Tổng quát về tích phân

  • View
    228

  • Download
    3

Embed Size (px)

Transcript

  • CHUYN :CC PH NG PHP TNH TCH PHN GV: NGUYN DUY KHI

    Trng THPT Nam H Bin Ha ng Nai Trang 1

    LI NI U Ngy nay php tnh vi tch phn chim mt v tr ht sc quan trng trong Ton hc,

    tch phn c ng dng rng ri nh tnh din tch hnh phng, th tch khi trn xoay,

    n cn l i tng nghin cu ca gii tch, l nn tng cho l thuyt hm, l thuyt

    phng trnh vi phn, phng trnh o hm ring...Ngoi ra php tnh tch phn cn c

    ng dng rng ri trong Xc sut, Thng k, Vt l, C hc, Thin vn hc, y hc...

    Php tnh tch phn c bt u gii thiu cho cc em hc sinh lp 12, tip theo

    c ph bin trong tt c cc trng i hc cho khi sinh vin nm th nht v nm th

    hai trong chng trnh hc i cng. Hn na trong cc k thi Tt nghip THPT v k

    thi Tuyn sinh i hc php tnh tch phn hu nh lun c trong cc thi mn Ton ca

    khi A, khi B v c khi D. Bn cnh , php tnh tch phn cng l mt trong nhng

    ni dung thi tuyn sinh u vo h Thc s v nghin cu sinh.

    Vi tm quan trng ca php tnh tch phn, chnh v th m ti vit mt s kinh

    nghim ging dy tnh tch phn ca khi 12 vi chuyn TNH TCH PHN

    BNG PHNG PHP PHN TCH - I BIN S V TNG PHN

    phn no cng c, nng cao cho cc em hc sinh khi 12 cc em t kt qu cao trong

    k thi Tt nghip THPT v k thi Tuyn sinh i hc v gip cho cc em c nn tng

    trong nhng nm hc i cng ca i hc.

    Trong phn ni dung chuyn di y, ti xin c nu ra mt s bi tp minh

    ha c bn tnh tch phn ch yu p dng phng php phn tch, phng php i bin s,

    phng php tch phn tng phn. Cc bi tp ngh l cc thi Tt nghip THPT v

    thi tuyn sinh i hc Cao ng ca cc nm cc em hc sinh rn luyn k nng tnh tch

    phn v phn cui ca chuyn l mt s cu hi trc nghim tch phn.

    Tuy nhin vi kinh nghim cn hn ch nn d c nhiu c gng nhng khi trnh by

    chuyn ny s khng trnh khi nhng thiu st, rt mong c s gp chn tnh ca

    qu Thy C trong Hi ng b mn Ton S Gio dc v o to tnh ng Nai. Nhn dp

    ny ti xin cm n Ban lnh o nh trng to iu kin tt cho ti v cm n qu thy c

    trong t Ton trng Nam H, cc ng nghip, bn b ng gp kin cho ti hon

    thnh chuyn ny. Ti xin chn thnh cm n./.

  • CHUYN :CC PH NG PHP TNH TCH PHN GV: NGUYN DUY KHI

    Trng THPT Nam H Bin Ha ng Nai Trang 2

    MC LC Li ni u 1

    Mc lc 2

    I. Nguyn hm:

    I.1. nh ngha nguyn hm 3

    I.2. nh l 3

    I.3. Cc tnh cht ca nguyn hm 3

    I.4. Bng cng thc nguyn hm v mt s cng thc b sung 4

    II. Tch phn:

    II.1. nh ngha tch phn xc nh 5

    II.2. Cc tnh cht ca tch phn 5

    II.3 Tnh tch phn bng phng php phn tch 5

    Bi tp ngh 1 9

    II.4 Tnh tch phn bng phng php i bin s 10

    II.4.1 Phng php i bin s loi 1 10

    nh l v phng php i bin s loi 1 13

    Mt s dng khc dng phng php i bin s loi 1 14

    Bi tp ngh s 2 14

    Bi tp ngh s 3 15

    Bi tp ngh s 4: Cc thi tuyn sinh i hc Cao ng 16

    II.4.2 Phng php i bin s loi 2 16

    Bi tp ngh s 5 21

    Cc thi Tt nghip trung hc ph thng 22

    Cc thi tuyn sinh i hc Cao ng 22

    II.5. Phng php tch phn tng phn 23

    Bi tp ngh s 6: Cc thi tuyn sinh i hc Cao ng 28

    III. Kim tra kt qu ca mt bi gii tnh tch phn bng my tnh

    CASIO fx570-MS 29

    Bi tp ngh s 7: Cc cu hi trc nghim tch phn 30

    Ph lc 36

  • CHUYN :CC PH NG PHP TNH TCH PHN GV: NGUYN DUY KHI

    Trng THPT Nam H Bin Ha ng Nai Trang 3

    I. NGUYN HM:

    I.1. NH NGHA NGUYN HM:

    Hm s F(x) c gi l nguyn hm ca hm s f(x) trn (a;b) nu vi mi

    x(a;b):

    F(x) = f(x)

    VD1: a) Hm s F(x) = x3 l nguyn hm ca hm s f(x) = 3x2 trn R

    b) Hm s F(x) = lnx l nguyn hm ca hm s f(x) = 1x

    trn (0;+)

    I.2. NH L:

    Nu F(x) l mt nguyn hm ca hm s f(x) trn (a;b) th:

    a) Vi mi hng s C, F(x) + C cng l mt nguyn hm ca f(x) trn khong .

    b) Ngc li, mi nguyn hm ca hm s f(x) trn khong (a;b) u c th vit

    di dng F(x) + C vi C l mt hng s.

    Theo nh l trn, tm tt c cc nguyn hm ca hm s f(x) th ch cn tm mt

    nguyn hm no ca n ri cng vo n mt hng s C.

    Tp hp cc nguyn hm ca hm s f(x) gi l h nguyn hm ca hm s f(x) v

    c k hiu: f(x)dx (hay cn gi l tch phn bt nh)

    Vy: f(x)dx = F(x)+C

    VD2: a) 22xdx= x +C b) sinxdx= - cosx+C c) 21 dx= tgx +C

    cosx

    I.3. CC TNH CH T CA NGUYN HM:

    1) ( ) f(x)dx f(x)'=

    2) ( ) = a 0a.f(x)dx a f(x)dx

    3) = f(x) g(x) dx f(x)dx g(x)dx

    4) ( ) ( ) =f(x)dx =F(x)+C f u(x) u'(x)dx F u(x) +C

    VD3: a) ( ) 4 2 5 3 2-6x + -2x + 4x5x 8x dx = x +Cb) ( ) 2x6cosx.sinxdx = -6 cosx.d cosx = -3cos +C

  • CHUYN :CC PH NG PHP TNH TCH PHN GV: NGUYN DUY KHI

    Trng THPT Nam H Bin Ha ng Nai Trang 4

    I.4. BNG CNG THC NGUYN HM:

    BNG CC NGUYN HM C BN

    NGUYN HM CC HM S CP THNG GP NGUYN HM CC HM S HP

    ( )

    ( )

    ( )

    +

    +1

    x x

    xx

    22

    22

    dx = x + C

    xx dx = + C ( -1)

    +1dx

    = ln x + C (x 0)x

    e dx = e + C

    aa dx = + C 0 < a 1

    lna

    cosx dx = sinx + C

    sinx dx = -cosx + C

    dx= 1+ tg x dx = tgx + C (x k )

    cos x 2dx

    = 1+ cotg x dxsi

    1/

    2/

    3/

    4/

    5/

    6/

    7/

    8/

    x/

    n9 = -cotgx + C (x k )

    ( )

    ( )

    +

    +1

    u u

    uu

    22

    2

    du = u+C

    uu du = +C ( -1)

    +1du

    = ln u +C (u = u(x) 0)u

    e du = e +C

    aa du = +C 0 < a 1

    lna

    cosu du = sinu+C

    sinu du = - cosu+C

    du= 1+ tg u du = tgu+C (u k

    1/

    2/

    3/

    4/

    5/

    6/

    7/

    8/

    9/

    )cos u 2du

    = 1+csin u

    ( ) 2otg u du = -cotgu+C (u k )

    CC CNG THC B SUNG

    CNG THC NGUYN HM TH NG GP:

    ( ) ( )

    ( )

    ( ) ( )

    ( )

    +1

    ax+b ax+b

    kxkx

    1dx = 2 x + C (x 0)

    x

    ax + b1ax + b dx = + C (a 0)

    a +11 1

    dx = ln ax + b + C (a 0)ax + b a

    1e dx = e + C (a 0)

    aa

    a dx = + C 0 k R, 0 < a 1k.lna

    1cos ax + b dx = sin ax + b

    1/

    2/

    3/

    4/

    5/

    6/

    7

    + C (a 0)a1

    sin ax + b dx = -/ cosa

    ( )

    +

    ax + b + C (a 0)

    tgx dx = - ln cosx + C (x k )2

    cotgx dx = ln sinx + C (9/ x

    /

    k

    8

    )

    CC CNG THC LY THA:

    m n m+n

    mm-n -n

    n n

    1 nnmm m m

    a . a = a

    a 1 = a ;

    1/

    2/

    3/

    = aa a

    a = a ; a = a

    CC CNG THC LNG GIC :

    a. CNG THC H BC:

    ( ) ( ) 2 21/ 21 1sin x = 1-cos2x cos x = 1+cos2x2 2

    /

    b. CNG THC BIN I TCH THNH T NG

    ( ) ( )

    ( ) ( )

    ( ) ( )

    1 cosa.cosb = cos a -b +cos a+b

    21

    sina.sinb = cos a -b - cos a+b21

    sina.cosb = sin a -b +sin a+b2

    1/

    2/

    3/

  • CHUYN :CC PH NG PHP TNH TCH PHN GV: NGUYN DUY KHI

    Trng THPT Nam H Bin Ha ng Nai Trang 5

    II. TCH PHN:

    II.1. NH NGHA TCH PHN XC NH:

    Gi s hm s f(x) lin tc trn mt khong K, a v b l hai phn t bt k ca K,

    F(x) l mt nguyn hm ca hm s f(x) trn K. Hiu F(b) F(a) c gi l tch phn t

    a n b ca f(x). K hiu:

    b

    a

    b

    a=f(x)dx =F(x) F(b)-F(a)

    II.2. CC TNH CH T CA TCH PHN:

    = ( ) 0/ 1a

    a

    f x dx

    = 2/ ( ) ( )a b

    b a

    f x dx f x dx

    = b b

    a a

    k f x dx k f x dx k . ( ) . ( ) (3/ 0)

    = [ ( ) ( )4 ]/ ( ) ( )b b b

    a a a

    f x g x dx f x dx g x dx

    = + b

    a

    f(x) ( ) )5/ (c b

    a c

    dx f x dx f x dx vi c(a;b)

    6/Nu f x x a b( ) 0, [ ; ] th a

    ( ) 0b

    f x dx .

    7/Nu f x g x x a b( ) ( ), [ ; ] th a

    ( ) ( )b b

    a

    f x dx g x dx .

    8/Nu m f x M x a b( ) , [ ; ] th a

    ( ) ( ) ( )b

    m b a f x dx M b a .

    9/ t bin thin trn [ ; ]a b = ( ) ( )t

    a

    G t f x dx l mt nguyn hm ca ( )f t v =( ) 0G a

    II.3. TNH TCH PHN B NG PHNG PHP PHN TCH:

    Ch 1: tnh tch phn = ( )b

    a

    I f x dx ta phn tch = + +1 1( ) ( ) ... ( )m mf x k f x k f x

    Trong : =ik i m0 ( 1,2, 3,..., )cc hm =if x i m( ) ( 1,2,3,..., ) c trong bng nguyn hm c bn.

    VD4: Tnh cc tch phn sau:

  • CHUYN :CC PH NG PHP TNH TCH PHN GV: NGUYN DUY KHI

    Trng THPT Nam H Bin Ha ng Nai Trang 6

    2

    2 3 2

    -1

    3 2 3 2

    2

    -1

    = (3x - 4x+3)dx =(x -2x +3x)

    =(2 -2.2 +3.2)-((-1) -2.(-1) +3.(-1))= 12

    1) I

    Nhn xt:Cu 1 trn ta ch cn p dng tnh cht 4 v s dng cng thc 1/ v 2/ trong bng nguyn hm.

    2 I 2 4 3 2

    2

    1

    3x -6x +4x -2x+4) = dx

    x

    Nhn xt:Cu 2 trn ta cha p dng ngay c cc cng thc trong bng nguyn hm, trc ht tch phn s trong du tch phn (ly t chia mu) ri p dng tnh cht 4 v s dng cng thc 1/, 2/, 3/ trong bng nguyn hm.

    I +

    = =

    2 24 3 2

    2

    2 2

    1 1

    3 22

    1

    3x -6x +4x -2x+4 2 4 = dx = (3x -6x+4- )dx

    x x x

    4(x -3x +4x -2ln |x |- ) 4-2ln2

    x

    3) I 2 2

    0

    x -5x+3= dx

    x+1

    Nhn xt: Cu 3 trn ta cng cha p dng ngay c cc cng thc trong bng nguyn hm, trc ht phn tch phn s trong du tch phn (ly t chia mu) ri p dng tnh cht 4 v s dng cng thc 1/, 2/ trong bng nguyn hm v cng thc 3/ b sung.

    I 6x +

    2 22

    0 0

    22

    0

    x -5x+3 9= dx = dx

    x+1 x+1

    x = -6x+9ln |x+1 | = 2 -12+9ln3 =9ln3 -10

    2

    ( )4) I 1x -x x -x -x

    0

    = e 2xe +5 e -e dx

    Nhn xt:Cu 4: biu thc trong du tch phn c dng tch ta cng cha p dng ngay c cc cng thc trong bng nguyn hm, trc ht nhn phn phi rt gn ri p dng tnh cht 4 v s