − 1 = 0
(1, 1)
y = f(x)
x2 cos y + x2 sen y − 2x+ 1 = 0.
(1,π/2)
Capı´tulo 5
Integral
∫
cos(x+ 3) dx
∫
sec(5x) tan(5x) dx
∫
senx cos x dx∫
sen θ cos3 θ dθ
∫
cos3 x dx
∫
3x
√
1− 2x2 dx
∫
x+ 3
(x2 + 6x)1/3
dx
∫
x(1 + x)2/3 dx
∫
x√
x− 1 dx∫
dx√
25− 16x2
∫
dx
4x2 + 9
∫
dx
x
√
4x2 − 9
∫
x2√
1− x6 dx∫
x
x4 + 3
dx
∫
dx√
28− 12x− x2
∫
dx
x2 − 4x+ 8
∫
x sen(5x) dx
∫
xe−2x dx
∫
log(1− x) dx
∫
e1/x
x3
dx∫
e2x senx dx
∫ √
x log x dx
∫
arcsen(x/2) dx
∫
xex
2
dx
∫
esen x cos x dx
∫
log x
x
dx∫
2x+ 1
x2 + x+ 1
dx
∫
cos(log x)
x
dx
∫
t2
4 + t3
dt
∫
ex
3 + 4ex
dx
∫
e
√
x
√
x
dx
∫
x
x2 + 4
dx∫
ex
e2x + 2ex + 1
dx
∫
ee
x
ex dx
∫
dx
x
√
log x
∫ 3
0
|x2 − 1| dx
∫ 1
0
2x
1 + 3x2
dx
g(x) = 2 +
∫ x
0
(ez + z)1/z dz lim
x→+∞
g′(x)
f(x) = 4+
∫ x
0
t+ et
2
3 + t4
dt. p(x) = ax2 + bx+ c
p(0) = f(0) p′(0) = f ′(0) p′′(0) = f ′′(0)
y = f(x) f(π) = 5
∫ x
0
t3f ′(t) dt = x4 senx
f(x) =
∫ x
0
eu
2
+ cos u
u5 + 3
du (f−1)′(0)
∫ +∞
−∞
xe−x
2
dx
∫ +∞
1
log x
x2
dx
∫ +∞
−∞
dx
1 + x2∫ +∞
1
dx√
x
∫ 0
−∞
e2x dx
∫
x
x2 + 4x− 5 dx
∫
x+ 1
x2 + 4x− 5 dx
∫
x2
x2 + 2x+ 1
dx∫
x
x2 − 2x− 3 dx
∫
dx
(x+ 1)(x2 + 1)
∫
x
(x+ 1)(x+ 2)(x+ 3)
dx
∫
2x− 7
x2 + 6x− 7 dx
∫
x+ 2
x2 + x
dx
∫
dx
x4 − 1
∫
dx
(x2 + 1)2∫
3x+ 5
x3 − x2 − x+ 1 dx
∫
sen θ
cos2 θ + cos θ − 2 dθ
∫
ex
e2x + 3ex + 2
dx
Capı´tulo 6
Aplicac¸o˜es da Integral
x
y
y = 1/x
1 2
x
y
y = 4/x
4
4
x
y
y = ex
y = log x
1 e
1
e
y = x3 y = x+ 6 2y + x = 0 2y2 = x+ 4 x = y2
y = 1/x y =
√
x y = 2 y = 1 + senx y = cos(2x) x ∈ [0,π]
y = xe−x y = 0 y = x2e−x y = 0
x = 3 y =
1
x2 − 1 x
[a, b] L ∈ R Ω
y = f(x) x = a x = b y = L 0 < L < f(x)
Ω y = L
y
y = ex x = 0 x = 1 y = 0
y = −1
f(x) = 2 + cos x g(x) = 2− cos x x ∈ [−π/2,π/2]
x
x2 = 4ay y = a a > 0 a 50π
n > 0 Ωn y = xn x x = 1
Wn Ωn x Vn
Ωn y lim
n→+∞
Vn
Wn
R y = e−x y =
1
x
x ≥ 1
R x
x
y = xe−x
A(−π/4, 0)
y = f(x) =
√
(π/4)2 − x2 y = g(x) = − log(√2 cos x)
x ∈ [−π/4,π/4]
B(π/4, 0)
Apeˆndice A
Respostas dos Exercı´cios
y = −4x+ 5 7y + 3x− 2 = 0
y = −4 x = 1 3y = −x− 2
2y = −x+ 11 3y − 2x− 17 = 0
(0, 1)
47 −3 2a2 − 3 − 52 3−1 1
g f
x
x
y
g(x) = |2x2 − 3|
f(x) = 2x2 − 3
R \ {√3} (1,+∞) [3,+∞)
(−∞,−4) ∪ [−1, 1] [2,+∞) [0, 3]
f(x) < 0 x ∈ (−√2,√2) f(x) > 0
x ≤ −√2 x ≥ √2
f(x) > 0 x ∈ R
f(x) > 0 x ∈ R
f(x) > 0 x ∈ (−2, 1) ∪ (2,+∞) f(x) < 0
x ∈ (−∞, 2] ∪ (1, 2]
f(x) < 0 x ∈ (−∞, 0) f(x) > 0
x ∈ [0,+∞)
f(x) < 0 x ∈ (2, 3] f(x) > 0 x ∈
(−∞, 2) ∪ (3,+∞)
2x2+3x+
√
3x2 + 4 (2x2+3x)
√
3x2 + 4
61 6x2 + 8 + 3
√
3x2 + 4
1
2
2(senx)2 + 3 senx 9
V (r) =
Ar
2
− πr3
A(x) = 9− 3x
a2−b2 = (a−b)(a+b) a =
√
x2 + 1 b = x√
x2 + 1 + x√
x2 + 1 + x
√
x2 + 1− x = (
√
x2 + 1)2 − x2√
x2 + 1 + x
=
x2 + 1− x2√
x2 + 1 + x
=
1√
x2 + 1 + x
.
lim
x→+∞
√
x2 + 1−x = lim
x→+∞
1√
x2 + 1 + x
= 0.
a3− b3 = (a− b)(a2+ ab+ b2) a =
3
√
x3 + 1 b = x
a2 + ab+ b2 = 3
√
(x3 + 1)2 + x 3
√
x3 + 1 + x2
3
√
x3 + 1− x = (
3
√
x3 + 1)3 − x3
3
√
(x3 + 1)2 + x 3
√
x3 + 1 + x2
=
(x3 + 1)− x3
3
√
(x3 + 1)2 + x 3
√
x3 + 1 + x2
=
1
3
√
(x3 + 1)2 + x 3
√
x3 + 1 + x2
.
x→ +∞ 0 √
x
√
x2 + 1−√x√
x
=
√
x+ 1/x− 1.
x→ +∞ √x+ 1/x− 1→ +∞
(2− a)x2 + (b− a)x+ 3
x+ 1
.
2−a = 0
2−a > 0 +∞ 2−a <
0 −∞ 2− a = 0
lim
x→+∞
(b − a)x+ 3
x+ 1
= 0
b− a b− a = 0 a = b = 2
c = 1
a = b = 8
g(1) = − 1
18
a = 2 b = 3
c = 4
c = −4
5
c =
1
2
f(1) = g(1) f ′(1) = g′(1)
a = −8
3
b =
11
3
c = 1
f(1−) = f(1+)
f ′(1−) = f ′(1+)
a = −1 b = 2
a = −7
6
b =
11
8
a = − 1
π
b =
π
4
f ′(x) = 3x2 sen2(5x) + 10x3 sen(5x) cos(5x)
f ′(x) = 3 tan2(senx) sec2(senx) cos x
f ′(x) = 2 sec(2x) tan(2x)esec(2x)
f ′(x) =
1
x log x
f ′(x) = 6e−x cos(e−x) sen(e−x)
f 1 f
1 lim
x→1
f(x) = f(1) x → 1
lim
x→1
f(x) = 2 = f(1)
x2 − 1
x− 1 = x+1 x+1 ≤ f(x) ≤
x2 + 3
2
f(x)− f(1)
x− 1
2
x− 1 ≤ f(x)− 2 ≤ x
2 + 3
2
− 2 = x
2 − 1
2
.
x−1 x−1 >
0 x− 1 < 0
1 ≤ f(x)− 2
x− 1 ≤
x+ 1
2
.
x→ 1
f ′(1) = 1
f(1) = 2 f ′(1) = 1
x = elog x
xarcsen x = elog x arcsen x.
xarcsen x
(
log x√
1− x2 +
arcsenx
x
)
.
f f
|x|
g(0) = 0 g′(0) = lim
h→0
g(h)
h
g(h)
h
=
|h|
h
f(h) h → 0+
f(0)
|h|
h
= 1 h > 0
h → 0− −f(0) |h|
h
= −1
h < 0
g′(0) f(0) = −f(0)
f(0) = 0
x = 0
a =
3
2
b = 4
f(x) =
3
√
tan2 x
f ′(a)
2
3
(tan a)−1/3
cos2 a
g′(x) =
f ′(x)x − f(x)
x2
g′ h(x) =
f ′(x)x − f(x) h′(x) = f ′′(x)x f ′
f ′′ > 0 x > 0 h′(x) > 0
h h(0) = f ′(0)0− f(0) = 0
h(x) > 0 x > 0 g′(x) = h(x) > 0
x > 0 g x > 0
f ′(x) = 5x4 + 3x2 + 3 > 0
x ∈ R
R
f(1) = 5 f−1(5) = 1
f ′(1) = 11
(f−1)′(5) =
1
f ′(1)
=
1
11
x > 0
(x5+7)1/3 = 2 x5 = 1 x = 1
f ′(1) (f−1)′(2) =
1
f ′(1)
=
12
5
y − 1 = 12
5
(x− 2)
x > 1
f−1(0) x > 1
f(x) = 0
x3
3
− x = 0
1
√
3 (f−1)′(0)
1
f ′(
√
3)
= 1/2
y =
1
2
x+
√
3
x > 1
f(x) = π/4
x2 − 1 = 1 x = √2
x > 0 (f−1)′(π/4) =
1
f ′(
√
2)
=
√
2/2
x
f(1) = e3 (f−1)′(e3) =
1
f ′(1)
=
1
9e3
x
(f−1)′(π/2) = −1
y = −x+ π + 2
2
f ′(2) = 0 x = 2
f g f ′(a) = g′(a) =
0 h′(x) = f ′(x)g(x) + f(x)g′(x) h′(a) = 0
f ′(a) = g′(a) = 0
x = a
x = a
x = a f g f ′(x)
g′(x) x < a a
f(a) g(a) x = a
h′(x) x < a
a
h′(x)
x > a a x = a
h
5/3 −∞ −∞ −1
2
2
π2
1
0
2
π
5/3 1/2 1 0
−1/6 0 −1/2 1/n
2 cos a 6a5/2
1 e e−2 e 1 ea
5 −1 2 0 0 0
+∞ π 0 1 e2 1
f ′ > 0
f
(−∞,−5) (−4,+∞) (−5,−4)
f ′ = 0
x = −4 x = −3/2
x = −4
x = −3/2
f ′′ > 0
f ′
f ′′ > 0 f ′
(−∞,−3) (−3/2, 0)
f ′′ < 0 (−3,−3/2) (0,+∞)
f ′′
x = −3 x = −3/2
y = 0 lim
x→+∞
f(x) =
lim
x→−∞
f(x) = 0 y = 0 x = 0
∞
x
y
−5 −4 −3 − 32
1
2
3
4
f(x)
f ′(x) =
x− 2
x3
f ′′(x) =
6− 2x
x4
x = 2
x = 0 (0, 2) (2,+∞)
x = 3
x = 0
±∞ y = 0
x
y
f(x) =
1− x
x2
2 3
f ′(x) =
x2(x− 3)
(x− 1)3 f
′′(x) =
6x
(x− 1)4
x = 3 x = 0
x = 1 (1, 3)
(3,+∞) x = 0
x = 1
±∞ ±∞
x
y
1 3
f(x) =
x3
(x− 1)2
f ′(x) =
2(x2 + 9)
(9− x2)2 f
′′(x) =
4x(x2 + 27)
(9− x2)3
f ′′
4x 9− x2
x = −3 (0, 3) (−3, 0)
x = 3 x = ±3
±∞
y = 0
x
y
f(x) =
2x
9− x2
−3 3
f ′(x) =
x− 2
(x+ 2)3
f ′′(x) =
2(4− x)
(x+ 2)4
x = −2 (−2, 2)
x = 2
x = 4
x = −2
±∞ y = 1
x
y
f(x) = 1− x
(x+ 2)2
−2 2 4
y = 1
h(x) =
x
(x − 1)2
h′(x) = − x+ 1
(x− 1)3 f
′′(x) =
2(x+ 2)
(x− 1)4
x = −1
x = −1 (−1, 1)
x = −2 (−2, 1)
x = 1
y = x2 − 1 x = 1
−∞
y = 0
x
y
f(x) =
⎧⎨
⎩
x
(x − 1)2 , x < 1,
x2 − 1, x ≥ 1.
y = x2 − 1
1−1−2
f ′ > 0
f
(−1/2,+∞) (−∞,−1/2)
f ′ = 0
x = −1/2
x = −1/2
f ′′ > 0
f ′
f ′′ > 0 f ′
(−∞, 0) (2,+∞)
f ′′ < 0 (0, 2)
f ′′
x = 0 x = 2
x
y
2
−2
2− 12
y = f(x)
f ′(x) = (1 − x)e−x f ′′(x) =
(x−2)e−x e−x f ′
(1−x) f ′′
x− 2 x = 1
lim
x→+∞
f(x) = 0 lim
x→−∞
f(x) = −∞
x = 2
f(0) = 0
x
y
f(x) = xe−x
1 2
g′(x) = (2 − x)xe−x g′′(x) =
(x2 − 4x + 2)e−x e−x
g′ (2− x)x g′′
x2 − 4x+ 2
g′ x = 0 x = 1
(2 − x)x g x = 0
(0, 2)
g′′ 2 ± √2
x2−4x+2 2−√2 ≈
0.58 2 +
√
2 ≈ 3.41
(0.58, 3.41)
lim
x→+∞
g(x) = 0 lim
x→−∞
g(x) = +∞
g(0) = 0
x
y
g(x) = x2e−x
20.5 3.4
x > 0 log
x < 0
lim
x→0+
h(x) = 0 h′(x) = 1 + log x
f ′′(x) = 1/x h′(c) = 0 = 1 + log c
log c = −1
c = e−1
0 < x < 1/e x > 1/e
f ′′(x) > 0 x > 0
lim
x→+∞
f(x) = +∞
x
y
h(x) = x log x
1
e
x(t)
y(t)
x2(t) + y2(t) = 52
t x(t)x′(t) + y(t)y′(t) = 0
x(t) = 3 x′(t) = 3
y(t) = 4 3 ·3+4y′(t) =
0 y′(t) = −9
4
PQR
x QR y PQ
x2 + y2 = 152 QR = x = 12 y = 9
xx′+ yy′ =
0 x′ = 1 x = 12 y = 9
y′ = −4/3
A = xy/2 A′ =
1
2
(x′y + xy′)
A′ =
−7
2
cm2/s
y =
√
152 − x2
A =
xy
2
=
x
2
√
152 − x2
A′ =
1
2
(x′
√
152 − x2 + x −2xx
′
2
√
152 − x2 ).
y
A′ =
1
2
(x′y − x2 x
′
y
) =
1
2
(1(9)− 122 1
9
) = −7/2.
x
40− x
40
=
y
60
y = 36 x = 16
−x′
40
=
y′
60
y′ = 0, 5 x′ = −1/3
A(t) = x(t)y(t)
A′ = x′y + xy′
t y′ = 0, 5, x′ = −1/3, y = 36, x =
16 A′(t) = −4cm/s
x P
tan θ(t) =
1
x(t)
θ′
cos2 θ
= − x
′
x2
.
x(t) = 1 θ =
π
4
x′ = −2
x
θ′
cos2(π/4)
= −−2
12
2θ′ = 2
θ′ = 1m/min
R(t) r(t) h(t)
R
r h−R
(h(t)−R(t))2 + r(t)2 = R(t)2.
R(t) = 1 h(t) = 4/3
r(t) = 2
√
2/3 r2(t) = 8/9
2
(h−R)(h′ −R′) + rr′ = RR′.
R′ = 0, 9 h′ = 0, 8 R = 1 h = 4/3 r =
2
√
2/3 r′ r′ = 7
5
√
2
rr′ =
14
15
V (t) =
π
3
r(t)2h(t)
V (t)
V ′(t) =
π
3
(2r′(t)r(t)h(t) + r2(t)h′(t)).
r(t)r′(t) =
14
15
r2(t) =
8/9 V ′(t) =
16π
15
3/
x y
x′ =
−20 y′ = −15
d
x2(t) + y2(t) = d2(t).
xx′ + yy′ =
dd′ x = 400 y = 300 d = 500
400(−20) + 300(−15) = 500d′ d′ = −25m/s
x(t) z(t)
d(t)
d2(t) = x2(t) + z2(t) x(0) = 0
z(0) = 48 x(4) = 4(20) = 80
z(4) = 48 + 3(4) = 60 t = 4
d(4) = 100
dd′ =
xx′ + zz′ t = 4 d = 100, x =
80, x′ = 20, z = 60, z′ = 3 100d′(4) =
80(20) + 60(3) d′(4) = 17, 8m/s
x(t) y(t)
d(t)
d2(t) = x2(t) + y2(t) + 102
x(0) = y(0 = 0 x(2) = 2(40) = 80
y(2) = 2(20) = 40 t = 2
d(2) = 90
dd′ =
xx′ + yy′ t = 2 d = 90, x =
80, x′ = 40, y = 40, y′ = 20 90d′(2) =
80(40) + 40(20) d′(2) =
400
9
x
y
s c(s) = (40s, 0, 10)
t(s) = (0, 20s, 0)
d(s) =
√
(40s)2 + (20s)2 + 10
d′(s) =
2 · 40s · 40 + 2 · 20s · 20
2
√
(40s)2 + (20s)2 + 10
.
d′(2) =
400
9
x(t)
y(t)
d(t) d2(t) = x2(t) + y2(t)
dd′ = xx′+
yy′ x = 120 − 2 · 30 = 60
y = 2 · 40 = 80 d = 100 x′ = −30
y′ = 40 100d′ = 60(−30) + 80(40)
d′ = −18 + 32 = 14
x y
s
f(s) = (−30s + 120, 0) c(s) =
(0, 40s)
d(s) =
√
(−30s+ 120)2 + (40s)2
d′(s) =
2(−30s+ 120)(−30) + 2(40s)(40)
2
√
(−30s+ 120)2 + (40s)2 .
d′(2) = 14
x y
x(t)y(t) = 50 x′y + xy′ = 0
x′ = 2 x = 5 y = 10
2 · 10 + 5y′ = 0, y′ = −4 P (t) =
2(x(t)+y(t)) P ′(t) = 2(x′(t)+y′(t))
P ′(t) = 2(2− 4) = −4m/s
d(t)
θ(t)
tan(θ(t)) =
1
d(t)
θ′(t)
cos2 θ(t)
= − d
′(t)
d2(t)
.
√
5
1 2 cos(θ) =
2√
5
d′(t) = −2
θ′(t) =
2
5
h b
h
10
=
b
8
h = 5 b = 4
h′ = 1/2
h′(t)
10
=
b′(t)
8
b′(t) = 2/5
V (t) = 50h(t)b(t)1/2
V ′ = 50(hb′ + h′b) =
50(5(2/5) + 1/2(4)) = 200cm3/min
x(t) s(t)
∆y
∆x
0
x x 40
1, 80− s(t)
x
=
3− 1, 80
40− x(t) =
1, 2
40− x(t) .
(1, 80 − s(t))(40 − x(t)) = 1, 2x(t)
x(t) = 20 s(t) = 3/5
(−s′)(40− x) + (1, 80− s)(−x′) = 1, 2x′.
x = 20, s = 3/5, x′ = −4
x
s′ =
12
25
= 0, 48m/s
P = (x0, y0) t
P y−y0 = −2x0(x−x0) y0 =
−x20+1 t y = 1−x20− 2x0(x−x0)
Q t x
y = 0 t
x Q x0 +
1− x20
2x0
M = (x0, 0) MQ =
1− x20
2x0
MQ′(t) =
−2x0(t)x0(t)′(2x0(t)) − (1− x20)2x′0(t)
4x0(t)2
MQ′(t) =
−4x20(t)x′0(t)− (1− x20)2x′0(t)
4x0(t)2
x0 = 1/
√
2 MQ′(t) = −3cm/min
P = (x0, y0) t
P y − y0 = − 1
x20
(x − x0)
y0 =
1
x0
t y =
1
x0
− 1
x20
(x − x0)
Q t x
y = 0 t
x Q 2x0
Q = (2x0, 0)
OP PQ√
x20 + y
2
0
OPQ y0
2x0 2x0y0/2 = x0y0
y0 =
1
x0
OPQ 1
P
2x+2y+2xy′4y2y′ = 0 2yy′+9− 3(x− 2)2 = 0.
x = 0 y = 1 m1 =
−2
3
m2 =
3
2
h′(x) = 2b sen(x/2) cos(x/2)(1/2) =
b sen(x/2) cos(x/2). h′(π/2) = b/2
y′ cosx+ y(− senx) + y + xy′ = 5π.
x = π/2 y = 5π y′ = 10
10 = y′ = h′(π/2) = b/2 b = 20
h(π/2) = a + b/2 = 5π
a+ 10 = 5/π
a = 5π − 10 b = 20
4y3y′ − 2yy′ + 4 cos(xy)(y + xy′) = 0.
x = 0 y = 1 y′(0) = −2
4xy2+4x2yy′+3y2 cos(πx)+y3(− sen(πx)(π)) = 0.
x = y = 1 y′ = −1/4
y = 5/4− x/4
2x(cos y + sen y) + x2(− sen y + cos y)y′ − 2 = 0.
x = 1 y = π/2
y′ = 0 y − π/2 = 0(x − 1)
y = π/2
ay′ = b a = 0 y′ = ±∞
x2(− sen y+ cos y) = 0 x = 0
sen y = cos y x = 0
1 = 0
sen y = cos y sen y =
cos y = ±
√
2
2
x ∈ R
2x2(
±√2
2
)− 2x+ 1 = 0 = ±√2x2 − 2x+ 1.
√
2x2 − 2x + 1 = 0
∆ < 0 −√2x2 −
2x + 1 = 0 x =
−2±
√
4 + 4
√
2
2
√
2
y
sen y = cos y = −
√
2
2
y = −3π
2
+ 2kπ
k ∈ Z
sen(x+ 3) + C
1
5
sec(5x) + C
1
2
sen2x+ C −1
4
cos4 θ + C
senx− 1
3
sen3x+ C −1
2
(1− 2x2)3/2 + C
3
4
(x2 + 6x)2/3 + C
3
8
(1 + x)8/3 − 3
5
(1 + x)5/3 + C
2
3
(x− 1)3/2 + 2(x− 1)1/2 + C
1
4
arcsen(4x/5) + C
1
6
arctan(2x/3) + C
1
3
arcsec(2x/3) + C
1
3
arcsen(x3) + C√
3
6
arctan(
√
3x2/3) + C
arcsen((x+ 6)/8) + C
1
2
arctan((x − 2)/2) + C
−x
5
cos(5x) +
1
25
sen(5x) + C
−e−2x(x/2 + 1/4) + C
(x− 1) log(1− x)− x+ C
e1/x(1− 1/x) + C
2
5
e2x senx− 1
5
e2x cosx+ C
2
3
x
√
x log x− 4
9
x
√
x+ C
x arcsen(x/2) +
√
4− x2 + C
u = x2
1
2
ex
2
+ C
u = senx esen x + C
u = log x
1
2
log2 x+ C
u = x2 + x+ 1 log |x2 + x+ 1|+ C
u = log x sen(log x) + C
u = 4 + t3
1
3
log |4 + t3|+ C
u = 3 + 4ex
1
4
log(3 + 4ex) + C
u =
√
x 2e
√
x + C
u = x2 + 4
1
2
log(x2 + 4) + C
u = ex u2 +2u+1 =
(u+ 1)2 v = u+ 1 − 1
ex + 1
+ C
u = ex ee
x
+ C
u = log x 2
√
log x+ C
22/3
2
3
log 2
g′(x) = (ex+x)1/x
y = (ex+x)1/x log y =
log(ex + x)
x
1 lim
x→+∞
log y(x) = 1
lim
x→+∞
y(x) = lim
x→+∞
g′(x) = e
f ′(x) =
x+ ex
2
3 + x4
a = 1/6 b = 1/3 c = 4
x3f ′(x) =
4x3 senx+x4 cosx f ′(x) = 4 senx+x cosx
f(x) = x senx−3 cosx+C f(π) = 5 = 3+C
C = 2 f(x) = x senx− 3 cosx+ 2
f ′(x) =
ex
2
+ cosx
x5 + 3
f ′(0) =
2
3
f(0) = 0 f−1(0) = 0
(f−1)′(0) =
1
f ′(0)
=
3
2
∫
xe−x
2
dx = −e−x2/2
0
∫
log x
x2
dx = −1 + log x
x
1∫
dx
1 + x2
= arctanx
π∫
dx√
x
= 2
√
x
+∞∫
e2x dx = e2x/2
1/2
1
6
log |(x+ 5)5(x− 1)|+ C
2
3
log |x+ 5|+ 1
3
log |x− 1|+ C
x− 1
x+ 1
− 2 log |x+ 1|+ C
3
4
log |x− 3|+ 1
4
log |x+ 1|+ C
1
4
log
(
(x+ 1)2
x2 + 1
)
+
1
2
arctanx+ C
−1
2
log |x+1|+2 log |x+2|− 3
2
log |x+3|+C
−5
8
log |x− 1|+ 21
8
log |x+ 7|+ C
− log |x+ 1|+ 2 log |x|+ C
−1
2
arctanx+
1
4
log |x− 1|− 1
4
log |x+ 1|+ C
1
2
(
arctanx+
x
1 + x2
)
+ C
− 4
x− 1 +
1
2
log
∣∣∣∣x+ 1x− 1
∣∣∣∣+ C
u = cos θ
1
3
log
∣∣∣∣ 2 + cos θ−1 + cos θ
∣∣∣∣+C
u = ex
log
(
1 + ex
2 + ex
)
+ C
(1, 1) y = x (2, 1/2) y = x/4
∫ 1
0
(x− x/4) dx+
∫ 2
1
(1/x− x/4) dx = log 2.
y = 4 y = 4/x
(1, 4)∫ 1
0
4 dx+
∫ 4
1
(4/x) dx = 4(1 + 2 log 2).
∫ 1
0
ex +
∫ e
1
(e− log x) dx = e2 − 2.
x3 = x + 6
x = 2 y = x + 6 2y + 2x = 0
(−4, 2)
∫ 0
−4
(x+6+x/2) dx+
∫ 2
0
(x+6−x3) dx = 12+10 = 22.
2y2 −
4 = y2 y = ±2∫ 2
−2
(y2 − (2y2 − 4)) dy = 32
3
.
√
x 1/x x = 1
y = 2 1/x x = 1/2
y = 2
√
x x = 4
∫ 1
1/2
(2− 1/x) dx+
∫ 4
1
(2 −√x) dx =
= 1− log 2 + 4
3
=
7
3
− log 2.
∫ π
0
(1 + senx− cos(2x)) dx = π + 2
∫
xe−x dx = (−x − 1)e−x
[0, +∞)∫
x2e−x dx = (−x2 − 2x− 2)e−x [0, +∞)
1
x2 − 1 =
1
2(x− 1)−
1
2(x+ 1)∫ +∞
3
dx
x2 − 1 =
log 4
2
− log 2
2
=
log 2
2
log 4 = log(22) = 2 log 2
π
∫ b
a
(L− f(x))2 dx
x = 1
y y ∈ [0, e]
x = log y
y y ∈ [1, e]
πe − π
∫ e
1
(log y)2 dy = 2π.
∫
(log y)2 dy = y(2+log2 y−2 log y)+C∫ e
1
(log y)2 dy = e− 2
π
∫ π/2
−π/2
((2 + cosx)− (−1))2 dx
π
∫ π/2
−π/2
((2 − cosx) − (−1))2 dx
19π2
2
+ 12π −
(
19π2
2
− 12π
)
= 24π
(3+cosx)2−(3−cosx)2 = 12 cosx
π
∫ π/2
−π/2
12 cosx dx = 24π
y = x2/(4a) = a x = ±2a
π
∫ 2a
−2a
a2 dx− π
∫ 2a
−2a
x4
16a2
dx =
= 4πa3 − 4πa
3
5
= 50π.
a = 5/2
Wn = π
∫ 1
0
(xn)2 dx =
π
2n+ 1
Vn = π
∫ 1
0
(1−(y1/n)2) dx = 2π
n+ 2
lim
n→+∞
Vn
Wn
=
lim
n→+∞
2(2n+ 1)
n+ 2
= 4
π
∫ +∞
1
((1/x)2 − e−2x) dx = π 2e
2 − 1
2e2
(xe−x)2 = x2e−2x∫
x2e−2x dx = −2x
2 + 2x+ 1
4
e−2x
π
∫ +∞
0
(xe−x)2 dx =
π
4
f ′(x) = − x√
(π/4)2 − x2
1 + [f ′(x)]2 =
(π/4)2
(π/4)2 − x2 r = π/4
∫ √
1 + [f ′(x)]2 dx =
∫
r√
r2 − x2 dx = r arcsen(x/r).
π2
4
≈ 2, 46
g′(x) = tanx∫ √
1 + tan2 x dx =
∫
sec dx = log(secx+ tanx).
log
(
2 +
√
2
2−√2
)
≈
1, 76