· Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu...

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Transcript of  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu...

Page 1:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log
Page 2:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log
Page 3:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log
Page 4:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log
Page 5:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log
Page 6:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log
Page 7:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log
Page 8:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log
Page 9:  · Question 11: xx2-3 + (x — 3) x2 for x > 3. Answer 11: and v = (x — 3)x2 therefore, y = Letu = Differentiating with respect to x, we have dy du dv dx dx dx x2-3 taking log

power integrals - uni-marburg.de · exp integrals Z expx = expx Z ax = ax loga Sub Z dx exp(1/x) x2 part Z dx ...

power integrals - uni-marburg.de · exp integrals Z expx = expx Z ax = ax loga Sub Z dx exp(1/x) x2 part Z dx ...

ln(x2 + 1)dx ∫ cos2 x dxusers.okan.edu.tr/sezgin.sezer/public_html/test2.pdf · 2015-11-24 · 29) 1 - x 9-2/3 A)1 + 2 27 x + 10 729 x2 + 40 19683 x3 B)1 - 2 27 x + 5 729 x2 - 40

ln(x2 + 1)dx ∫ cos2 x dxusers.okan.edu.tr/sezgin.sezer/public_html/test2.pdf · 2015-11-24 · 29) 1 - x 9-2/3 A)1 + 2 27 x + 10 729 x2 + 40 19683 x3 B)1 - 2 27 x + 5 729 x2 - 40

matx.dk · Integralregning Bestem R f(x)dx (polynomium) MATX.DK LYTL˛SLEGL˛R 521 Bestem Z x2 2x 1dx 1=3x3 x2 x + C 522 Bestem Z 4x2 2x 1dx 4=3x3 x2 x + C 523 Bestem Z 4x3 +x 1dx

matx.dk · Integralregning Bestem R f(x)dx (polynomium) MATX.DK LYTL˛SLEGL˛R 521 Bestem Z x2 2x 1dx 1=3x3 x2 x + C 522 Bestem Z 4x2 2x 1dx 4=3x3 x2 x + C 523 Bestem Z 4x3 +x 1dx

AP Calculus AB Indefinite Integrals (U-Sub & a couple ... 2 + ex -2x dx dx x In (3x) sinxcosx 1 + cos2 x xcsc x 12. 15. 18. xtan2 x2 dx sin (3x) 1 + cos (3x) xcsc 3x2 cot 3x2 dx sin

AP Calculus AB Indefinite Integrals (U-Sub & a couple ... 2 + ex -2x dx dx x In (3x) sinxcosx 1 + cos2 x xcsc x 12. 15. 18. xtan2 x2 dx sin (3x) 1 + cos (3x) xcsc 3x2 cot 3x2 dx sin

s3-ap-southeast-1.amazonaws.com · x.e dx dx x Bài 16 Tính tích phân: dx dx 4-3x 0 In 2 dx x2 dx Bài 15: Tính tích phân: tan xdx sin x dx I + cos x Bài 14: Tính tích phân:

s3-ap-southeast-1.amazonaws.com · x.e dx dx x Bài 16 Tính tích phân: dx dx 4-3x 0 In 2 dx x2 dx Bài 15: Tính tích phân: tan xdx sin x dx I + cos x Bài 14: Tính tích phân:

Universidade Federal de Vi˘cosa Centro de Ci^encias … 146/2017-I/listas/Gabarito_Lista3... · s) dy dx = 2e2x t) dy dx = 1 2 x 1 2 e p x u) dy dx = 2x+2 x2+2 v) dy dx = 3sec2(3x)

Universidade Federal de Vi˘cosa Centro de Ci^encias … 146/2017-I/listas/Gabarito_Lista3... · s) dy dx = 2e2x t) dy dx = 1 2 x 1 2 e p x u) dy dx = 2x+2 x2+2 v) dy dx = 3sec2(3x)

Pauta - WordPress.com · Pauta 1.- (a) Use integración directa para calcular! 3 p x x 3e x +1 dx (b) Resuelva usando sustitución simple (2x+1) x2 +x+3 3 2 dx

Pauta - WordPress.com · Pauta 1.- (a) Use integración directa para calcular! 3 p x x 3e x +1 dx (b) Resuelva usando sustitución simple (2x+1) x2 +x+3 3 2 dx

1.- · 2021. 2. 24. · Bloque 4) 3 In x2 +1 + k dx Determina estas integrales de funciones racionales. dx dx=41n x —2 + k dx x dx In x dx = dx In x x + dx x x — arc tg x dx =

1.- · 2021. 2. 24. · Bloque 4) 3 In x2 +1 + k dx Determina estas integrales de funciones racionales. dx dx=41n x —2 + k dx x dx In x dx = dx In x x + dx x x — arc tg x dx =

MÉTODOS DE INTEGRACIÓN, ESTADÍSTICA APLICADA · 11/2dx 1 x2 arctg 2221(x2)/2 2 2 x1 1 3 x 2 Resulta dx L(x 4x 8) arctg C x4x8 2 2 2 Calcular 2 2 x dx x2x1 Numerador y denominador

MÉTODOS DE INTEGRACIÓN, ESTADÍSTICA APLICADA · 11/2dx 1 x2 arctg 2221(x2)/2 2 2 x1 1 3 x 2 Resulta dx L(x 4x 8) arctg C x4x8 2 2 2 Calcular 2 2 x dx x2x1 Numerador y denominador

POLYMER PHYSICS & DNA - physique.cuso.ch€¦ · POLYMER PHYSICS & DNA Bertrand Duplantier Institut de Physique Théorique, Saclay, ... ∂x2 x2 t ∞ ∞ x2P x t dx 2Dt ... Polymer

POLYMER PHYSICS & DNA - physique.cuso.ch€¦ · POLYMER PHYSICS & DNA Bertrand Duplantier Institut de Physique Théorique, Saclay, ... ∂x2 x2 t ∞ ∞ x2P x t dx 2Dt ... Polymer

Answers - Nulake · 36 99 36 13 Page 38 186. dy dx = 2x cos x – x2 sin x 187. dy dx = 3 tan 3x + (9x – 6) sec2 3x 188. g’(x) = dy dx = 2x cot(5x – 1)

Answers - Nulake · 36 99 36 13 Page 38 186. dy dx = 2x cos x – x2 sin x 187. dy dx = 3 tan 3x + (9x – 6) sec2 3x 188. g’(x) = dy dx = 2x cot(5x – 1)

x V=x(1 +m?)[vex1)(x x2)dx= zt+m?)(x2x1)?.

x V=x(1 +m?)[vex1)(x x2)dx= zt+m?)(x2x1)?.

B. C´alculo de primitivas. - ulpgc.es · Tabla de integrales inmediatas ... cosh2 x =tghx+C U√ dx 1−x2 =arcsenx+C dx senh2 x = −cothx+C U √ dx a2 −x2 =arcsen x a +C Aunque

B. C´alculo de primitivas. - ulpgc.es · Tabla de integrales inmediatas ... cosh2 x =tghx+C U√ dx 1−x2 =arcsenx+C dx senh2 x = −cothx+C U √ dx a2 −x2 =arcsen x a +C Aunque

Tiwari Academy · 2020. 1. 14. · dx dx dx dx dx dx dx = 5x4 - 20x3 + 45x2 - 52x + 11 (iii) log ààtR, logy = log(x2 - 5x + 8) + log(x3 + 7x + 9) 45 1 dy y. dx — (x2 1 dy dx x

Tiwari Academy · 2020. 1. 14. · dx dx dx dx dx dx dx = 5x4 - 20x3 + 45x2 - 52x + 11 (iii) log ààtR, logy = log(x2 - 5x + 8) + log(x3 + 7x + 9) 45 1 dy y. dx — (x2 1 dy dx x

abasu9/papers/FOLDE.pdfIn Exercises 5—10, find the indefinite integral. 10. ex(2 + e-2X)dx e2x(xex + 1) dx 2x+5 x2 + 2x 3 x(l dx — x2)2 dx EXERCISES EIEæzis es 1-6, write the

abasu9/papers/FOLDE.pdfIn Exercises 5—10, find the indefinite integral. 10. ex(2 + e-2X)dx e2x(xex + 1) dx 2x+5 x2 + 2x 3 x(l dx — x2)2 dx EXERCISES EIEæzis es 1-6, write the

Ejercicio 2.- - yoquieroaprobar.es · 2020. 5. 10. · Calcula: cos x ax —In Isenxl sen x dx a) a) cotg x dx cotg x d.x 5x dx — 2 5 2 2x clx 1 + (x2)2 arc tg (x2) + k Calcula:

Ejercicio 2.- - yoquieroaprobar.es · 2020. 5. 10. · Calcula: cos x ax —In Isenxl sen x dx a) a) cotg x dx cotg x d.x 5x dx — 2 5 2 2x clx 1 + (x2)2 arc tg (x2) + k Calcula:

12 CÁLCULO DE PRIMITIVAS...1 Página 337 REFLEXIONA Y RESUELVE Concepto de primitiva NÚMEROS Y POTENCIAS SENCILLAS 1 a) ∫1 dx = x b) ∫2 dx = 2x c) ∫ dx = x 2 a) ∫2xdx= x2

12 CÁLCULO DE PRIMITIVAS...1 Página 337 REFLEXIONA Y RESUELVE Concepto de primitiva NÚMEROS Y POTENCIAS SENCILLAS 1 a) ∫1 dx = x b) ∫2 dx = 2x c) ∫ dx = x 2 a) ∫2xdx= x2

power integrals - uni-marburg.de · x3 +x2 +x+1 x2 = 1 2 x2 +x− 1 x +logx Z dx ... rational integrals substitution Z dx x x2 −1 3 = 1 8 x2 −1 4 Z dx x2 ... dx 1 x2 +x+1 = 2

power integrals - uni-marburg.de · x3 +x2 +x+1 x2 = 1 2 x2 +x− 1 x +logx Z dx ... rational integrals substitution Z dx x x2 −1 3 = 1 8 x2 −1 4 Z dx x2 ... dx 1 x2 +x+1 = 2

COMPUTER BASED TEST (CBT) Memory Based Questions & … · x2= 4b(y + b), tgk¡ b ,d izkpy gS] dk vody lehdj.k gksxk % (1) 2 2 x dx dy 2 dx dy x (2) x dx dy 2y dx dy x 2 (3) 2 2 x

COMPUTER BASED TEST (CBT) Memory Based Questions & … · x2= 4b(y + b), tgk¡ b ,d izkpy gS] dk vody lehdj.k gksxk % (1) 2 2 x dx dy 2 dx dy x (2) x dx dy 2y dx dy x 2 (3) 2 2 x

e dA e dxdyjessada/CALIII/exam/f103105-1-50.pdf · 2009-07-30 · (1) (x2 2xy)dx x2dy 0 (2) (32 2) (3 2) 0 2 x xdy xy ydx (3) (x2 2y)dx 2x dy 0 (4) xdy (x y)dx 0 (5) xdy (y 2xex)dx

e dA e dxdyjessada/CALIII/exam/f103105-1-50.pdf · 2009-07-30 · (1) (x2 2xy)dx x2dy 0 (2) (32 2) (3 2) 0 2 x xdy xy ydx (3) (x2 2y)dx 2x dy 0 (4) xdy (x y)dx 0 (5) xdy (y 2xex)dx