Leyanlstore.leya.com/asa/2017/novos_projetos/images/Primitivas_e_Calculo...Β Β· 1 1 = π + ,...
Transcript of Leyanlstore.leya.com/asa/2017/novos_projetos/images/Primitivas_e_Calculo...Β Β· 1 1 = π + ,...
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1ππ₯ = π₯ + π, π β β
π₯πΌππ₯ =π₯πΌ+1
πΌ+1+ π,
π β β , πΌ β β\{0,β1}
1
π₯ππ₯ = ln |π₯| + π, π β β
ππ₯ππ₯ = ππ₯ + π, π β β
sin π₯ ππ₯ = βcos π₯ + π, π β β
cos π₯ ππ₯ = sin π₯ + π, π β β
π
πΌπΉ
ππΌ πΉ
πΌπΉβ² π₯ = π π₯ , βπ₯ β πΌ
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πΌ
π
π π₯ = 1 , π₯ β€ 00 , π₯ > 0
0 β πΌ ππΌ
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π₯πΌππ₯πΌ
β’ πΌ = 0
1ππ₯ = π₯ + π, π β β
β’ πΌ = β1
1
π₯ππ₯ = ln|π₯| + π, π β β
β’
π₯πΌππ₯ =π₯πΌ+1
πΌ + 1+ π, π β β , πΌ β β\{0, β1}
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1
π₯ππ₯
1
π₯ππ₯ =
πππ₯ + π, π β β
πΌ β β+ 1
π₯ππ₯ = πππ₯ + π, π β β
πΌ β ββ 1
π₯ππ₯ = ln βπ₯ + π, π β β
π π π₯ =1
π₯
β β 0 π π π₯ = ππ π₯0,+β .
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ππ π₯ ππ₯ = π π(π₯)ππ₯
π π₯ + π(π₯) ππ₯ = π(π₯)ππ₯ + π(π₯)ππ₯
π’β² π₯ π π’ π₯ ππ₯ = πΉ π’ π₯ + π, π β β
πΉ π
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a) 3π₯ππ₯ = 3 π₯ππ₯ = 3π₯2
2+ π, π β β
b) 3(2π₯ β 1)2ππ₯ = 3
22(2π₯ β 1)2ππ₯ =
3
2 2(2π₯ β 1)2ππ₯ =
=3
2Γ(2π₯β1)3
3+ π, π β β =
2π₯β1 3
2+ π, π β β
c) tgπ₯ππ₯ = sinπ₯
cosπ₯ππ₯ = β
βsinπ₯
cosπ₯ππ₯ = β ln cosπ₯ + π, π β β
d) π₯
2ππ₯ =
π₯
2ππ₯ =
1
2 π₯
1
2ππ₯ =1
2Γπ₯32
3
2
+ π, π β β =2π₯3
3+ π,
π β β
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a) π₯3 + π₯2 + 1 ππ₯
b) π3π₯ππ₯
c) 2π₯ππ₯2ππ₯
d) π₯3 π₯4 + 1 8ππ₯
e) sinπ₯ cos2π₯ππ₯
f) β cos3π₯ππ₯
g) ββ ln (3π₯)
π₯ππ₯
a)π₯4
4+π₯3
3+ π₯ + π, π β β
b)π3π₯
3+ π, π β β
c) ππ₯2+ π, π β β
d)π₯4+1
9
36+ π, π β β
e) βcos3π₯
3+ π, π β β
f) sinπ₯ βsin3π₯
3+ π, π β β
g)ln (3π₯) 2
2+ π, π β β
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π, π , (π β€ π) π
π π π π₯ ππ₯π
π
π₯ = π π₯ = ππ
π π₯ ππ₯3
1
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β’ π < π π π₯ ππ₯π
π= β π π₯ ππ₯
π
π
β’ π(π₯) β€ 0 π, π , (π β€ π)
π π₯ ππ₯π
π
π₯ = ππ₯ = π π
β π π₯ ππ₯3
1
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π π, π ,
(π < π) πΉπ π, π
πΉπ π₯ = π π¦ ππ¦π₯
π
π π, π πΉπ π = 0
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π π, π ,
(π < π)
π π₯ ππ₯π
π
= πΉ(π₯) ππ = πΉ π β πΉ(π)
πΉ π π, π
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π π, π ,
(π < π) πΉ π π, π
π π₯ ππ₯π
π= πΉπ π
πΉ π πΉ = πΉπ + π
π β β
πΉ π β πΉ π = πΉπ π + π β πΉπ π + π = πΉπ π =
π π₯ ππ₯π
π
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a) 3π₯ππ₯ =3π₯2
2 1
3
=3Γ32
2β3Γ12
2
3
1=24
2= 12
b) (2π₯ + π2π₯+1)ππ₯0
2=
= 2 π₯ππ₯0
2+1
2 2π2π₯+1ππ₯0
2
= 2π₯2
2 2
0
+1
2π2π₯+1 2
0
= 2(0 β 2) +1
2π1 β π5
= β4 +πβπ5
2
0 < 2
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16
3
1
2
5
ππ₯ + πβπ₯ ππ₯ln3
βln3
sin(2π₯) ππ₯
π3
π6
β π₯ + 1 ππ₯2
β2
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π2π¦ππ¦π₯
0
π π₯ = π2π₯ π2π¦ππ¦π₯
0= πΉ0(π₯)
πΉ0β²(π₯)
πΉ0(π₯) π2π¦ππ¦
π₯
0
=π2π¦
20
π₯
=π2π₯
2β1
2
πΉ0β²(π₯) π2π₯
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tan π¦3 ππ¦0
2π₯
π π₯ = tan π₯3 tan π¦3 ππ¦π₯
0= πΉ0(π₯)
πΉ0β² π₯ = tan π₯3
πΉ0 π₯ = tan π¦3 ππ¦
π₯
0
β βπΉ0 π₯ = tan π¦3 ππ¦
0
π₯
β
β βπΉ0 2π₯ = tan π¦3 ππ¦
0
2π₯
tan π¦3 ππ¦0
2π₯
β²
= βπΉ0 2π₯ β² = βπΉ0β² 2π₯ Γ 2 = β2tan 2π₯
3
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a) cos(π¦2) ππ¦π₯2
0
b) ββ 1 β π¦2ππ¦cos π₯
sin π₯
a) 2π₯ cos(π₯4)
b) βsin π₯ | sin π₯ | β cos π₯ | cos π₯ |
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[π΄π΅πΆπ·]
π΄ 0, 2 , π΅ 3, 1 , πΆ 5, 5
π· 0, 4
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π·πΆ π¦ =π₯
5+ 4
π΄π΅ π¦ = βπ₯
3+ 2
π₯
5+ 4 ππ₯
3
0
β βπ₯
3+ 2 ππ₯
3
0
=42
5
π΅πΆ
π¦ = 2π₯ β 5
18
5
12
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π π₯ = β π₯ β 2 + 4
π π₯ = β π₯ β 1 + 2
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Filipe Carvalho