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Page 1: SplitPDF - AUT · 2021. 8. 15. · ,Ansys 10.0 Jl 4-5 Pennes L La e) (Cryosurgery) 44 0.5 k OJ-.a 4.4.' J (MRI CT) Jl y; 0) = To(x) d2T0(x) dx 2 dT0(x) = dx 2 ô(ph) ôt 6) Cb ) dJJ(-a-a

Page 2: SplitPDF - AUT · 2021. 8. 15. · ,Ansys 10.0 Jl 4-5 Pennes L La e) (Cryosurgery) 44 0.5 k OJ-.a 4.4.' J (MRI CT) Jl y; 0) = To(x) d2T0(x) dx 2 dT0(x) = dx 2 ô(ph) ôt 6) Cb ) dJJ(-a-a

Page 3: SplitPDF - AUT · 2021. 8. 15. · ,Ansys 10.0 Jl 4-5 Pennes L La e) (Cryosurgery) 44 0.5 k OJ-.a 4.4.' J (MRI CT) Jl y; 0) = To(x) d2T0(x) dx 2 dT0(x) = dx 2 ô(ph) ôt 6) Cb ) dJJ(-a-a

Page 4: SplitPDF - AUT · 2021. 8. 15. · ,Ansys 10.0 Jl 4-5 Pennes L La e) (Cryosurgery) 44 0.5 k OJ-.a 4.4.' J (MRI CT) Jl y; 0) = To(x) d2T0(x) dx 2 dT0(x) = dx 2 ô(ph) ôt 6) Cb ) dJJ(-a-a

Page 5: SplitPDF - AUT · 2021. 8. 15. · ,Ansys 10.0 Jl 4-5 Pennes L La e) (Cryosurgery) 44 0.5 k OJ-.a 4.4.' J (MRI CT) Jl y; 0) = To(x) d2T0(x) dx 2 dT0(x) = dx 2 ô(ph) ôt 6) Cb ) dJJ(-a-a

Page 6: SplitPDF - AUT · 2021. 8. 15. · ,Ansys 10.0 Jl 4-5 Pennes L La e) (Cryosurgery) 44 0.5 k OJ-.a 4.4.' J (MRI CT) Jl y; 0) = To(x) d2T0(x) dx 2 dT0(x) = dx 2 ô(ph) ôt 6) Cb ) dJJ(-a-a

Page 7: SplitPDF - AUT · 2021. 8. 15. · ,Ansys 10.0 Jl 4-5 Pennes L La e) (Cryosurgery) 44 0.5 k OJ-.a 4.4.' J (MRI CT) Jl y; 0) = To(x) d2T0(x) dx 2 dT0(x) = dx 2 ô(ph) ôt 6) Cb ) dJJ(-a-a

Tema 5...5.4.2.- Integración por simple inspección Ejemplo 3: Calcular (ln )x2 dx x Utilizando la fórmula anterior: 2 3 (ln ) 1 1(ln ) (ln )2 3 x dx x dx x K x x Ejemplo 4: Calcular

DX)|6«5/ JÔ V p·5d 6«[ Y¶Z~p¸ h B DX(+X DÇO...DX)|6«5/ JÔ V p·5d 6«[ Y Z~p¸ h B DX(+X DÇO ë6¿"èDX(+X DÇ_N 3Uý DX(+X DÇ 3" _Ú+ý'Ù#ò p· . 6 5 Q . 6 5 p¸ ë6¿"èeÚ

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RELACIÓN DE EJERCICIOS DE INTEGRALESINTEGRALES INDEFINIDAS RESUELTAS 4 18. dx x x ³ s 2 tg Sol: k x 2 tg2, k x dx x dx x x x f f ³ ³ 2 tg s 1 tg s tg 2 2 1 19. dx x x ³ n 2 g

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1 INTEGRALI UTILI lunedì 20 aprile 2020 · 2020. 4. 20. · dx =. u=sin2(x), dv =1⁄x2. − sin2(x) x |0 ∞+ . 0 ∞ 2sin(x)cos(x) x dx = = . 0 ∞ sin(2x) x dx =. u=2x. 0 ∞

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1. Reglas de integración √3x 3 dx √3x 1 2 1 2 x dx x2 + 9 2–4x 4 L 2 dx x – 1 3x 2 2 3 3x 1 4(x – 1)4 dx (x – 1)5 (4x – 1)6 6 112. ∫ Solución: arc sen x + k 113

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