• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
formulas calculo integral y diferencial
 

formulas calculo integral y diferencial

on

  • 750 views

aqui les dejo un formulario que en lo personal me ha servido mucho..

aqui les dejo un formulario que en lo personal me ha servido mucho..

Statistics

Views

Total Views
750
Views on SlideShare
750
Embed Views
0

Actions

Likes
0
Downloads
51
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    formulas calculo integral y diferencial formulas calculo integral y diferencial Presentation Transcript

    • Formulario de Cálculo Diferencial e Integral Jesús Rubí M.Formulario de ( a + b ) ⋅ ( a 2 − ab + b2 ) = a3 + b3 θ sen cos tg ctg sec csc Gráfica 4. Las funciones trigonométricas inversas arcctg x , arcsec x , arccsc x : sen α + sen β 1 1 = 2sen (α + β ) ⋅ cos (α − β ) ( a + b ) ⋅ ( a3 − a 2 b + ab2 − b3 ) = a 4 − b4 0 0 ∞ ∞ 0 2 2Cálculo Diferencial 1 1 1 1 30 12 3 2 3 2 sen α − sen β = 2 sen (α − β ) ⋅ cos (α + β ) 4 3 2 1 3 ( a + b ) ⋅ ( a 4 − a3b + a 2 b2 − ab3 + b4 ) = a5 + b5e Integral VER.4.3 45 1 2 1 2 1 1 2 2 3 2 2 ( a + b ) ⋅ ( a5 − a 4 b + a3b2 − a 2 b3 + ab4 − b5 ) = a 6 − b6 1 1 60 3 2 12 3 1 3 2 2 3 cos α + cos β = 2 cos (α + β ) ⋅ cos (α − β )Jesús Rubí Miranda (jesusrubim@yahoo.com) 2 2 2 90 1 0 ∞ 0 ∞ 1http://mx.geocities.com/estadisticapapers/ ⎛ n ⎞ cos α − cos β 1 1 = −2sen (α + β ) ⋅ sen (α − β ) ( a + b ) ⋅ ⎜ ∑ ( −1) a n− k b k −1 ⎟ = a n + b n ∀ n ∈ k +1 impar ⎡ π π⎤ 1http://mx.geocities.com/dicalculus/ ⎝ k =1 ⎠ y = ∠ sen x y ∈ ⎢− , ⎥ 2 2 ⎣ 2 2⎦ sen (α ± β ) 0 VALOR ABSOLUTO ⎛ n ⎞ ( a + b ) ⋅ ⎜ ∑ ( −1) y ∈ [ 0, π ] k +1 a n − k b k −1 ⎟ = a n − b n ∀ n ∈ par y = ∠ cos x tg α ± tg β = ⎧a si a ≥ 0 ⎝ k =1 ⎠ -1 cos α ⋅ cos β a =⎨ arc ctg x π π arc sec x ⎩− a si a < 0 y = ∠ tg x y∈ − arc csc x SUMAS Y PRODUCTOS , 1 ⎡sen (α − β ) + sen (α + β ) ⎤ -2 2 2 -5 0 5 sen α ⋅ cos β = a = −a n 2⎣ ⎦ a1 + a2 + + an = ∑ ak 1 IDENTIDADES TRIGONOMÉTRICASa ≤ a y −a≤ a y = ∠ ctg x = ∠ tg y ∈ 0, π 1 k =1 x sen α ⋅ sen β = ⎡cos (α − β ) − cos (α + β ) ⎤ sen θ + cos2 θ = 1 2⎣ ⎦ 2 n a ≥0y a =0 ⇔ a=0 ∑ c = nc y = ∠ sec x = ∠ cos 1 y ∈ [ 0, π ] 1 + ctg 2 θ = csc2 θ 1 k =1 cos α ⋅ cos β = ⎡cos (α − β ) + cos (α + β ) ⎤ 2⎣ ⎦ n n x ab = a b ó ∏a = ∏ ak n n ⎡ π π⎤ tg 2 θ + 1 = sec2 θ ∑ ca = c ∑ ak 1 k k =1 k =1 k y = ∠ csc x = ∠ sen y ∈ ⎢− , ⎥ tg α + tg β n n k =1 k =1 x ⎣ 2 2⎦ sen ( −θ ) = − sen θ tg α ⋅ tg β = a+b ≤ a + b ó ∑a ≤ ∑ ak n n n ctg α + ctg β k =1 k k =1 ∑(a k =1 k + bk ) = ∑ ak + ∑ bk k =1 k =1 Gráfica 1. Las funciones trigonométricas: sen x , cos ( −θ ) = cosθ FUNCIONES HIPERBÓLICAS cos x , tg x : EXPONENTES n tg ( −θ ) = − tg θ e x − e− xa p ⋅ a q = a p+q ∑(a − ak −1 ) = an − a0 senh x = sen (θ + 2π ) = sen θ k 2 k =1 2 ap 1.5 e x + e− x = a p−q n n cos (θ + 2π ) = cosθ cosh x = aq ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤ ⎣ ⎦ ⎣ ⎦ 1 2 k =1 tg (θ + 2π ) = tg θ( a p ) = a pq senh x e x − e − x q 0.5 n tgh x = = (a + l ) = 0 sen (θ + π ) = − sen θ cosh x e x + e − x(a ⋅ b) = ap ⋅bp p 2 -0.5 cos (θ + π ) = − cosθ 1 e x + e− x n 1 − r n a − rl ctgh x = =⎛a⎞ ap p ∑ ar k −1 =a = -1 tg (θ + π ) = tg θ tgh x e x − e − x⎜ ⎟ = p k =1 1− r 1− r⎝b⎠ sen x b -1.5 1 2 sen (θ + nπ ) = ( −1) sen θ sech x = = cos x n ( n + n) n 1 2 ∑k = tg xa p/q = a q p -2 -8 -6 -4 -2 0 2 4 6 8 cosh x e x + e − x 2 k =1 cos (θ + nπ ) = ( −1) cos θ n 1 2 LOGARITMOS csch x = = ∑ k 2 = 6 ( 2n3 + 3n2 + n ) n 1 Gráfica 2. Las funciones trigonométricas csc x ,log a N = x ⇒ a x = N tg (θ + nπ ) = tg θ senh x e x − e − x k =1 sec x , ctg x :log a MN = log a M + log a N senh : → sen ( nπ ) = 0 ∑ k 3 = 4 ( n 4 + 2n3 + n 2 ) n 1 M 2.5 cosh : → [1, ∞ = log a M − log a N cos ( nπ ) = ( −1) nlog a k =1 2 N tgh : → −1,1 ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n ) n 1 tg ( nπ ) = 0 1.5log a N r = r log a N 1 ctgh : − {0} → −∞ , −1 ∪ 1, ∞ k =1 ⎛ 2n + 1 ⎞ + ( 2n − 1) = n 2 π ⎟ = ( −1) → 0,1] 0.5 n log b N ln N 1+ 3 + 5 + sen ⎜log a N = = sech : ⎝ 2 ⎠ 0 log b a ln a n − {0} → − {0} n! = ∏ k -0.5 csch :log10 N = log N y log e N = ln N ⎛ 2n + 1 ⎞ k =1 -1 cos ⎜ π⎟=0 ALGUNOS PRODUCTOS -1.5 ⎝ 2 ⎠ Gráfica 5. Las funciones hiperbólicas senh x , ⎛n⎞ n! csc xa ⋅ ( c + d ) = ac + ad ⎜ ⎟= , k≤n -2 sec x ⎛ 2n + 1 ⎞ cosh x , tgh x : ⎝ k ⎠ ( n − k )!k ! ctg x tg ⎜ π⎟=∞ ⎝ 2 ⎠ -2.5( a + b) ⋅ ( a − b) = a − b -8 -6 -4 -2 0 2 4 6 8 2 2 5 n ⎛n⎞ ( x + y ) = ∑ ⎜ ⎟ xn−k y k π⎞ n Gráfica 3. Las funciones trigonométricas inversas ⎛( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b2 4 sen θ = cos ⎜θ − ⎟ 2 k =0 ⎝ k ⎠ arcsen x , arccos x , arctg x : ⎝ 2⎠ 3( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2 2 2 ( x1 + x2 + + xk ) n =∑ n! x1n1 ⋅ x2 2 ⎛ π⎞ cosθ = sen ⎜θ + ⎟ n nk x 4( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd 1 k n1 !n2 ! nk ! 3 ⎝ 2⎠ 0( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd CONSTANTES sen (α ± β ) = sen α cos β ± cos α sen β -1( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd π = 3.14159265359… 2 cos (α ± β ) = cos α cos β ∓ sen α sen β -2 se nh x e = 2.71828182846… -3 co sh x( a + b ) = a3 + 3a 2b + 3ab2 + b3 tg α ± tg β 3 1 tgh x tg (α ± β ) = -4 -5 0 5 TRIGONOMETRÍA 1 ∓ tg α tg β( a − b ) = a3 − 3a 2b + 3ab2 − b3 3 0 CO 1 FUNCIONES HIPERBÓLICAS INV sen θ = cscθ = sen 2θ = 2sen θ cosθ( a + b + c ) = a 2 + b2 + c 2 + 2ab + 2ac + 2bc 2 HIP CA sen θ 1 -1 arc sen x arc cos x arc tg x cos 2θ = cos 2 θ − sen 2 θ ( senh −1 x = ln x + x 2 + 1 , ∀x ∈ ) cosθ = secθ =( a − b ) ⋅ ( a + ab + b ) = a − b ( ) -2 2 tg θ 2 2 3 3 cosθ -3 -2 -1 0 1 2 3 HIP tg 2θ = cosh −1 x = ln x ± x 2 − 1 , x ≥ 1 sen θ CO 1 − tg 2 θ( a − b ) ⋅ ( a3 + a 2 b + ab2 + b3 ) = a 4 − b4 tg θ = = ctg θ = 1 1 ⎛1+ x ⎞ cosθ CA tg θ 1 tgh −1 x = ln ⎜ ⎟, x < 1( a − b ) ⋅ ( a 4 + a3b + a 2 b2 + ab3 + b 4 ) = a5 − b5 sen 2 θ = (1 − cos 2θ ) 2 ⎝ 1− x ⎠ 2 ⎛ ⎞ π radianes=180 1 ⎛ x +1 ⎞ ctgh −1 x = ln ⎜ n 1( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n ∀n ∈ cos 2 θ = (1 + cos 2θ ) ⎟, x > 1 2 ⎝ x −1 ⎠ ⎝ k =1 ⎠ 2 1 − cos 2θ ⎛ 1 ± 1 − x2 ⎞ HIP tg 2 θ = sech −1 x = ln ⎜ ⎟, 0 < x ≤ 1 CO 1 + cos 2θ ⎜ x ⎟ ⎝ ⎠ θ ⎛1 x2 + 1 ⎞ csch −1 x = ln ⎜ + ⎟, x ≠ 0 CA ⎜x x ⎟ ⎝ ⎠
    • Formulario de Cálculo Diferencial e Integral Jesús Rubí M. ∫ tgh udu = ln cosh u IDENTIDADES DE FUNCS HIP d dv du DERIVADA DE FUNCS HIPERBÓLICAS INTEGRALES DE FUNCS LOG & EXP ( uv ) = u + vcosh 2 x − senh 2 x = 1 d du ∫ e du = e u u dx dx dx senh u = cosh u1 − tgh 2 x = sech 2 x d dw dv du dx dx ∫ ctgh udu = ln senh u ( uvw) = uv + uw + vw a u ⎧a > 0 ∫ a du = ln a ⎨a ≠ 1 ∫ sech udu = ∠ tg ( senh u ) d du u dx dx dx dx cosh u = senh uctgh x − 1 = csch x 2 ⎩ d ⎛ u ⎞ v ( du dx ) − u ( dv dx ) dx dxsenh ( − x ) = − senh x ∫ csch udu = − ctgh ( cosh u ) −1 ⎜ ⎟= d du au ⎛ 1 ⎞ dx ⎝ v ⎠ v2 tgh u = sech 2 u ∫ ua du = ln a ⋅ ⎜ u − ln a ⎟ ucosh ( − x ) = cosh x dx dx ⎝ ⎠ 1 = ln tgh u d n ( u ) = nu n−1 dutgh ( − x ) = − tgh x d du ∫ ue du = e ( u − 1) 2 u u dx dx ctgh u = − csch 2 u dx dx INTEGRALES DE FRACsenh ( x ± y ) = senh x cosh y ± cosh x senh y dF dF du ∫ ln udu =u ln u − u = u ( ln u − 1) = ⋅ (Regla de la Cadena) d sech u = − sech u tgh u du du 1 ucosh ( x ± y ) = cosh x cosh y ± senh x senh y dx du dx dx dx 1 u ∫ u 2 + a 2 = a ∠ tg a du 1 ∫ log ( u ln u − u ) = ( ln u − 1) udu = tgh x ± tgh y = a d du ln a ln a 1 utgh ( x ± y ) = dx dx du csch u = − csch u ctgh u = − ∠ ctg 1 ± tgh x tgh y dx dx u 2 ∫ u log a udu = 4 ⋅ ( 2log a u − 1) a a dF dF du DERIVADA DE FUNCS HIP INV 1 u−asenh 2 x = 2senh x cosh x ∫ u 2 − a 2 2a u + a ( u > a ) = du = ln 2 2 dx dx du d 1 du u2cosh 2 x = cosh x + senh x senh −1 u = ⋅ ∫ u ln udu = 4 ( 2ln u − 1) 2 2 dy dy dt f 2′ ( t ) ⎪ x = f1 ( t ) ⎧ 1 + u 2 dx 1 a+u ∫ a 2 − u 2 = 2a ln a − u ( u < a ) dx du 2 2tgh 2 x = 2 tgh x = = donde ⎨ 1 + tgh 2 x dx dx dt f1′( t ) ⎪ y = f2 ( t ) ⎩ d ±1 du ⎧+ si cosh -1u > 0 ⎪ INTEGRALES DE FUNCS TRIGO cosh −1 u = ⋅ , u >1 ⎨ 1 DERIVADA DE FUNCS LOG & EXP dx u 2 − 1 dx ⎪− si cosh u < 0 ⎩ -1 ∫ sen udu = − cos u INTEGRALES CONsenh 2 x = ( cosh 2 x − 1) 2 d ( ln u ) = du dx 1 du = ⋅ d tgh −1 u = 1 du ⋅ , u <1 ∫ cos udu = sen u ∫ du = ∠ sen u 1 dx u u dx dx 1 − u 2 dx a2 − u2 acosh x = ( cosh 2 x + 1) 2 d log e du d 1 du ∫ sec udu = tg u 2 2 ( log u ) = ⋅ ctgh −1 u = ⋅ , u >1 = −∠ cos u cosh 2 x − 1 1 − u 2 dx ∫ csc udu = − ctg u 2 dx u dx dx atgh 2 x = ⎧− ( ) −1 cosh 2 x + 1 d log e du ( log a u ) = a ⋅ a > 0, a ≠ 1 ∓1 du ⎪ si sech u > 0, u ∈ 0,1 du ∫ sec u tg udu = sec u d senh 2 x dx u dx dx sech −1 u = ⋅ ⎨ −1 u 1 − u 2 dx ⎪ + si sech u < 0, u ∈ 0,1 ∫ u 2 ± a2 = ln u + u 2 ± a 2tgh x = ⎩ cosh 2 x + 1 d u ( e ) = eu ⋅ du d 1 du ∫ csc u ctg udu = − csc u du 1 u dx dx −1 csch u = − ⋅ , u≠0 ∫u = lne x = cosh x + senh x dx u 1 + u 2 dx ∫ tg udu = − ln cos u = ln sec u a2 ± u2 a a + a2 ± u 2e − x = cosh x − senh x d u ( a ) = au ln a ⋅ du du 1 a dx dx INTEGRALES DEFINIDAS, PROPIEDADES ∫ ctg udu = ln sen u ∫ u u 2 − a 2 = a ∠ cos u OTRAS d v ( u ) = vu v−1 du + ln u ⋅ u v ⋅ dv ∫ { f ( x ) ± g ( x )} dx = ∫ f ( x ) dx ± ∫ g ( x ) dx b b bax + bx + c = 0 2 dx dx dx a a a ∫ sec udu = ln sec u + tg u 1 = ∠ sec u ∫ cf ( x ) dx = c ⋅ ∫ f ( x ) dx c ∈ ∫ csc udu = ln csc u − ctg u b b DERIVADA DE FUNCIONES TRIGO a a −b ± b 2 − 4ac ⇒ x= d du a a u 2 a2 u ( sen u ) = cos u ∫ a − u du = 2 a − u + 2 ∠ sen a ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx b c b 2 2 2 2a u 1 ∫ sen udu = − sen 2u 2 dx dx b 2 − 4ac = discriminante a a c 2 4 ( ) 2 d du ( cos u ) = − sen u ∫ f ( x ) dx = −∫ f ( x ) dx b a u 2 aexp (α ± iβ ) = eα ( cos β ± i sen β ) si α , β ∈ ∫ u ± a du = 2 u ± a ± 2 ln u + u ± a 2 2 2 2 2 u 1 ∫ cos udu = 2 + 4 sen 2u 2 dx dx a b ∫ f ( x ) dx = 0 a LÍMITES d du ( tg u ) = sec2 u MÁS INTEGRALES ∫ tg udu = tg u − u a 2 e au ( a sen bu − b cos bu ) 1lim (1 + x ) x = e = 2.71828... dx dx m ⋅ ( b − a ) ≤ ∫ f ( x ) dx ≤ M ⋅ ( b − a ) b ∫e sen bu du = au ∫ ctg udu = − ( ctg u + u )x →0 d du ( ctg u ) = − csc2 u a 2 + b2 2 a x ⎛ 1⎞lim ⎜1 + ⎟ = e dx dx ⇔ m ≤ f ( x ) ≤ M ∀x ∈ [ a, b ] , m, M ∈ e au ( a cos bu + b sen bu ) ⎝ x⎠ ∫ u sen udu = sen u − u cos u ∫ e cos bu du =x →∞ au d du ( sec u ) = sec u tg u ∫ f ( x ) dx ≤ ∫ g ( x ) dx b b a 2 + b2 sen x dx dx =1 ∫ u cos udu = cos u + u sen u a alim ALGUNAS SERIESx →0 d du ⇔ f ( x ) ≤ g ( x ) ∀x ∈ [ a, b ] x ( csc u ) = − csc u ctg u INTEGRALES DE FUNCS TRIGO INV f ( x0 )( x − x0 ) 2 1 − cos x dx dx f ( x ) = f ( x0 ) + f ( x0 )( x − x0 ) + =0 ∫ f ( x ) dx ≤ ∫ f ( x ) dx si a < b b blim ∫ ∠ sen udu = u∠ sen u + 1 − u 2! 2x →0 d du x ( vers u ) = sen u a a f( n) ( x0 )( x − x0 ) n ex −1 dx dx INTEGRALES =1 ∫ ∠ cos udu = u∠ cos u − 1 − u + + 2lim : Taylor ∫ adx =axx →0 x DERIV DE FUNCS TRIGO INVER n! x −1 ∫ ∠ tg udu = u∠ tg u − ln 1 + u f ( 0 ) x 2 2 d 1 dulim x →1 ln x =1 ( ∠ sen u ) = ⋅ ∫ af ( x ) dx = a ∫ f ( x ) dx f ( x ) = f ( 0) + f ( 0) x + dx 1 − u 2 dx 2! ∫ ∠ ctg udu = u∠ ctg u + ln 1 + u 2 ∫ ( u ± v ± w ± ) dx = ∫ udx ± ∫ vdx ± ∫ wdx ± ( 0) x DERIVADAS ( n) n d 1 du f ∫ ∠ sec udu = u∠ sec u − ln ( u + u ) df f ( x + ∆x ) − f ( x ) ∆y ( ∠ cos u ) = − ⋅ 2 −1 + + : MaclaurinDx f ( x ) = dx 1 − u 2 dx ∫ udv = uv − ∫ vdu ( Integración por partes ) = lim = lim n! dx ∆x→0 ∆x ∆x → 0 ∆x d 1 du = u∠ sec u − ∠ cosh u x 2 x3 xn ( ∠ tg u ) = ⋅ u n +1 ex = 1 + x + + + + + ∫ ∠ csc udu = u∠ csc u + ln ( u + )d (c) = 0 dx 1 + u 2 dx ∫ u du = n + 1 n n ≠ −1 u 2 −1 2! 3! n!dx d 1 du x3 x 5 x 7 x 2 n −1 ( ∠ ctg u ) = − 2 ⋅ sen x = x − + − + + ( −1) n −1d ( cx ) = c dx 1 + u dx du ∫ u = ln u = u∠ csc u + ∠ cosh u 3! 5! 7! ( 2n − 1)!dx d 1 du ⎧ + si u > 1 INTEGRALES DE FUNCS HIP x 2n−2 ( ∠ sec u ) = ± 2 ⋅ ⎨ cos x = 1 − x2 x4 x6 + − + + ( −1) n −1ddx ( cx n ) = ncxn−1 dx u u − 1 dx ⎩ − si u < −1 ∫ senh udu = cosh u 2! 4! 6! ( 2n − 2 )! d 1 du ⎧− si u > 1 ∫ cosh udu = senh u ( ∠ csc u ) = ∓ x 2 x3 x 4 nd du dv dw n −1 x (u ± v ± w ± ) = ± ± ± ⋅ ⎨ ln (1 + x ) = x −+ − + + ( −1) dx u u 2 − 1 dx ⎩+ si u < −1 2 3 4 n ∫ sech udu = tgh udx dx dx dx 2 2 n −1d du d 1 du x3 x5 x7 ( cu ) = c ( ∠ vers u ) = ⋅ ∠ tg x = x − + − + + ( −1) n −1 x ∫ csch udu = − ctgh u 2dx dx dx 2u − u 2 dx 3 5 7 2n − 1 ∫ sech u tgh udu = − sech u ∫ csch u ctgh udu = − csch u
    • Formulario de Cálculo Diferencial e Integral Jesús Rubí M. ÁLGEBRA LINEALDef. El determinante de una matriz ⎡a a ⎤ A = ⎢ 11 12 ⎥ ⎣ a21 a22 ⎦está dado por a11 a12 det A = = a11a22 − a12 a21 . a21 a22Def. El determinante de una matriz ⎡ a11 a12 a13 ⎤ ⎢ ⎥ A = ⎢ a21 a22 a23 ⎥ ⎢ ⎣ a31 a32 ⎥ a33 ⎦está dado por a11 a12 a13 a11 ⋅ a22 ⋅ a33 + a12 ⋅ a23 ⋅ a31det A = a21 a22 a23 = + a13 ⋅ a21 ⋅ a32 − a11 ⋅ a23 ⋅ a32 . a31 a32 a33 −a12 ⋅ a21 ⋅ a33 − a13 ⋅ a22 ⋅ a31