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c ∈ [a, b]
R2
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x = a
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f : [a, b] → R
f (x) ≥ 0 x ∈
[a, b]
y = f (x)
x
x = a
x = b
a y b
x
f
R
×
R
[a, b]
2
∆x = b−a
2
xi
i ∈ {0, 1, 2} x0 = a
x1 = c x2 = b
∆x
M i = M ax{f (x) : x ∈ [xi−1,
xi]}.
a y c b
x
f
R
2i=1
M i∆x, M i = M ax{f (x) :
x ∈ [xi−1, xi]}.
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R
[a, b]
n
n = 3, 4, 5, 6, · · ·?
R
y
xa b ba x
y
n
∆x → 0
n
→ ∞).
[a, b]
[a, b]
P = {x0, x1, x2,...,xi,...,xn}
a = x0 < x1 < x2 < ... <
xn = b
[a, b]
[x0, x1] , [x1, x2] , [x2, x3] , ..., [xi−1, xi] ,..., [xn−1,
xn] ,
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|[x0, x1]| = x1 − x0 = ∆x1|[x1, x2]| = x2 − x1
= ∆x2|[x2, x3]| = x3 − x2 = ∆x3· · ·|[xi−1, xi]|
= xi − xi−1 = ∆xi
· · ·|[xn−1, xn]| = xn − xn−1 = ∆xn.
[1, 12]
P = {1, 2, 4, 8, 12}
[1, 12].
[1, 2], [2, 4], [4, 8] e [8, 12].
1 = x0 < 2 = x1
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y
x
f(x)=xsen x
S (f, P ), P
A = 1, 863
y
x
f(x)=xsen x
S (f, P ), P
A = 1, 746
f : [a, b] → R
P = {x0, x1, x2,...,xi,...,xn}
[a, b],
a = x0 < x1 < x2
< ... < xn = b. mi
f
[xi−1, xi] i = 1, 2, 3,...,n
f
P
S (f, P )
S (f, P ) = m1(x1 − x0) + m2(x2 − x1)
+ ... + mn(xn − xn−1) =n
i=1
mi(xi − xi−1).
f : [0, 2] → R
f (x) = xsenx
[a, b]
S (f, P )
f (x) = x sin x
[0, 2]
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y
x
f(x)=xsen x
S (f, P ), P
A = 1, 642
y
x
f(x)=xsen x
S (f, P ), P
A = 1, 718
f : [a, b] → R
f
limn→+∞
S (f, P ) = limn→+∞
S (f, P )
limn→+∞
ni=1
mi(xi − xi−1) = limn→+∞
ni=1
M i(xi − xi−1),
P = {x0, x1, x2, · · · , xn}
[a, b].
f
a
b
ba
f (x) dx = limn→+∞
ni=1
f (wi) (xi − xi−1), onde wi ∈ [xi−1, xi] .
S =
ni=1
f (wi) ∆xi,
wi ∈ [xi−1, xi] f
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∆xi
f (x)
S (f, P )
S (f, P )
x
f ≥ 0)
f ≤ 0).
1 + 1 + 1 + ... + 1 = k
k vezes
1 + 2 + 3 + ... + k = (1
+ k)k
2
12 + 22 + 32 + ... + k2 =
k (k + 1) (2k + 1)
6
13 + 23 + 33 + ... + k3 =
k2 (k + 1)2
4
14 + 24 + 34 + ... + k4 = k (k + 1)
(6k3 + 9k2 + k − 1)
30
y = x2
+ 1, x = 0, x = 4
y = 0
P = {x0,x1, x2,...,xn} [0,
4]
y
x
f (x) = x2 + 1
∆x = ∆x1 = ∆x2 = ... =
∆xn
∆x = 4 − 0
n =
4
n
xi ∈ P
x0 = 0, x1 = ∆x, x2 = 2∆x, x3 = 3∆x,...,
xn = n∆x.
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M i f (x) = x
2 + 1
[xi−1, xi].
f
M i = f (xi). f
S (f, P ) =
M 1∆x + M 2∆x + M 3∆x + .... + M n∆x=
f (x1)∆x + f (x2)∆x + f (x3)∆x + ... + f (xn)∆x
=
f (∆x)∆x + f (2∆x)∆x + f (3∆x)∆x + ... + f (n∆x)∆x
= ∆x[(∆x)2 + 1 + (2∆x)2 + 1 + (3∆x)2 + 1 + ... +
(n∆x)2 + 1]
= ∆x[1 + 1 + ... + 1 + (∆x)2 + 4(∆x)2 + 9(∆x)2
+ ... + n2(∆x)2]
= ∆x[n + ∆x2(1 + 22 + 32 + ... + n2)]
= ∆x
n + ∆x2
n(n + 1)(2n + 1)
6
=
4
n n + 42
n2
n(n + 1)(2n + 1)
6 = 4 +
64
6
(n + 1)(2n + 1)
n2
= 4 + 32
3
2 +
3
n +
1
n2
= 4 +
64
3 +
32
n +
32
3n2.
40
(x2 + 1)dx = limn→+∞
4 +
64
3 +
32
n +
32
3n2
=
76
3 .
f,
y = 16 − x2, x = 1, x = 4 y =
0
P = {x0,x1, x2,...,xn} [1,
4]
y
x
f (x) = 16 − x2
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∆x = ∆x1 = ∆x2 = ... =
∆xn
∆x = 4 − 1
n =
3
n
xi ∈ P
x0 = 1, x1 = 1 + ∆x, x2 = 1 + 2∆x, x3 = 1 +
3∆x, · · · , xn = 1 + n∆x.
mi f (x) = 16 − x2 [xi−1, xi].
[1, 4]
f
mi = f (xi). f
S (f, P ) =
m1∆x + m2∆x + m3∆x + .... + mn∆x
=
f (x1)∆x + f (x2)∆x + f (x3)∆x + ... + f (xn)∆x
= f (1 + ∆x)∆x + f (1 +
2∆x)∆x + f (1 +
3∆x)∆x + ... + f (1 + n∆x)∆x
= [16
−(1 + ∆x)2 + 16
−(1 + 2∆x)2 + 16
−(1 + 3∆x)2 +
· · ·+ 16
−(1 + n∆x)2]∆x
= 16n∆x − [1 + 2∆x + (∆x)2 + 1 + 2 · 2∆x + (2∆x)2 + 1
+ 2 · 3∆x + (3∆x)2 ++ · · · + 1 + 2 · n∆x + (n∆x)2]∆x
= 16n∆x − n∆x − 2(1 + 2 + 3 + · · · + n)(∆x)2 − (12 + 22 +
32 + · · · + n2)(∆x)3
= 15n∆x − 2 · n(n + 1)2
· (∆x)2 − n(n + 1)(2n + 1)6
· (∆x)3
= 15n · 3n − 9 · n
2 + n
n2 − 9 · 2n
3 + 3n2 + n
2n3
= 45 − 9 − 9n − 9 − 27
2n − 9
2n2 = 27 − 45
2n − 9
2n2
41
(16 − x2)dx = limn→+∞
27 − 45
2n − 9
2n2
= 27.
f, g : [a, b] → R
f (x)
f (x) = c,
ba
cdx = c(b − a).
k
ba
kf (x) dx = k
ba
f (x) dx.
b
a
[f (x) + g (x)]dx = b
a
f (x) dx + b
a
g (x) dx.
f (x) ≤ g (x)
x ∈ [a, b] ,
ba
f (x) dx ≤ b
a
g (x) dx.
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m ≤ f (x) ≤ M
x ∈ [a, b] ,
m (b − a) ≤
ba
f (x) dx ≤ M (b − a) .
c
∈[a, b] ,
b
a
f (x) dx = c
a
f (x) dx + b
c
f (x) dx.
ba
f (x) dx = − a
b
f (x) dx.
aa
f (x)dx = 0.
f (x) = x2 − 2x + 2
[
−1, 2].
f y = 0, x = −1
x = 2.
f
15
[−1, 1]
[1, 2].
f
[−1, 1]
[1, 2].
f (x) = x2 − 2x + 2
[−1, 1]
P = {x0,x1, x2,...,xn} [−1,
1]
P
∆x = ∆x1 = ∆x2 = · · ·
= ∆xn.
∆x =
1 − (−1)n
= 2
n
xi ∈ P
x0 = −1, x1 = −1 + ∆x, x2 = −1 + 2∆x, x3 =
−1 + 3∆x, · · · , xn = −1 + n∆x.
M i f (x) = x2 − 2x + 2
[xi−1, xi].
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f
M i = f (xi−1).
f
S (f, P ) =
M 1∆x + M 2∆x + M 3∆x + · ·
· + M n∆x=
f (x0)∆x + f (x1)∆x + f (x2)∆x +
· · ·+ f (xn
−1)∆x
= f (−1)∆x + f (−1 +
∆x)∆x + f (−1 + 2∆x)∆x + · · ·
+ f (−1 + (n − 1)∆x)∆x= ∆x{5 + (−1 + ∆x)2 − 2(−1 + ∆x) +
2+ (−1 + 2∆x)2 − 2(−1 + 2∆x) + 2+
+ · · · + (−1 + (n − 1)∆x)2 − 2(−1 + (n − 1)∆x) + 2}= ∆x{5 + (1
− 2∆x + (∆x)2) + 2 − 2∆x + 2+ 1 − 4∆x + 22(∆x)2 + 2
− 4∆x + 2+
+ · · · + 1 − 2(n − 1)∆x + (n − 1)2(∆x)2 + 2 − 2(n −
1)∆x + 2}= ∆x{5 + 5 − 4∆x + (∆x)2+ 5 − 8∆x +
22(∆x)2+
+ · · · + 5 − 4(n − 1)∆x + (n − 1)2(∆x)2}= ∆x 5n − 4∆x (1 +
2 + · · · + (n − 1)) + (∆x)
2
1 + 22 +
· · ·+ (n
−1)2
= 2
n · 5n − 4 · 2
n · n(n − 1)
2 +
2n
2 · (n − 1)n (2n − 1)6
=
2
n ·
5n − 4(n − 1) + 23 ·
2n2 − 3n + 1n
= 2 +
8
n +
4
3 ·
2 − 3n
+ 1
n2
=
14
3 +
4
n +
4
3n2.
[1, 2]
Q = {x0,x1, x2,...,xn} [1, 2]
Q
∆x = ∆x1 = ∆x2 = · · ·
= ∆xn.
∆x =
2 − 1n
= 1
n
xi ∈ Q
x0 = 1, x1 = 1 + ∆x, x2 = 1 + 2∆x, x3 = 1 +
3∆x, · · · , xn = 1 + n∆x.
M i,
M i = f (xi).
[1, 2]
Q
S (f, Q) =
M 1∆x + M 2∆x + M 3∆x + · ·
· + M n∆x=
f (x1)∆x + f (x2)∆x + f (x3)∆x +
· · · + f (xn)∆x= [f (1 + ∆x) + f (1 +
2∆x) + f (1 + 3∆x) + · · · + f (1
+ n∆x)]∆x= {[(1 + ∆x)2 − 2(1 + ∆x) + 2] + [(1 + 2∆x)2 −
2(1 + 2∆x) + 2] +
+[(1 + 3∆x)2 − 2(1 + 3∆x) + 2] + · · · + [(1 + n∆x)2 − 2(1
+ n∆x) + 2]}∆x= {[1 + (∆x)2] + [1 + (2∆x)2] + [1 +
(3∆x)2] + · · · + [1 + (n∆x)2]}∆x= n∆x + (12 + 22 + 32
+ · · · + n2)(∆x)3
= n
·
1
n
+ n(n + 1)(2n + 1)
6 ·1
n3
= 4
3
+ 1
2n
+ 1
6n2
f
[−1, 2]
S (f, P ∪ Q) = 143
+ 4
n +
4
3n2 +
4
3 +
1
2n +
1
6n2 = 6 +
9
2n +
3
2n2.
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f,
f
15
[−1, 1]
[1, 2].
f
[−1, 1] [1, 2].
y
x
f (x) = x2 − 2x + 2
[−1, 1]
P
mi,
[xi−1, xi], mi = f (xi).
f
[−1, 1], P, S (f, P ) =
m1∆x + m2∆x + m3∆x + · · ·
+ mn∆x
=
f (x1)∆x + f (x2)∆x + f (x3)∆x +
· · · + f (xn)∆x= f (−1 +
∆x)∆x + f (−1 + 2∆x)∆x + f (−1 +
3∆x)∆x + · · · + f (−1 + n∆x)∆x= ∆x
(−1 + ∆x)2 − 2(−1 + ∆x) + 2+ (−1 + 2∆x)2 − 2(−1 + 2∆x) + 2+
+ · · · + (−1 + n∆x)2 − 2(−1 + n∆x) + 2 = ∆x 1 −
2∆x + (∆x)2 + 2 − 2∆x + 2+ 1 − 4∆x + 22(∆x)2 + 2 −
4∆x + 2+
+ · · · + 1 − 2n∆x + n2(∆x)2 + 2 − 2n∆x + 2 =
∆x
5 − 4∆x + (∆x)2+ 5 − 8∆x + 22(∆x)2+ · · · + 5 −
4n∆x + n2(∆x)2
= ∆x
5n − 4∆x (1 + 2 + · · · + n) + (∆x)2 1 + 22 + · · ·
+ n2=
2
n ·
5n − 4 · 2n · (n + 1)n
2 +
2
n
2· n(n + 1) (2n + 1)
6
= 2
n ·
5n − 4(n + 1) + 23 ·
2n2 + 3n + 1
n
= 2 − 8
n + 4
3 ·2 + 3
n + 1
n2 = 14
3 − 4
n + 4
3n2.
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[1, 2]
Q
mi,
[xi−1, xi], mi = f (xi−1).
f
[1, 2],
Q,
S (f, Q) =
m1∆x + m2∆x + m3∆x + · · · + mn∆x=
f (x0)∆x + f (x1)∆x + f (x2)∆x +
· · · + f (xn−1)∆x=
f (1)∆x + f (1 + ∆x)∆x + f (1 +
2∆x)∆x + · · · + f (1 + (n − 1)∆x)∆x= ∆x{1 + (1 +
∆x)2 − 2(1 + ∆x) + 2+ (1 + 2∆x)2 − 2(1 + 2∆x) + 2+
+ · · · + (1 + (n − 1)∆x)2 − 2(1 + (n − 1)∆x) + 2}= ∆x{1 + [1 +
(∆x)2] + [1 + (2∆x)2] + · · · + [1 + ((n − 1)∆x)2]}=
n∆x + [12 + 22 + · · · + (n − 1)2](∆x)3
= n · 1n
+ (n − 1)n(2n − 1)6
· 1n3 = 4
3 − 1
2n + 1
6n2.
f
[−1, 2]
S (f, P ∪ Q) = 143 − 4
n +
4
3n2 +
4
3 − 1
2n +
1
6n2 = 6 − 9
2n +
3
2n2.
f,
A = 1−1(x2 − 2x + 2)dx +
2
1(x2 − 2x + 2)dx
= limn→+∞
14
3 +
4
n +
4
3n2
+ lim
n→+∞
4
3 +
1
2n +
1
6n2
=
14
3 +
4
3 = 6.
f.
R
f (x) = 9 g(x) = x2,
x ≤ 0,
f
g
R
R
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R
R1
R2, R1 y = g(x), y
= 0, x = −3
x − 0 R2 y = f (x),
y = 0, x = −3 x − 0.
R1 : AR1 = 0
−39dx = 9[0
−(−
3)] = 27
R2 : R2
R2
R2 AR2 =
0
−3
x2dx
P = {x0, x1, x2, · · · , xn}
[−3, 0]
P
∆x = ∆x1 = ∆x2 = ...
= ∆xn.
∆x =
0 − (−3)n
= 3
n
xi ∈ P
x0 = −3, x1 = −3 + ∆x, x2 = −3 + 2∆x, · · ·
, xn = −3 + n∆x.
xi, i = 1, 2, · · · , n,
g(xi) = x
2i . g
P
S (g, P ) =n
i=1
g(xi)∆x =n
i=1
x2i ∆x = (x21 + x22 + · · ·
+ x2n)∆x
= [(−3 + ∆x)2 + (−3 + 2∆x)2 + · · · + (−3 + n∆x)2]∆x=
9 − 6∆x + (∆x)2+ 9 − 6 · 2∆x + (2∆x)2+ · · · + 9 − 6 ·
n∆x + (n∆x)2∆x
= 9n∆x − 6(∆x)2(1 + 2 + · · · + n) + (∆x)3(12 + 22 + · · ·
+ n2)= 27 − 54
n2n(n + 1)
2 +
27
n3n(n + 1)(2n + 1)
6
= 27 − 271 + 1n+ 9
22 + 3
n + 1
n2
= 9 + 27
2n +
9
2n2
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AR2 = limn→+∞
9 +
27
2n +
9
2n2
= 9
.
R
AR = AR1 − AR2 = 27 − 9 = 18 .
40
(−x2 − 1)dx.
f (x) = −x2 − 1
R
R
40
(−x2 − 1)dx
P = {x0, x1, x2, · · · , xn} [0,
4]
P
∆x = ∆x1 = ∆x2 = ... = ∆xn.
∆x = 4−(0)n
= 4n
xi ∈ P x0 = 0, x1 = ∆x,
x2 = 2∆x, · · · , xn = n∆x.
xi−1,
i = 1, 2, · · · , n, f (xi−1).
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f
P
S (f, P ) =n
i=1
f (xi−1)∆x
= [f (x0) + f (x1) + f (x2) + · · ·
f (xn−i)]∆x=
−1 + [−(∆x)2 − 1] + [−(2∆x)2 − 1] + · · · + [−((n − 1)∆x)2 −
1]∆x= −n∆x − [12 + 22 + · · · + (n − 1)2](∆x)3
= −n · 4n − (n − 1)n(2n − 1)
6 ·
4
n
3= −4 − 32(2n
2 − 3n + 1)3n2
= −4 − 643
+ 32
n − 32
3n2
4
0 (−x2
− 1)dx = limn→+∞−763 + 32n − 323n2
= −763 .
f : [a, b] → R c ∈
[a, b]
ba
f (x)dx =
f (c)(b − a).
40
(x2 + 1)dx = 76
3 .
f (x) = x2+ 1
[0, 4]
c ∈ (0, 4) 4
0
(x2 + 1)dx = f (c)(4 − 0).
c2 + 1 = 76
4 · 3 ⇒ c2 =
16
3 ⇒ c = ±4
√ 3
3 .
c = −4
√ 3
3
c = 4
√ 3
3
f (x) ≥ 0 [a, b].
ba
f (x)dx
f a
b,
f (c)
P
f
c
P
x
x = a
x = b
f (c)(b −a)
f
a
b.
c
f c
1
b − a b
a
f (x)dx
f
[a, b].
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F (x) = lim∆x→0
F (x + ∆x) − F (x)∆x
= lim∆x→0 1∆x x+∆xa f (t) dt −
x
af (t) dt
= lim∆x→0
1
∆x
xa
f (t) dt +
x+∆xx
f (t) dt − x
a
f (t) dt
= lim∆x→0
1
∆x
x+∆xx
f (t) dt,
c ∈ [x, x + ∆x]
x+∆x
x
f (t) dt = f (c) (x +
∆x
−x) = f (c)∆x
F (x) = lim∆x→0
f (c)
∆x → 0 c → x
f f (c) →
f (x)
F (x) = lim∆x→0
F (x + ∆x) − F (x)∆x
= f (x) .
f : [a, b] → R
[a, b]
F : [a, b] → R
(a, b)
F (x) = f (x) .
F : [a, b] → R,
f : [a, b] → R
[a, b]
[a, b]
f : [a, b] → R
[a, b] ,
b
a f (x)dx = G(b) − G(a)
G
f,
G = f .
F (x) =
xa
f (t)dt.
F (x) = f (x),
F
f.
G
f
[a, b],
G(x) = F (x) + c.
G(b) − G(a) = [F (b) + c] − [F (a) + c]
= b
a
f (t)dt − a
a
f (t)dt =
ba
f (t)dt
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t
x
ba
f (x)dx = G(b) − G(a)
ba
f (x)dx = G(x)
b
a
.
f : [0, 4] → R f (x)
= x2 + 1.
A =
40
(x2 + 1)dx = x3
3 + x
40
= 64
3 + 4 =
76
3 .
x
f (x) = 1
8(x2−2x+8),
x
[−2, 4].
f (x) = 1
8(x2 − 2x + 8)
A =
4−2
1
8(x2 − 2x + 8)dx = 1
8 (
x3
3 − x2 + 8x)
4
−2
=
1
8 433 − 42 + 8(4) − (−2)33 − (−2)2 + 8(−2)=
1
8
64
3 − 16 + 32 + 8
3 + 4 + 16
=
60
8 =
15
2 u.a.
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51
√ x − 1
x dx =
20
t
t2 + 12tdt = 2
20
t2
t2 + 1dt = 2
20
t2 + 1 − 1t2 + 1
dt =
= 2 20
t2 + 1t2 + 1
− 1t2 + 1
dt = 2 20
dt − 2 20
dtt2 + 1
=
= 2t
2
0
− 2 arctan t2
0
= 4 − 2arctan2.
f, g : [a, b] → R
ba
f (x)g(x)dx = f (x)g(x)b
a− b
af (x)g(x)dx.
u = f (x) ⇒ du =
f (x)dxdv = g (x)dx ⇒ v =
g(x)
b
a udv = uvb
a−
b
a vdu.
π3
0
sin3 xdx.
u = sin2 x ⇒ du = 2 sin x cos
xdxdv = sin xdx ⇒ v = sin
xdx = − cos x
π3
0
sin3 xdx = sin2 x(− cos x)π3
0
− π
3
0
− cos x(2 sin x cos x)dx
= − sin2 x cos xπ3
0
+ 2
π3
0
cos2 x sin xdx
= (− sin2 x cos x − 23
cos3 x)
π3
0
= −34 · 12 −
112 + 23 = 524 .
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f : [a, ∞) → R x ∈
[a, +∞)
+∞a
f (x) dx = limb→+∞ ba
f (x) dx,
+∞0
1
1 + x2dx.
y
x
f (x) = 1
1+x2
f
+∞
0
1
1 + x2dx = lim
b→+∞ b
0
1
1 + x2dx = lim
b→+∞arctan x
b
0= lim
b→+∞(arctan b − arctan 0) = lim
b→+∞arctan b =
π2
.
f : (−∞, b] → R
x ∈ (−∞, b]
b−∞
f (x) dx = lima→−∞
ba
f (x) dx,
0
−∞
1
1 + x2dx
0−∞
1
1 + x2dx = lim
a→−∞
0a
1
1 + x2dx = lim
a→−∞arctan x
0
a
= lima→−∞
[arctan 0 − arctan a] = − lima→−∞
arctan a = −−π
2
=
π
2.
f : (−∞, ∞) → R x ∈
(−∞, +∞)
+∞−∞
f (x) dx = lima→−∞
ca
f (x) dx + limb→+∞
bc
f (x) dx,
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+∞−∞
1
1 + x2dx
c = 0,
+∞−∞
1
1 + x2dx = lim
a→−∞ 0
a
1
1 + x2dx + lim
b→+∞ b0
1
1 + x2dx
= lima→−∞
arctan x
0
a
+ limb→+∞
arctan x
b
0
= lima→−∞
(arctan 0 − arctan a) + limb→+∞
(arctan b − arctan 0)= lim
a→−∞arctan a + lim
b→+∞arctan b
= −−π
2+
π
2 = π.
c ∈[a, b]
f : [a, b] → R
[a, b]
c ∈ [a, b]
ba
f (x) dx = limα→c−
αa
f (x) dx + limβ→c+
bβ
f (x) dx,
1−1
1
x2dx.
y
x
f (x) = 1
x2
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[−1, 1] ,
x = 0
1
−1
1
x2dx = lim
α
→0−
α
−1
1
x2dx + lim
β
→0+ 1
β
1
x2dx
= limα→0−
−1x
α
−1+ lim
β→0+−1x
1
β
= limα→0−
−1α
−−1
−1
+ limβ→0+
−1 −
−1β
= [+∞ − 1] + [−1 + ∞] = +∞
f (x) =
1
x2
[−1, 1].
+4−∞
|x|exdx;
π0
sin x
cos2 x dx.
+4−∞
|x|exdx = lima→−∞
0a
−xexdx + 40
xexdx
= lima→−∞
−xex0
a
− 0
a
−exdx + xex
4
0
− 40
exdx
= lima→−∞
0 + aea + e0 − ea+ 4e4 − 0 − (e4 − 1)
= lima→−∞
aea − lima→−∞
ea + 3e4 + 1
= lima→−∞
a
e−a
+ 3e4 + 1 = lima→−∞
1
−e−a
+ 3e4 + 1 = 3e4 + 1
π0
sin x
cos2 x dx = lim
a→π2
−
a0
sin x
cos2 x dx + lim
b→π2
+
πb
sin x
cos2 x dx
= lima→π
2
−
1
cos x
a
0
+ lim
b→π2
+
1
cos x
π
b
= lima→π2
− 1cos a − 1+ limb→π2
+ −1 − 1cos b= +∞ − 2 + ∞ = +∞
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f
f (x) ≥ 0
x
[a, b]
x = a x = b
y = 0
y = f (x)
A = b
a
f (x) dx.
f
g,
f (x) ≥ g(x)
x ∈ [a, b],
y
xba
y=f(x)
y=g(x)
A =
ba
f (x)dx − b
a
g(x)dx =
ba
[f (x) − g(x)] dx.
f (x) − g(x)
dx.
g
g(x) < 0
x ∈ [a, b]
x = a
x = b
y = 0
y = g (x)
A = ba
[0 − g(x)] dx = − ba
g(x)dx.
x
f (x) =
2x,
x
[−2, 2] .
f
[−2, 0] [0, 2]
A = 0
−2(0
−2x)dx +
2
0
(2x−
0)dx = 0
−2−2xdx +
2
0
2xdx =−
x20
−2+ x22
0
= 8 u.a.
f (x) = 2x,
[−2, 2] , 8
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x
y
x
f (x) = 2x
y = x2
y =
√ x
y = x2
y =√
x
x2 =
√ x
x
x = 0
x = 1
y = x2
y =
√ x
x ∈ [0, 1] y
x
y = x2
y =
√ x.
y = √ x
y = x2,
x ∈ [0, 1] .
A =
10
√ x − x2 dx = 2
3x
3
2 − 13
x3
1
0
= 2
3 − 1
3 =
1
3 u.a.
1
3
(0, 0)
(2, 12
).
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x
y
y = x
y = x
4,
A =
10
x − 1
4x
dx +
21
1
x − 1
4x
dx
= 3
4
10
xdx +
21
1
xdx − 1
4
21
xdx
= 3
8x2
1
0
+
ln |x| − 1
8x2
2
1
= 3
8 + ln(2)
− 1
2 −ln(1) − 18=
4
8 − 1
2 + ln(2) = ln(2) u.a.
ln(2)
y + x2 = 6
y + 2x = 3.
y = 6 − x2y = 3
−2x
⇒ 6 − x2 = 3 − 2x ⇒ x2 − 2x − 3 = 0
⇒ x = −1
x = 3.
[−1, 3]
A =
3−1
[(6 − x2) − (3 − 2x)]dx
=
3−1
(3 − x2 + 2x)dx
= 3x − x33
+ x23−1
= 9 − 273
+ 9 − (−3 + 13
+ 1) = 32
3
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y
x
y + x2 = 6
y + 2x = 3.
32
3
y = x2, y = 2 − x2
y = 2x + 8.
y = x2
, y = 2 − x2
y = 2x + 8
y = x2
y = 2x + 8
x = 4, y =
16 x = −2, y = 4.
y = x2
y = 2 − x2 x = 1, y =1
x = −1, y = 1.
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A1 =
−1−2
(2x + 8) − (x2)dx = −1−2
(2x + 8 − x2)dx = 83
,
A2 = 1−1
(2x + 8) − (2 − x2)dx = 1−1
(2x + 6 + x2)dx = 383
,
A3 =
41
(2x + 8) − (x2)dx = 18.
A = A1 + A2 + A3
= 8
3 +
38
3 + 18 =
100
3 u.a.
x = y + 1 x = y2 − 1.
x
y
x = y + 1
x = y2 − 1
y2 − 1 = y + 1 ⇒ y2 − y − 2 = 0
⇒ y = −1 y = 2
y = −1 ⇒ x = 0 y = 2
⇒ x = 3.
x
y
x,
y = x − 1 y = ±√ x +
1.
x
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A = 0
−1
√ x + 1
−(
−
√ x + 1)dx +
3
0
√ x + 1
−(x
−1)dx
=
0−1
2√
x + 1dx +
30
(√
x + 1 − x + 1)dx
= 4
3
(x + 1)3
0
−1+
2
3
(x + 1)3 − x
2
2 + x
3
0
= 4
3 +
16
3 − 9
2 + 3 − 2
3 =
9
2 u.a.
y
y.
x = y + 1 x = y2−1.
A =
2−1
(y + 1) − (y2 − 1)dy
=
2−1
(y − y2 + 2)dy = y2
2 − y3
3 + 2y
2
−1
= 2 − 83
+ 4 − 12 − 1
3 − 2 = 9
2 u.a.
x
y
y =
√ x − 2, x + y = 2
x + 2y = 5,
x.
y.
y =
√ x − 2 ≥ 0.
x + y = 2x + 2y = 5
,
x = −1, y = 3.
y =
√ x − 2
x + y = 2 ,
x = 2
y = 0.
y =
√ x − 2
x + 2y = 5
,
x = 3,
y = 1.
x,
y
x,
y =
5 − x2
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x
y
y =√
x
−2
x + y = 2
x + 2y = 5
y = 2−x y = √ x − 2
x = 2
A =
2−1
5 − x
2
− (2 − x)
dx +
32
5 − x
2
− √ x − 2 dx
=
2−1
1 + x
2 dx +
32
5 − x
2 − √ x − 2
dx.
y
x
y,
x = 5−2y
x = 2 − y x = y2 + 2
y = 1,
A =
10
(y2 + 2) − (2 − y) dy + 3
1
[(5 − 2y) − (2 − y)] dy
=
10
(y2 + y)dy +
31
(3 − y) dy.
y
R
A =
21
(2x2) − (2√ x) dx + 4
2
(−2x + 12) − (2√ x) dx.
R.
R
y
x
y = 2x2
y = 2
√ x
x
y = −2x + 12
y = 2
√ x.
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R
R
y = 2x2, y = −2x + 12
y = 2
√ x
A :
y = 2x2
y = 2√
x ⇒ (1, 2); B :
y = 2x2
y =
−2x + 12
⇒ (2, 8)
C : y = −2x + 12y =
2
√ x
⇒ (4, 4).
y
R
A =
42
y2
4 −
y
2
dy +
84
12 − y
2 −
y
2
dy.
y = f (x)
[a, b]
R
R,
x = φ (t)
y = ψ (t)
t ∈ [α, β ] , f.
A =
ba
f (x) dx =
ba
ydx.
y = ψ (t)
dx = φ(t)dt
a = φ(α)
b = φ(β )
A = β
α
ψ(t)φ(t)dt.
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t = π2
t = 0.
E 1 E 2
A = 4 0π2
[4 sin t(−2sin t)dt − 4 0π2
sin t(−2sin t)]dt
=
0π2
(−32 sin2 t + 8 sin2 t)dt = 0
π2
−24 sin2 tdt
= 24
π2
0
1
2(1 − cos2t)dt =
12t − 12
2 sin2t
π2
0
= 6π u.a.
r = f (θ)
[α, β ]
r = f (θ)
[α, β ]
∆θi = θi − θi−1
ri = f (θi).
X = {θ0, θ1, θ2, θ3,...,θn} [α,
β ]
α = θ0 < θ1 < θ2 < θ3
< ... < θn = β.
∆θ1, ∆θ2, ∆θ3,..., ∆θn
X ri
ξ i ∈ ∆θi θi−1 ≤
ξ i ≤ θi
ri ∆θi
Ai = 1
2 (ri)2∆θi
r = f (θ)
An =n
i=1
12 (ri)2∆θi.
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|∆θ| X n
|∆θ|
A = limn→∞
An = lim|∆θ|→0
n
i=11
2
(ri)2∆θi =
1
2 β
α
r2dθ
A = 1
2
βα
r2dθ,
r =
1 − cos θ
r = 1.
x,
xy.
θ
[0, π
2].
A = 2
2
π2
0
12dθ − 22
π2
0
(1 − cos θ)2dθ = π
2
0
(2cos θ − cos2 θ)dθ
= π20
2cos θ − 12
(1 + cos 2θ)dθ = 2 sin θ − 12
θ − 14
sin 2θπ
2
0
= 2 − π4
.
2 − π
4
r = 1
r = 2cos(2θ).
2 cos(2θ) = 1 ⇒ cos2θ = 12 ⇒ 2θ =
π
3 ⇒ θ = π
6 (
1o
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θ
[0, π
6]
A = 8 · 12
π6
0
[(2 cos(2θ))2 − (1)2]dθ = 4 π
6
0
(4cos2(2θ) − 1)dθ.
r = 5 cos θ
r = 5
√ 3sin θ.
x = r cos θ, y = r sin θ r2
= x2 + y2,
r = 5 cos θ ⇒ r2
= 5r cos θ ⇒ x2
+ y
2
= 5x ⇒ x − 522
+ y
2
=
25
4
r = 5√
3sin θ ⇒ r2 = 5√
3r sin θ ⇒ x2 + y2 = 5√
3y ⇒ x2 + (y − 5√
3
2 )2 =
75
4
5√ 3sin θ = 5 cos θ ⇒ √ 3tan θ = 1
⇒ tan θ =
√ 3
3 ⇒ θ = π
6 .
r =
5√
3sin θ
θ ∈ [0, π6
] r = 5cos θ
θ ∈ [ π
6, π
2].
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A = 1
2 π6
0
(5√
3sin θ)2dθ + 1
2 π2
π6
(5cos θ)2dθ
= 1
2
π6
0
75sin2 θdθ + 1
2
π2
π6
25cos2 θdθ.
R
I = 2
1
2
π4
0
(2
(θ))2 dθ + 1
2
π2
π4
(√
2)2 dθ
.
R.
x;
y.
R.
R
θ = π4
ρ = 2
θ
ρ =
√ 2.
x2+ (y − 1)2 = 1
x2+ y2 =
√ 2.
R
R
ρ = 2 sin θ
ρ =
√ 2
ρ = 2 sin θ
ρ =√
2 =⇒ θ = π
4
3π
4
I
(−1, 1)
(1, 1).
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x :
I = 1
−1(√
2
−x2
−1 +
√ 1
−x2) dx
I = 2
1
0
(√
2
−x2
−1 +
√ 1
−x2) dx
y :
I = 2
10
2y − y2 dy + 2
√ 21
2 − y2 dy
R
I
A = 2
1
2
π4
0
(2
(θ))2 dθ + 1
2
π2
π4
(√
2)2 dθ
=
π4
0
4
2θ dθ +
π2
π4
2 dθ
= 4
π4
0
1 − (2θ)
2 dθ + 2θ
π2
π4
= 2
θ − (2θ)
2
π4
0
+ π
2 = (π − 1)
y = f (x)
[a, b] ,
AB,
a b xi
M n
xi-1 x1
Δ s
M 0
Δ x
f(xi )
Δ y
y
x
f(xi-1 ) M 1
M i-1
M i
AB
X = {M 0, M 1, M 2, ..., M n}
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∆yi∆xi
= f (ξ i) . (III )
(II )
(I )
ln =n
i=1 1 + ∆yi∆xi2∆xi (IV )
(III )
(IV )
ln =n
i=1
1 + (f (ξ i))
2∆xi.
|∆x| AB. n → ∞,
|∆x| → 0 (ξ i) → x.
l = limn→∞ln = lim|∆x|→0
ni=1 1 + (f (ξ i))2∆xi =
b
a 1 + (f (x))2dx.
AB
[a, b]
l =
ba
1 + (f (x))2dx.
y = √
x,
x
[0, 4] .
y
x
f (x) =
√ x
y = f (x) = √ x
f (x) = 12√ x
l =
ba
1 + (f (x))2dx =
40
1 +
1
2√
x
2dx
=
40
1 +
1
4xdx =
40
4x + 1
4x dx =
1
2
40
√ 4x + 1√
x dx.
x = 0.
t2 = x,
dx = 2tdt
x ∈ [0, 4], t ∈ [0, 2] .
l = 1
2
20
√ 4t2 + 1√
t22tdt =
20
√ 4t2 + 1dt.
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l = 12
t√ 4t2 + 1 + 14
ln 2t + √ 4t2 + 1 2
0
=√
17 + 1
4 ln(4 +
√ 17) u.c.
x = φ (t)
y = ψ (t) ,
t ∈ [α, β ] ,
y = f (x) .
dx = φ (t) dt
dy = ψ (t) dt,
f (x) = dy
dx = ψ (t) dtφ (t) dt =
ψ (t)φ (t) .
l =
ba
1 + (f (x))2dx
=
βα
1 +
(ψ (t))2
(φ (t))2φ (t) dt
= β
α (φ (t))2 + (ψ (t))2
φ (t)2 φ (t) dt
=
βα
(φ (t))2 + (ψ (t))2
φ (t) φ (t) dt
=
βα
(φ (t))2 + (ψ (t))2dt.
l = β
α (φ (t))2 + (ψ (t))2dt.
r
2πr.
x(t) = r cos ty(t) = r sin t
t ∈ [0, 2π].
l = 2π0
(−r sin t)2 + (r cos t)2dt = 2π0
r2(sin2 t + cos2 t)dt = 2π0
rdt = rt|2π0 = 2πr.
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y
x
3
3-3
-3
φ (t) = 3 cos3 t, ψ(t) = 3 sin3 t t ∈ [0, 2π].
t ∈ [0, π
2]
φ (t) = −9cos2 sin t
ψ (t) = 9 sin2 t cos t,
l = 4
π2
0
(−9cos2 t sin t)2 + 9sin2 t cos t2dt = 4 ·
9 π2
0
cos4 t sin2 t + sin4 t cos2 tdt
= 36
π2
0
cos2 t sin2 t
cos2 t + sin2 t
dt = 36
π2
0
cos t sin tdt = 18 sin2 t
π2
0
= 18 u.c.
18
x = 3t
y = 2t
3
2 .
t = 0 t = 1?
x = φ(t) = 3t y = ψ(t) = 2t
3
2
t
∈[0, 1],
l =
10
32 + (3t
1
2 )2dt =
10
√ 9 + 9tdt
= 3
10
√ 1 + tdt = 2(1 + t)
3
2
1
0
= 2(2)3
2 − 2(1) 32 = 4√
2 − 2
t = 0
t = 1
4√
2 − 2
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φ (θ) = r cos θ
ψ (θ) = r sin θ
r = f (θ)
θ ∈
[α, β ]
r
f (θ)
φ (θ) = f (θ)cos θ ψ (θ)
= f (θ)sin θ
φ (θ) = f (θ)cos θ − f (θ)sin
θ = r cos θ − r sin θψ (θ)
= f (θ) senθ + f (θ)cos θ
= r senθ + r cos θ.
(φ (t))2 + (ψ (t))2 = (r cos θ − rsenθ)2 +
(rsenθ + r cos θ)2
(φ (t))2 + (ψ (t))2 = (r)2 + r2.
l =
βα
(r)2 + r2dθ.
r = a (1 + cos θ)
r = a (1 + cos θ)
r = −a sin θ.
l =
βα
(r)2 + r2dθ
= 2
π0
(−a sin θ)2 + (a (1 + cos θ))2dθ
= 2a
π0
sin2 θ + 1 + 2 cos θ + cos2 θdθ
= 2a π0
√ 2 + 2 cos θdθ
= 2a · 2 π0
cos θ
2dθ
= 4a · 2sin 12
θ
π
0
= 8a u.c.
r = a (1 + cos θ)
8a u.c.
r = 2e2θ
θ ≥ 0
r = a,
a > 2.
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2e2θ = a ⇒ e2θ = a2 ⇒ 2θ =
ln a
2 ⇒ θ = 1
2 ln
a
2
θ ∈ 0, 1
2 ln a2
.
r = 2e2θ
r = 4e2θ
l =
12 ln a
2
0
(4e2θ)2 + (2e2θ)2dθ =
12 ln a
2
0
√ 20e4θdθ
=
12 ln a
2
0
2√
5e2θdθ =√
5e2θ
1
2 ln a
2
0
=√
5a
2 − 1
u.c.
T
y = f (x)
x,
[a, b]
x
y
z
a b
y=f(x)
r=f(x)
dx
Cálculo do elemento de volume
dV= r dx
dV= f(x) dx
π ²
π ²[ ]
x
y
a b
y=f(x)
Área plana
x
P = {x0, x1, · · · , xn}
[a, b] ∆x1, ∆x2, · · · ,
∆xn
ξ i ∈ ∆xi f (ξ i)
∆xi V i = π [f (ξ i)]
2∆xi
n − cilindros
V n =n
i=1
π [f (ξ i)]2∆xi.
|∆θ|
n → ∞,
|∆θ| → 0
ξ i → x
V
T
V = limn→∞
V n = lim|∆θ|→0
ni=1
π [f (ξ i)]2∆xi = π
ba
[f (x)]2 dx.
x
V = π
ba
[f (x)]2 dx.
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2m
4m
6m
cilindro
cone
V 1 V 2
V = V 1 + V 2.
V 1 y = 2 x
x
y
-2
y
z
x
V 1 = π
40
22dx = 4π · 4 = 16π.
r = 2
h = 6
y = 1
3x
y
x
y
z
x
V 2 = π 60
19
x2dx = 127
πx36
0
= 63π27
= 8π.
V = 16π + 8π = 24π u.v.
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f (x) = x3,
x
x.
x
y
x
r
y
z
y = x3
x
V = π
21
x32
dx = π
21
x6dx = πx7
7
21
= 127π
7 u.v.
y = x2
y = x + 2
x
x
y
x
y
z
x
x2 = x + 2 ⇒ x2 − x − 2 = 0 ⇒
x = −1 x = 2.
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V = π
2−1
(x + 2)2dx − π 2−1
x2
2
dx
= π 2−1
(x2 + 4x + 4 − x4)dx
= π
1
3x3 + 2x2 + 4x − 1
5x5 2
−1=
72
5 π u.v.
(x − 2)2 + y2 = 1
y.
y
1
1 2 3-1
-1
x
(x − 2)2 + y2 = 1
x
(x − 2)2 = 1 − y2 ⇒ x = 2 ±
1 − y2
x = 2 +
1 − y2
y,
x = 2 −
1 − y2.
V = V 1 − V 2,
V 1 = π 1−1(2 + 1 −
y2)2dy
V 2 = π
1−1
(2 −
1 − y2)2dy
V = π
1−1
[(2 +
1 − y2)2 − (2 −
1 − y2)2]dy = π 1−1
8
1 − y2dy.
y = sin θ,
dy =
cos θdθ
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V = π
π2
−π2
8
1 − sin2 θ cos θdθ
= 8π π2−π
2
cos2 θdθ = 4π π2−π
2
(1 + cos 2θ)dθ
= π[4θ + 2 sin (2θ)]
π2
−π2
= 4π2.
4π2
y = f (x)
x = g(y)
y = f (x),
x ∈ [a, b],
y = c,
