Tich phan (Ly thuyet va bai tap)

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Tran S TungTch phan Trang 1 Nhac lai Gii han ao ham Vi phan 1.Cac gii han ac biet: a)=x 0sinxlim 1x He qua: =x 0xlim 1sinx

=u(x) 0sinu(x)lim 1u(x) =u(x) 0u(x)lim 1sinu(x) b) xx1lim 1 e, x Rx + = He qua:1xx 0lim(1 x) e.+ =x 0ln(1 x)lim 1x+=xx 0e 1lim 1x-=2.Bang ao ham cac ham so s cap c ban va cac he qua:(c) = 0 (c la hang so) 1(x )' xa a-= a1(u )' u u'a a-= a21 1'x x = - 21 u''u u = - ( )1x '2 x=( )u'u '2 u=x x(e )' e =u u(e )' u' .e =x x(a )' a . lna =u u(a )' a . lna . u' =1(ln x )'x=u'(ln u )'u=a1(log x ')x. lna=au'(log u )'u. lna=(sinx) = cosx(sinu) = u.cosu 221(tgx)' 1 tg xcos x= = +22u'(tgu)' (1 tg u).u'cos u= = +221(cot gx)' (1 cot g x)sin x-= = - +22u'(cot gu)' (1 cot g u).u'sin u-= = - +3.Vi phan:Cho ham so y = f(x) xac nh tren khoang (a ; b) va co ao ham taix (a; b) . Cho so gia Dx tai x sao chox x (a; b) + D . Ta goi tch y.Dx (hoac f(x).Dx) la vi phan cua ham so y = f(x) tai x, ky hieu la dy (hoac df(x)). dy = y.Dx(hoac df(x) = f(x).Dx Ap dung nh ngha tren vao ham so y = x, thdx = (x)Dx = 1.Dx = Dx V vay ta co:dy = ydx (hoac df(x) = f(x)dx) Tch phanTran S TungTrang 2 NGUYEN HAM VA TCH PHAN 1.nh ngha:Ham so F(x) c goi la nguyen ham cua ham so f(x) tren khoang (a ; b) neu moi x thuoc (a ; b), ta co: F(x) = f(x). Neu thay cho khoang (a ; b) la oan [a ; b] th phai co them: F'(a ) f(x) va F'(b ) f(b)+ -= = 2.nh ly:Neu F(x) la mot nguyen ham cua ham so f(x) tren khoang (a ; b) th : a/Vi moi hang so C, F(x) + C cung la mot nguyen ham cua ham so f(x) tren khoang o. b/Ngc lai, moi nguyen ham cua ham so f(x) tren khoang (a ; b) eu co the viet di dang: F(x) + C vi C la mot hang so. Ngi ta ky hieu ho tat ca cac nguyen ham cua ham so f(x) laf(x)dx. Do o viet:f(x)dx F(x) C = + Bo e: Neu F(x) = 0 tren khoang (a ; b) th F(x) khong oi tren khoang o. 3.Cac tnh chat cua nguyen ham: ( )f(x)dx ' f(x) = af(x)dx a f(x)dx (a 0) = [ ] f(x) g(x) dx f(x)dx g(x)dx + = + [ ] [ ] f(t)dt F(t) C f u(x) u'(x)dx F u(x) C F(u) C (u u(x)) = + = + = + = 4.S ton tai nguyen ham:nh ly: Moi ham so f(x) lien tuc tren oan [a ; b] eu co nguyen ham tren oan o. 8o| 1. NGUYEN HAM Tran S TungTch phan Trang 3 BANG CAC NGUYEN HAM Nguyenhamcuaca chamsoscap thng gap Nguyenhamcuacachamsohp (di ay u = u(x)) dx x C = +du u C = + 1xx dx C ( 1)1a+a= + a -a + 1uu du C ( 1)1a+a= + a -a + dxln x C (x 0)x= + duln u C (u u(x) 0)u= + = x xe dx e C = + u ue du e C = + xxaa dx C (0 a 1)lna= + < uuaa du C (0 a 1)lna= + < cosxdx sinx C = +cos udu sin u C = + sinxdx cosx C = - +sin udu cos u C = - + 22dx(1 tg x)dx tgx Ccos x = + = + 22du(1 tg u)du tgu Ccos u = + = + 22dx(1 cot g x)dx cot gx Csin x = + = - + 22du(1 cot g u)du cot gu Csin u = + = - + dxx C (x 0)2 x= + > duu C (u 0)2 u = + > 1cos(ax b)dx sin(ax b) C (a 0)a+ = + + 1sin(ax b)dx cos(ax b) C (a 0)a+ = - + + dx 1ln ax b Cax b a= + ++ ax b ax b1e dx e C (a 0)a+ += + dx 2ax b C (a 0)a ax b= + + + Tch phanTran S TungTrang 4 Von de 1:Ik0 0I0h 080IE0 hkN 8k08 0I0h 08hk Bai toan 1: CMR F(x) la mot nguyen ham cua ham so f(x) tren (a ; b) PHNG PHAP CHUNG Ta thc hien theo cac bc sau:+Bc 1: Xac nh F(x) tren (a ; b) +Bc 2: Chng to rangF'(x) f(x) vi x (a; b) = " Chu y: Neu thay (a ; b) bang [a ; b]th phai thc hien chi tiet hn, nh sau:+Bc 1: Xac nh F(x) tren (a ; b) Xac nh F(a+) Xac nh F(b) +Bc 2: Chng to rang F'(x) f(x), x (a ; b)F'(a ) f(a)F'(b ) f(b)+-= " == V du 1: CMR ham so:2F(x) ln(x x a) = + +vi a > 0la mot nguyen ham cua ham so 21f(x)x a=+ tren R. Giai: Ta co: 2222 22x1(x x a)'2 x aF'(x) [ln(x x a)]'x x a x x a++ ++= + + = =+ + + + 22 2 2x a x 1f(x)x a(x x a) x a+ += = =+ + + + Vay F(x) vi a > 0 la mot nguyen ham cua ham so f(x) tren R. V du 2: CMR ham so:x2e khi x 0F(x)x x 1 khi x 0 = + + la mot nguyen ham cua ham so: 2x khi x 1f(x)2 khi x 1 = > tren R. Giai: e tnh ao ham cua ham so F(x) ta i xet hai trng hp:a/Vix 1 , ta co:2x khi x 1F'(x)2 khi x 1< = > b/Vi x = 1, ta co: e ham so F(x) co ao ham tai iem x = 1, trc het F(x) phai lien tuc tai x = 1, do o :x 1 x 1limF(x) limF(x) f(1) a b 1 b 1 a (1)- + = = + = = -ao ham ben trai cua ham so y = F(x) tai iem x = 1. 2x 1 x 1f(x) F(1) x 1F'(1) = lim lim 2.x 1 x 1- - -= =- - ao ham ben phai cua ham so y = f(x) tai iem x0 = 0. x 1 x 1 x 1F(x) F(1) ax b 1 ax 1 a 1F'(1 ) lim lim lim a.x 1 x 1 x 1+ + ++ - + - + - -= = = =- - - Ham so y = F(x) co ao ham tai iem x = 1F'(1 ) F'(1 ) a 2.- + = = (2) Thay (2) vao (1), ta c b = 1. Vay ham so y = F(x) co ao ham tai iem x = 1, neu va ch neu a = 2, b = 1. Khi o: F(1) = 2 = f(1) Tom lai vi a = 2, b = 1 th F(x) la mot nguyen ham cua ham so f(x). V du 4: Xac nh a , b , c e ham so:-= + +2 2xF(x) (ax bx c)ela mot nguyen ham cua 2 2xF(x) (2x 8x 7)e-= - - +tren R. Giai: Ta co: 2x 2 2xF'(x) (2ax b)e 2(ax bx c)e- -= + - + +2 2x2ax 2(a b)x b 2c e- = - + - + - Do o F(x) la mot nguyen ham cua f(x) tren R F'(x) f(x), x R = "- + - + - = - + - " 2 22ax 2(a b)x b 2c 2x 8x 7, x R a 1 a 1a b 4 b 3b 2c 7 c 2= = - = = - - = - = Vay -= - +2 2xF(x) (x 3x 2)e . Tran S TungTch phan Trang 7 BAI TAP Bai 1.Tnh ao ham cua ham so xF(x) ln tg2 4p = + T o suy ra nguyen ham cua ham so 1f(x)cos x= . Bai 2.Chng to rang ham so 2ln(x 1), x 0F(x)x0 , x 0 + = = la mot nguyen ham cua ham so 22 22 ln(x 1), x 0f(x)x 1 x1 , x 0 +- =+ = Bai 3.Xac nh a, b, c sao cho ham so 2 xF(x) (ax bx c).e-= + +la mot nguyen ham cua ham so 2 xf(x) (2x 5x 2)e-= - +tren R. S:a = 2 ; b = 1 ; c = 1. Bai 4. a/Tnh nguyen ham 3 22x 3x 3x 7F(x) cua f(x) va F(0) 8.(x 1)+ + -= =+ b/Tm nguyen ham F(x) cua 2xf(x) sin va F .2 2 4p p = = S:a/2x 8F(x) x ;2 x 1= + ++b/ 1F(x) (x sinx 1)2= - +Bai 5. a/Xac nh cac hang so a, b, c sao cho ham so:

2F(x) (ax bx c) 2x 3 = + + -la mot nguyen ham cua ham so: 220x 30x 7 3f(x) tre n khoang ;2 2x 3- + = + - b/Tm nguyen ham G(x) cua f(x) vi G(2) = 0. S:a/ a 4; b 2; c 1; = = - = b/ 2G(x) (4x 2x 10) 2x 3 22. = - + - - Tch phanTran S TungTrang 8 Von de 2:Ik0 0I0h 080IE0 hkN 8k08 IE0 80 0008 8k080k0 080IE0 hkN 00 8k0 V du 1: CMR , neuf(x)dx F(x) C = + th 1f(ax b)dx F(ax b) C v i a 0.a+ = + + Giai: Ta luon co: 1f(ax b)dx f(ax b)d(ax b) vi a 0.a+ = + + Ap dung tnh chat 4, ta c:1 1f(ax b)dx (ax b)d(ax b) F(ax b) C (pcm)a a+ = + + + + . Ghi chu: Cong thc tren c ap dung cho cac ham so hp:f(t)dt F(t) C f(u)du F(u) C, vi u u(x) = + = + = V du 2: Tnh cac tch phan bat nh sau:a/ 3(2x 3) dx +b/4cos x.sinxdxc/xx2edxe 1 +d/2(2lnx 1)dxx+ Giai: a/Ta co: 4 43 31 1 (2x 3) (2x 3)(2x 3) dx (2x 3) d(2x 3) . C C.2 2 4 8+ ++ = + + = + = + b/Ta co: 54 4cos xcos x.sinxdx cos xd(cos x) C5= - = - + c/Ta co: x xxx x2e d(e 1)dx 2 2ln(e 1) Ce 1 e 1+= = + ++ + d/Ta co: 22 3(2lnx 1) 1 1dx (2lnx 1) d(2lnx 1) (2lnx 1) C.x 2 2+= + + = + + V du 3: Tnh cac tch phan bat nh sau:a/ 2x2sin dx2b/2cot g xdxc/ tgxdxd/3tgxdxcos x Giai: a/Ta co: 2x2sin dx (1 cosx)dx x sinx C2= - = - + b/Ta co: 221cot g xdx 1 dx cot gx x Csin x = - = - - + c/Ta co: sinx d(cosx)tgxdx dx ln cosx Ccosx cosx= = - = - + Tran S TungTch phan Trang 9 d/Ta co: 33 4 4 3tgx sinx d(cosx) 1 1dx dx cos x C C.cos x cos x cos x 3 3cos x-= =- = - + = - + V du 4: Tnh cac tch phan bat nh sau:a/ 2xdx1 x +b/ 21dxx 3x 2 - + Giai: a/Ta co: 222 2x 1 d(1 x ) 1dx ln(1 x ) C1 x 2 1 x 2+= = + ++ + b/Ta co: 21 1 1 1dx dx dxx 3x 2 (x 1)(x 2) x 2 x 1 = = - - + - - - - x 2ln x 2 ln x 1 C ln C.x 1-= - - - + = +- BAI TAP Bai 6.Tm nguyen ham cua cac ham so:a/ 2xf(x) cos ;2= b/ 3f(x) sin x.S: a/1(x sinx) C ;2+ + b/ 31cos x cos x C.3- + +Bai 7.Tnh cac tch phan bat nh : a/ x xe (2 e )dx;-- b/ xxedx ;2c/ 2x x xx2 .3 .5dx10. d/ 2 5xxe 1dx;e-+e/ xxedxe 2 + S: a/ x2e x C; - + b/ xxeC;(1 ln2)2+-c/ x6Cln6 +d/ 2 6x x1e e C;6- -- - + e/ xln(e 2) C + + . Bai 8.Tnh cac tch phan bat nh : a/ 4 4x x 2 dx-+ +;b/ 3 5x xdx ;c/ 2x x 1dx +; d/ 2001(1 2x) dx; - e/ 3 4lnxdxx- S: a/ 3x 1C;3 x- + b/ 5 75x C;7+ c/ 2 21(x 1) x 1 C3+ + +; d/ 20021 (1 2x). C;2 2002-- + e/ 1(3 4lnx) 3 4lnx C.6+ + +Tch phanTran S TungTrang 10 Von de 3:Ik0 0I0h 080IE0 hkN 8k08 Ph0008 PhkP Phk0 0h Phngphapphantchthcchatlaviecs dungcacongnhatthcebienoibieu thc di dau tch phan thanh tong cac bieu thc ma nguyen ham cua moi bieu thc o co the nhan c t bang nguyen ham hoac ch bang cac phep bien oi n gian a biet. Chuyquantrong:iemmauchotlaphepphantchlacothe rutraytngchorieng mnh t mot vai minh hoa sau: Vi 3 2 6 3f(x) (x 2) th vie t lai f(x) x 4x 4. = - = - +Vi 2x 4x 5 2f(x) th viet lai f(x) x 3x 1 x 1- += = - +- -. Vi 21 1 1f(x) th vie t la i f(x)x 5x 6 x 3 x 2= = -- + - - Vi 1 1f(x) th vie t la i f(x) ( 3 2x 2x 1)2 2x 1 3 2x= = - - ++ + - Vi x x 2 x x xf(x) (2 3 ) th vie t la i f(x) 4 2.6 9 . = - = - +Vi 3f(x) 8cos x.sinx th vie t la i f(x) 2(cos3x 3cosx).sinx = = +2cos3x.sinx 6cosx.sinx sin4x sin2x 3sin2x sin4x 2sin2x. = + = - + = + 2 2tg x (1 tg x) 1 = + - 2 2cot g x (1 cot g x) 1 = + - n 2n2 2x (1 x ) 1 1x1 x 1 x+ += ++ +. o ch la mot vai minh hoa mang tnh ien hnh. V du 1: Tnh tch phan bat nh: 2002I x(1 x) dx. = - Giai: S dung ong nhat thc :x = 1 (1 x)ta c:2002 2002 2002 2003x(1 x) [1 (1 x)](1 x) (1 x) (1 x) . - = - - - = - - -Khi o: 2002 2003 2002 20032003 2004I (1 x) dx (1 x) dx (1 x) d(1 x) (1 x) d(1 x)(1 x) (1 x)C.2003 2004= - - - = - - - + - -- -= - + + Tong qua t: Tnh tch phan bat nh:I x(ax b) dx, vi a 0a= + S dung ong nhat thc: 1 1x .ax [(ax b) b]a a= = + -Tran S TungTch phan Trang 11 Ta c: 11 1x(ax b) [(ax b) b)(ax b) [ (ax b) d(ax b) (ax b) d(ax d)]a aa a a+ a+ = + - + = + + - + + Ta xet ba trng hp : Vi a = 2, ta c: 1 221I [ (ax b) d(ax b) (ax b) d(ax b)]a- -= + + - + +

21 1[ln ax b ] C.a ax b= + + ++ Vi a = 1, ta c: 12 21 1I [ d(ax b) (ax b) d(ax b)] [ax b ln ax b ] C.a a-= + - + + = + - + + ViR \ { 2; 1}, a - -ta c:2 121 (ax b) (ax b)I [ ] C.a 2 1a+ a++ += + +a + a+ V du 2: Tnh tch phan bat nh: 2dxIx 4x 3=- + Giai: Ta co:21 1 1 (x 1) (x 3) 1 1 1. .x 4x 3 (x 3)(x 1) 2 (x 3)(x 1) 2 x 3 x 1- - - = = = - - + - - - - - - Khi o:- - = - = - = - - - + - - - - 1 dx dx 1 d(x 3) d(x 1) 1I . [ ' .(ln x 3 ln x 1) C2 x 3 x 1 2 x 3 x 1 2 -= +-1