Bài tập toán cao cấp tập 3 part 2 pdf

´ ’ ıch o a a a 10.2 C´c l´.p h`m kha t´ l´.p c´c h`m so cˆp a o a a a o Phu.o.ng ph´p II Thay x = v`o (10.7) ta c´ = A · ⇒ A = ´ Tiˆp theo, thay x = −1 v`o (10.7) ta thu du.o.c: −1 = −B2 · hay e a ’ ´ l` B2 = Dˆ t`m B1 ta thˆ gi´ tri x = v`o (10.7) v` thu du.o.c a e ı e a a a = A − B1 − B2 hay l` B1 = A − B2 = − a Do d´ o dx dx dx − + I= x−1 x+1 (x + 1)2 x−1 1 + ln + C =− 2(x + 1) x+1 3x + dx x(1 + x2)2 ’ ’ ´ ’ o a ı a a o a a Giai Khai triˆn h`m du.´.i dˆu t´ch phˆn th`nh tˆng c´c phˆn e a ’ th´.c co ban u 3x + A Bx + C Dx + F = + + )2 x(1 + x x 1+x (1 + x2 )2 T` d´ u o V´ du T´ I = ı ınh 3x + ≡ (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + F )x + A a ’ Cˆn b˘ng c´c hˆ sˆ cua c´c l˜y th`.a c`ng bˆc cua x ta thu du.o.c a ` a a e o ’ a u u u ´   A+B =0     C =0   2A + B + D = ⇒ A = 1, B = −1, C = 0, D = −1, F =    C +F =3     A = ` T` d´ suy r˘ng u o a xdx dx − − x + x2 = ln |x| − ln(1 + x2 ) − = ln |x| − ln(1 + x2 ) + I= xdx dx +3 (1 + x2)2 (1 + x2)2 dx (1 + x2 )−2 d(1 + x2) + (1 + x2 )2 + 3I2 2(1 + x2 ) 33 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh 34 dx ` ` o b˘ng cˆng th´.c truy hˆi thu du.o.c a o u (1 + x2 )2 Ta t´ I2 = ınh 10.1 Ta c´ o x 1 x · + + I1 = 2) 1+x 2(1 + x x + arctgx + C = 2) 2(1 + x I2 = dx + x2 ´ Cuˆi c`ng ta thu du.o.c o u 3x + + arctgx + C I = ln |x| − ln(1 + x2 ) + 2(1 + x2) ` ˆ BAI TAP T´ c´c t´ phˆn (1-12) ınh a ıch a xdx (x + 1)(x + 2)(x − 3) (DS 2x4 + 5x2 − dx 2x3 − x − DS x2 + ln |x − 1| + ln(2x2 + 2x + 1) + arctg(2x + 1)) 2x3 + x2 + 5x + dx (x2 + 3)(x2 − x + 1) DS ln |x + 1| − ln |x + 2| + |x − 3|) 20 x 2x − 1 √ arctg √ + ln(x2 − x + 1) + √ arctg √ ) 3 3 x4 + x2 + dx x(x − 2)(x + 2) (DS 21 21 x2 − ln |x| + ln |x − 2| + ln |x + 2|) 8 ´ ’ ıch o a a a 10.2 C´c l´.p h`m kha t´ l´.p c´c h`m so cˆp a o a dx x(x − 1)(x2 − x + 1)2 (DS 10 2x − 1 2x − − √ arctg √ ) − x2 − x + 3 x−1 x ln x4 − x2 + dx (x2 − 1)(x2 + 4)(x2 − 2) (DS √ x−1 x x− √ ) ln + arctg + √ ln − 10 x+1 20 x+ 3x2 + 5x + 12 dx (x2 + 3)(x2 + 1) √ x 5 arctg √ + ln(x2 + 1) + arctgx) − ln(x + 3) − 2 (DS (x4 + 1)dx x5 + x4 − x3 − x2 (DS ln |x| + x3 + x + dx x4 − (DS 10 1 ln |x − 1| + ln |x + 1| − arctgx) 4 x4 dx − x4 (DS 11 1 1 + ln |x − 1| − ln |x + 1| + ) x 2 x+1 (x2 − x + ln x+1 + arctgx) x−1 3x + dx + 2x + 2)2 (DS 2x − + arctg(x + 1)) + 2x + 2) 2(x2 35 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh 36 12 x4 − 2x2 + dx (x2 − 2x + 2)2 (DS x + 13 x2 + 2x + dx (x − 2)(x2 + 1)3 (DS 14 3−x + ln(x2 − 2x + 2) + arctg(x − 1)) x2 − 2x + 11 3 1−x ln |x2 − 2| − ln |x2 + 1| + − arctgx) 10 x +1 x2 dx (x + 2)2 (x + 1) (DS 15 x2 + dx (x − 1)3 (x + 3) (DS 16 x5 − x−1 + ln ) − 4(x − 1) 8(x − 1) 32 x+3 dx − x2 (DS 17 + ln |x + 1|) x+2 (x − 1)2 2x + 1 + ln + √ arctg √ ) x x +x+1 3 3x2 + dx x3 + 4x2 + 4x (DS ln |x| + ln |x + 2| + 18 10 ) x+2 2x5 + 6x3 + dx x4 + 3x2 (DS x2 − x − √ arctg √ ) 3x 3 ´ ’ ıch o a a a 10.2 C´c l´.p h`m kha t´ l´.p c´c h`m so cˆp a o a 19 x3 + 4x2 − 2x + dx x4 + x (DS 20 37 ln 2x − |x|(x2 − x + 1) + √ arctg √ ) (x + 1) 3 x3 − dx x4 + 10x2 + 25 (DS x 25 − 3x ln(x2 + 5) + − √ arctg √ ) + 5) 10(x 10 5 ’ ˜ Chı dˆ n x4 + 10x2 + 25 = (x2 + 5)2 a 10.2.2 ´ ’ T´ phˆn mˆt sˆ h`m vˆ ty do.n gian ıch a o o a o ’ ’ o a o e ı a Mˆt sˆ t´ phˆn h`m vˆ ty thu.`.ng g˘p c´ thˆ t´nh du.o.c b˘ng phu.o.ng o o ıch a a o ’ ` ´ u ty h´a h`m du.´.i dˆu t´ch phˆn Nˆi dung cua phu.o.ng ph´p ´ ’ ph´p h˜ ’ o a a u o a ı a o a a t´ch phˆn d˜ cho cua h`m vˆ ty vˆ ´ ’ ’ a n`y l` t` mˆt ph´p biˆn dˆi du ı a a ım o e e o a a o ’ ` e ’ ´ e a ınh a u e o t´ phˆn h`m h˜.u ty Trong tiˆt n`y ta tr` b`y nh˜.ng ph´p dˆi ıch a a u ’ ´ ´ ’ o o o biˆn cho ph´p h˜.u ty h´a dˆi v´.i mˆt sˆ l´.p h`m vˆ ty quan e e u o o o a o ’ ´ ´.c k´ hiˆu R(x1 , x2, ) hay r(x1 , x2, ) l` h`m h˜.u ´ nhˆt Ta quy u o y e a a a u ˜ ´ ´ ’ o o ty dˆi v´.i mˆ i biˆn x1, x2 , , xn o e ´ I T´ phˆn c´c h`m vˆ ty phˆn tuyˆn t´ ıch a a a o ’ a e ınh T´ phˆn dang ıch a R x, ax + b cx + d p1 , , ax + b cx + d pn dx (10.8) d´ n ∈ N; p1 , , pn ∈ Q; a, b, c ∈ R; ad − bc = du.o.c h˜.u ty o u ’ ’ ´ h´a nh` ph´p dˆi biˆn o o e o e ax + b = tm cx + d ´ ’ a ’ a o u ’ o dˆy m l` mˆ u sˆ chung cua c´c sˆ h˜.u ty p1 , , pn a ˜ o a ´ II T´ phˆn dang ıch a R(x, √ ax2 + bx + c)dx, a = 0, b2 − 4ac = (10.9) 38 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh ’ ´ o e e c´ thˆ h˜.u ty h´a nh` ph´p thˆ Euler: o e u ’ o √ √ ´ e (i) ax2 + bx + c = ± ax ± t, nˆu a > 0; √ √ ´ e (ii) ax2 + bx + c = ±xt ± c, nˆu c > 0; √ (iii) ax2 + bx + c = ±(x − x1 )t √ ax2 + bx + c = ±(x − x2 )t ’ d´ x1 v` x2 l` c´c nghiˆm thu.c kh´c cua tam th´.c bˆc hai a u a o a a a e c´c vˆ phai cua d˘ng th´.c c´ thˆ lˆy theo tˆ ’ ’ ´ ’ ´ ’ ’ a ´ ax + nbx + c (Dˆu o a e o a ’ u o e a ho.p t`y y) u ´ III T´ phˆn cua vi phˆn nhi th´.c D´ l` nh˜.ng t´ phˆn dang ıch a ’ a ıch a o a u u xm (axn + b)pdx (10.10) ’ d´ a, b ∈ R, m, n, p ∈ Q v` a = 0, b = 0, n = 0, p = 0; biˆu th´.c o a e u m n p o.c goi l` vi phˆn nhi th´.c x (zx + b) du a a u T´ phˆn vi phˆn nhi th´.c (10.10) du.a du.o.c vˆ t´ch phˆn h`m ıch a a e a a u ` ı u ty ba tru.`.ng ho.p sau dˆy: h˜ ’ u o a ´ 1) p l` sˆ nguyˆn, a o e m+1 ´ l` sˆ nguyˆn, a o e 2) n m+1 ´ 3) + p l` sˆ nguyˆn a o e n - ’ ˜ Dinh l´ (Trebu.s´p) T´ phˆn vi phˆn nhi th´.c (10.10) biˆu diˆn y e ıch a a e e u ´ du.o.c du.´.i dang h˜.u han nh` c´c h`m so cˆp (t´.c l` du.a du.o.c vˆ o u o a a a u a e ` u ty hay h˜.u ty h´a du.o.c) v` chı ´ nhˆt mˆt ´ u ’ o a ’ ıt a o t´ch phˆn h`m h˜ ’ ı a a u m+1 m+1 ´ ´ , + p l` sˆ nguyˆn a o e ba sˆ p, o n n ´ ´ 1) Nˆu p l` sˆ nguyˆn th` ph´p h˜.u ty h´a s˜ l` e a o e ı e u ’ o e a x = tN ’ a a d´ N l` mˆ u sˆ chung cua c´c phˆn th´.c m v` n o a ˜ o a ´ a u m+1 ´ ´ l` sˆ nguyˆn th` d˘t a o e ı a 2) Nˆu e n axn + b = tM ´ ’ ıch o a a a 10.2 C´c l´.p h`m kha t´ l´.p c´c h`m so cˆp a o a 39 d´ M l` mˆ u sˆ cua p o a ˜ o ’ a ´ m+1 ´ ´ + p l` sˆ nguyˆn th` d˘t a o e ı a 3) Nˆu e n a + bx−n = tM d´ M l` mˆ u sˆ cua p o a ˜ o ’ a ´ ´ CAC V´ DU I V´ du T´ ı ınh √ √ x + x2 + x √ dx , x(1 + x) 1) I1 = 2) I2 = dx (2 + x)(2 − x)5 · ’ Giai 1) T´ phˆn d˜ cho c´ dang I, d´ p1 = 1, p2 = , ıch a a o o ˜ ´ ’ a o p3 = Mˆ u sˆ chung cua p1 , p2 , p3 l` m = Do d´ ta d˘t x = t6 a o a Khi d´: o I=6 =6 t + t3 + t + t4 + t t dt = dt t6(1 + t2) + t2 √ 3√ dt = x + 6arctg x + C t3dt + 1+t ` ´ ’ ´ 2) B˘ng ph´p biˆn dˆi so cˆp ta c´ a e e o a o I2 = − x dx · + x (2 − x)2 D´ l` t´ phˆn dang I Ta d˘t o a ıch a a 2−x = t3 2+x v` thu du.o.c a x=2 − t3 , + t3 dx = −12 t2dt · (1 + t3)2 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh 40 T` d´ u o I2 = −12 t3 (t3 + 1)2 dt =− (t3 + 1)2 16t dt = t 2+x 2−x + C V´ du T´ c´c t´ phˆn ı ınh a ıch a 1) I1 = 3) I3 = dx √ , 2) I2 = x x2 + x + dx √ ,· (x + 1) + x − x2 dx , (x − 2) −x2 + 4x − √ ’ ’ Giai 1) T´ phˆn I1 l` t´ch phˆn dang II v` a = > nˆn ta su ıch a a ı a a e ´ dung ph´p thˆ Euler (i) e e √ x2 + x + = x + t, x2 + x + = x2 + 2tx + t2 √ t2 − −t2 + t − , x= x2 + x + = x + t = − 2t − 2t 2(−t + t − 1) dx = dt (1 − 2t)2 T` d´ u o √ 1−t + x − x2 + x + dt √ = ln + C = ln + C I1 = t2 − 1+t − x + x2 + x + ´ 2) Dˆi v´.i t´ phˆn I2 (dang II) ta c´ o o ıch a o −x2 + 4x − = −(x − 1)(x − 3) ´ ’ e e o v` d´ ta su dung ph´p thˆ Euler (iii): a √ −x2 + 4x − = t(x − 1) o Khi d´ −(x − 1)(x − 3) = t2(x − 1)2 , −(x − 3) = t2 (x − 1), t= 3−x , x−1 √ t2 + 2t , −x2 + 4x − = t(x − 1) = 2+1 t t +1 −4tdt dx = (t + 1)2 x= ´ ’ ıch o a a a 10.2 C´c l´.p h`m kha t´ l´.p c´c h`m so cˆp a o a v` thu du.o.c a √ √ 1−t x−1− 3−x dt √ I2 = = ln + C = ln √ + C t2 − 1+t x−1+ 3−x ´ ’ 3) Dˆi v´.i t´ phˆn I3 (dang III) ta c´ C = > Ta su dung o o ıch a o ´ ph´p thˆ Euler (ii) v` e e a √ + x − x2 = tx − 1, + x − x2 = t2x2 − 2tx + 1, √ 2t + t2 + t − , , x= + x − x2 = tx − = t +√ t2 + −2(t2 + t − 1) + + x − x2 , dx = t= · x (t2 + 1)2 o Do d´ d(t + 1) dt = −2 = −2arctg(t + 1) + C t2 + 2t + + (t + 1)2 √ + x + + x − x2 + C = −2arctg x V´ du T´ c´c t´ phˆn ı ınh a ıch a √ √ x √ dx, x 0; 2) I2 = dx; x 1−√ 1) I1 = (1 + x) x3 I3 = −2 3) I3 = dx x2 (1 + x3)5 · ’ Giai 1) Ta c´ o I1 = 1 x2 + x3 −2 dx, 1 ˜ o ` ’ , n = , p = −2, mˆ u sˆ chung cua m v` n b˘ng a ´ a a ’ ´ ´ V` p = −2 l` sˆ nguyˆn, ta ´p dung ph´p dˆi biˆn x = t6 v` thu du.o.c ı a o e a e o e a o d´ m = I1 = t8 dt = (1 + t2)2 = t5 − 4t3 + 18t − 18 4t2 + dt (1 + t2)2 t2 dt (1 + t2 )2 t4 − 2t2 + − dt −6 + t2 41 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh 42 V` ı t2 dt =− )2 (1 + t td 1 + t2 =− t + arctgt 2) 2(1 + t ´ nˆn cuˆi c`ng ta thu du.o.c e o u 3x1/6 − 21arctgx1/6 + C I1 = x5/6 − 4x1/2 + 18x1/6 + + x1/3 ´ o 2) Ta viˆt I2 du.´.i dang e I2 = x − x− dx m+1 ’ a ´ a = −1 l` sˆ nguyˆn v` ta a o e a O dˆy m = , n = − , p = v` 2 n ’ ´ ’ c´ tru.`.ng ho.p th´ hai Ta su dung ph´p dˆi biˆn o o u e o e 1 − √ = t4 x3 Khi d´ x = (1 − t4)− , dx = (1 − t4)− t3dt v` vˆy o a a 2 t4 t td = dt = − )2 (1 − t 1−t − t4 2t 1 − dt = + 4) 3(1 − t 1−t + t2 1+t 2t − ln − arctgt + C, = 3(1 − t4 ) 1−t I2 = dt − t2 1/4 o d´ t = − x−3/2 ´.i dang ´ 3) Ta viˆt I3 du o e I3 = x−2 (1 + x3 )− dx m+1 ’ a ´ a + p = −2 l` sˆ nguyˆn a o e O dˆy m = −2, n = 3, p = − v` n ’ ´ Do vˆy ta c´ tru.`.ng ho.p th´ ba Ta thu.c hiˆn ph´p dˆi biˆn a o o u e e o e + x−3 = t3 ⇒ + x3 = t3x3 ´ ’ ıch o a a a 10.2 C´c l´.p h`m kha t´ l´.p c´c h`m so cˆp a o a V´ du T´ ı ınh J= sin x + cos x dx sin2 x cos x + cos3 x ´ ´ ’ Giai H`m du.´.i dˆu t´ phˆn c´ t´ chˆt l` a o a ıch a o ınh a a R(− sin x, − cos x) = R(sin x, cos x) π π ’ ´ ´ ’ ’ o Do d´ ta su dung ph´p dˆi biˆn t = tgx, x ∈ − , Chia tu sˆ e o e o 2 ˜ ´ ’ ´ o a ıch a v` mˆ u sˆ cua biˆu th´.c du.´.i dˆu t´ phˆn cho cos3 x ta c´ a a o ’ e u o 2t + 2tgx + d(tgx) = dt tg x + t2 + t = ln(t2 + 9) + arctg +C tgx + C = ln(tg2 x + 9) + arctg J= V´ du T´ ı ınh J= dx sin x + cos6 x ’ ´ Giai Ap dung cˆng th´.c o u cos2 x = (1 + cos 2x), sin2 x = (1 − sin 2x) ta thu du.o.c cos6 x + sin6 x = (1 + cos2 2x) D˘t t = tg2x, ta t` du.o.c a ım dt 4dx =2 2x 2+4 + cos t tg2x t + C = arctg + C = arctg 2 J= 51 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh 52 V´ du T´ ı ınh J= sin x cos xdx ’ a Giai D˘t z = sin2 x ta thu du.o.c J= z 1/4(1 − z)− dx o a ıch a ’ a D´ l` t´ phˆn cua vi phˆn nhi th´.c v` u a +1 m+1 +p = − = n ’ ´ Do vˆy ta thu.c hiˆn ph´p dˆi biˆn a e e o e − = t4, z − dz = 4t3 dt, z2 z2 = (t4 + 1)2 v` d´ a o J = −2 t2 dt (t4 + 1)2 a D˘t t = ta thu du.o.c y J=2 y4 dy (1 + y 4)2 ` ` e e ıch a u a a a Thu.c hiˆn ph´p t´ phˆn t`.ng phˆn b˘ng c´ch d˘t a u = y, dv = y3 dy ⇒ du = dy, (1 + y 4)2 v=− ta thu du.o.c y + 4) 4(1 + y y + J1 =− 4) 2(1 + y J=2 − dy + y4 4(1 + y 2) ´ ’ ıch o a a a 10.2 C´c l´.p h`m kha t´ l´.p c´c h`m so cˆp a o a ’ ’ ’ ˜ ´ ´ ’ o ’ e u o a ı a Dˆ t´ J1 ta biˆu diˆn tu sˆ cua biˆu th´.c du.´.i dˆu t´ch phˆn e ınh e e sau: nhu 1= (y + 1) − (y − 1) o v` d´ a y2 + y2 − dy − dy y4 + y4 + 1 1 + dy − dy 1 y y = − 1 2 y2 + y2 + y y 1 d y+ d y+ 1 y y = − 2 2 y− y+ +2 −2 y y 1 √ y− y+ − 1 y y + C = √ arctg √ − √ ln √ 2 yb + + y J1 = ´ Cuˆi c`ng ta thu du.o.c o u √ y+ − y− y y +C + √ arctg √ − √ ln J=− 4) √ 2(1 + y 2 y+ + y d´ o y= , t t= −1, z z = sin2 x ` ˆ BAI TAP ’ a o u a T´ c´c t´ phˆn b˘ng c´ch su dung c´c cˆng th´.c lu.o.ng gi´c ınh a ıch a ` a a ’ ´ ’ ´ dˆ biˆn dˆi h`m du.´.i dˆu t´ phˆn o a ıch a e e o a 53 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh 54 sin xdx cos3 x ) (DS − cos x + cos4 xdx (DS 3x sin 2x sin 4x + + ) 32 sin5 xdx (DS cos5 x cos3 x − − cos x) cos7 xdx (DS sin x − sin3 x + cos2 x sin2 xdx sin3 x cos2 xdx (DS cos5 x cos3 x − ) cos3 x sin5 xdx (DS sin6 x sin8 x − ) 8 dx sin 2x 3 sin5 x sin7 x − ) x sin 4x ) (DS − 32 (DS ln |tgx|) dx π x + ) (DS ln tg x cos x π x sin x + cos x dx (DS ln tg + ln tg + 10 sin 2x 2 11 tg5 x tg3 x sin2 x dx (DS + ) cos6 x ’ ˜ Chı dˆ n D˘t t = tgx a a (DS − (cos 4x + cos 2x)) 12 sin 3x cos xdx 13 sin 14 cos3 x dx sin2 x (DS − 15 sin3 x dx cos2 x (DS 2x x cos dx 3 (DS x cos − cos x) − sin x) sin x + cos x) cos x ) ´ ’ ıch o a a a 10.2 C´c l´.p h`m kha t´ l´.p c´c h`m so cˆp a o a 16 cos3 x dx sin5 x cotg4x ) (DS − 17 sin5 x dx cos3 x (DS 18 tg5 xdx (DS cos2 x + ln | cos x| − ) cos2 x tg4x tg2 x − − ln | cos x|) ’ ´ Trong c´c b`i to´n sau dˆy h˜y ´p dung ph´p dˆi biˆn a a a a a a e o e 2t x − t2 2dt t = tg , sin x = , cos x = , x = 2arctgt, dx = 2 1+t 1+t + t2 x + tg (DS ln x ) − tg x π (DS √ ln tg + 19 dx + cos x 20 dx sin x + cos x 21 sin x + cos x dx sin x + cos x (DS 22 23 ) (12x − ln |2tgx + 3| − ln | cos x|) 13 x dx (DS ln + tg ) + sin x + cos x dx (2 − sin x)(3 − sin x) (DS x x 2tg − 3tg − 1 2 √ arctg √ √ − √ arctg ) 3 2 T´ c´c t´ phˆn dang ınh a ıch a sinm x cosn xdx, 24 sin3 x cos5 xdx (DS m, n ∈ N cos x − cos6 x) 55 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh 56 25 sin2 x cos4 xdx 26 1 x − sin 4x + sin2 2x ) 16 sin4 x cos6 xdx (DS (DS 1 sin 8x − sin 4x + sin5 2x + x) 211 · 26 sin 4x sin2 2x x − − ) 16 64 48 (DS sin5 x − sin7 x + sin9 x) 28 sin4 x cos5 xdx 1 (DS sin7 x − sin9 x) 29 sin6 x cos3 xdx T´ c´c t´ phˆn dang ınh a ıch a 27 sin4 x cos2 xdx (DS sinα x cosβ xdx, α, β ∈ Q √ 3 sin3 x √ dx (DS cos x cos2 x + √ ) 30 cos x cos x cos x ˜ ’ a Chı dˆ n D˘t t = cos x a 31 32 33 34 35 √ dx sin11 x cos x ’ ˜ a Chı dˆ n D˘t t = tgx a (DS − 3(1 + 4tg2 x) 8tg2x · tg2 x ) √ sin3 x √ dx (DS 3 cos x cos2 x − ) 2x cos √ 3 11 cos x) cos2 x sin3 xdx (DS − cos5/3 x + 11 √ dx √ (DS 4 tgx) 5x sin x cos 5 sin3 x 14 √ dx (DS cos x − cos x) cos x 14 Chu.o.ng 11 T´ phˆn x´c dinh Riemann ıch a a ’ ıch 11.1 H`m kha t´ Riemann v` t´ phˆn x´c a a ıch a a dinh 58 - 11.1.1 Dinh ngh˜ 58 ıa ’ - ` ’ ıch 11.1.2 Diˆu kiˆn dˆ h`m kha t´ 59 e e e a ´ ’ ’ ıch a a 11.1.3 C´c t´ chˆt co ban cua t´ phˆn x´c dinh 59 a ınh a a ınh t´ phˆn x´c d inh 61 ıch a a 11.2 Phu.o.ng ph´p t´ ’ ıch a a 11.3 Mˆt sˆ u.ng dung cua t´ phˆn x´c d inh 78 o o´ ´ ’ ’ ’ 11.3.1 Diˆn t´ h` ph˘ng v` thˆ t´ vˆt thˆ 78 e ıch ınh a a e ıch a e a e ıch a o 11.3.2 T´ dˆ d`i cung v` diˆn t´ m˘t tr`n xoay 89 ınh o a 11.4 T´ phˆn suy rˆng 98 ıch a o 11.4.1 T´ phˆn suy rˆng cˆn vˆ han 98 ıch a o a o ’ a 11.4.2 T´ phˆn suy rˆng cua h`m khˆng bi ch˘n 107 ıch a o o a ıch a a Chu.o.ng 11 T´ phˆn x´c dinh Riemann 58 11.1 ’ ıch H`m kha t´ Riemann v` t´ phˆn a a ıch a x´c dinh a 11.1.1 - Dinh ngh˜ ıa ’ ’ a a u Gia su h`m f(x) x´c dinh v` bi ch˘n trˆn doa n [a, b] Tˆp ho.p h˜.u a a e a n ’ han diˆm xk k=0 : e a = x0 < x1 < x2 < · · · < xn−1 < xn = b a a du.o.c goi l` ph´p phˆn hoach doan [a, b] v` du.o.c k´ hiˆu l` T [a, b] hay a e y e a n gian l` T ’ a - ’ ’ Dinh ngh˜ 11.1.1 Gia su [a, b] ⊂ R, T [a, b] = {a = x0 < x1 < ıa ˜ · · · < xn = b} l` ph´p phˆn hoach doan [a, b] Trˆn mˆ i doan [xj−1 , xj ], a e a e o ’m ξj v` lˆp tˆng ’ j = 1, , n ta chon mˆt c´ch t`y y diˆ o a u ´ e a a o n S(f, T, ξ) = f (ξj )∆xj , ∆xj = xj − xj−1 j=1 ’ ’ a goi l` tˆng t´ phˆn (Riemann) cua h`m f (x) theo doan [a, b] tu.o.ng ıch a a o ng v´.i ph´p phˆn hoach T v` c´ch chon diˆm ξj , j = 1, n Nˆu gi´.i ’ ´ u ´ o e a a a e e o han n lim S(f, T, ξ) = lim d(T )→0 d(T )→0 f (ξj )∆xj (11.1) j=1 ` o o a e a a a tˆn tai h˜.u han khˆng phu thuˆc v`o ph´p phˆn hoach T v` c´ch o u ’ chon c´c diˆm ξj , j = 1, n th` gi´.i han d´ du.o.c goi l` t´ch phˆn x´c a a e ı o o a a ı ’ a dinh cua h`m f(x) ’ ı Tˆp ho.p moi h`m kha t´ch Riemann trˆn doa n [a, b] du.o.c k´ hiˆu a e a y e l` R[a, b] a ’ ıch 11.1 H`m kha t´ Riemann v` t´ch phˆn x´c dinh a a ı a a 11.1.2 59 - ` ’ ’ ıch Diˆu kiˆn dˆ h`m kha t´ e e e a - ´ Dinh l´ 11.1.1 Nˆu h`m f (x) liˆn tuc trˆn doan [a, b] th` f ∈ R[a, b] y e a e e ı ´ e ´ ’ ı ’ a ` e a y ` a Hˆ qua Moi h`m so cˆp dˆu kha t´ch trˆn doan bˆt k` n˘m tron e a p x´c dinh cua n´ ’ tˆp ho a a o - ’ ’ a a Dinh l´ 11.1.2 Gia su f : [a, b] → R l` h`m bi ch˘n v` E ⊂ [a, b] y a a p c´c diˆm gi´n doan cua n´ H`m f (x) kha t´ch Riemann ’ ’ ’ ı l` tˆp ho a a a e a o a p E c´ dˆ - khˆng, t´.c l` E trˆn doan [a, b] v` chı tˆp ho e o o a ’ a o u a ` ` e e ’ thoa m˜n diˆu kiˆn: ∀ ε > 0, tˆn tai hˆ dˆm du.o.c (hay h˜.u han) c´c a u a e e o ´ ’ khoang (ai , bi ) cho ∞ N ∞ (ai , bi ), E⊂ i=1 (bi − ) = lim i=1 N →∞ (bi − ) < ε i=1 ’ ´ ’ ’ ı Nˆu c´c diˆu kiˆn cua dinh l´ 11.1.2 (goi l` tiˆu chuˆn kha t´ch e a ` e e y a a e b ’ a ı a ’ ıch a Lo.be (Lebesgue)) du.o.c thoa m˜n th` gi´ tri cua t´ phˆn f (x)dx a ’ khˆng phu thuˆc v`o gi´ tri cua h`m f (x) tai c´c diˆm gi´n doan v` o o a a ’ a e a a a o.c bˆ sung mˆt c´ch t`y y nhu.ng phai ’ ’ ’ tai c´c diˆm d´ h`m f (x) du o o a u ´ e o a a ’ bao to`n t´ bi ch˘n cua h`m trˆn [a, b] a ınh a ’ a e 11.1.3 ´ ’ ’ C´c t´ a ınh chˆt co ban cua t´ phˆn x´c a ıch a a dinh a 1) f (x)dx = a b 2) a f(x)dx = − f(x)dx a b ´ 3) Nˆu f, g ∈ R[a, b] v` α, β ∈ R th` αf + βg ∈ R[a, b] e a ı ıch a a Chu.o.ng 11 T´ phˆn x´c dinh Riemann 60 ´ 4) Nˆu f ∈ R[a, b] th` |f(x)| ∈ R[a, b] v` e ı a b b f(x)dx |f (x)|dx, a a < b a ´ 5) Nˆu f, g ∈ R[a, b] th` f (x)g(x) ∈ R[a, b] e ı ´ 6) Nˆu fg ∈ D[a, b] v` ]c, d] ⊂ [a, b] th` f (x)g(x) ∈ R[c, d] e a ı ’ ´ o e o 7) Nˆu f ∈ R[a, c], f ∈ R[c, b] th` f ∈ R[a, b], d´ diˆm c c´ e ı ’ a ´ ’ thˆ s˘p xˆp t`y y so v´.i c´c diˆm a v` b e ´ e u ´ o a e a ´ a o o Trong c´c t´ chˆt sau dˆy ta luˆn luˆn xem a < b a ınh a b ´ 8) Nˆu f ∈ R[a, b] v` f e a th` ı f (x)dx a ´ 9) Nˆu f, g ∈ R[a, b] v` f (x) e a g(x) ∀ x ∈ [a, b] th` ı b b f (x)dx g(x)dx a a ´ 10) Nˆu f ∈ C[a, b], f(x) e 0, f (x) ≡ trˆn [a, b] th` ∃ K > e ı cho b f (x)dx K a ´ 11) Nˆu f, g ∈ R[a, b], g(x) e trˆn [a, b] e M = sup f (x), m = inf f (x) [a,b] [a,b] th` ı b m b g(x)dx a b f (x)g(x)dx ≤ M a g(x)dx a a ınh ıch a a i 11.2 Phu.o.ng ph´p t´ t´ phˆn x´c d nh 11.2 61 Phu.o.ng ph´p t´ a ınh t´ ıch phˆn x´c a a dinh ’ ıch e ’ ’ a Gia su h`m f(x) kha t´ trˆn doan [a, b] H`m a x f(x)dt, F (x) = a x b a ´ e e e du.o.c goi l` t´ phˆn v´.i cˆn trˆn biˆn thiˆn a ıch a o a - Dinh l´ 11.2.1 H`m f(x) liˆn tuc trˆn doan [a, b] l` c´ nguyˆn h`m y a e e a o e a ’ a trˆn doan d´ Mˆt c´c nguyˆn h`m cua h`m f (x) l` h`m e o o a e a a a x f (t)dt F (x) = (11.2) a ´ ´ e e e o o T´ phˆn v´.i cˆn trˆn biˆn thiˆn du.o.c x´c dinh dˆi v´.i moi h`m ıch a o a a a ’ ’ ıch e f (x) kha t´ trˆn [a, b] Tuy nhiˆn, dˆ h`m F (x) dang (11.2) l` nguyˆn e e a a e ` o e a ’ ’ e h`m cua f(x) diˆu cˆt yˆu l` f (x) phai liˆn tuc a e ´ ´ ` e e a a a ıa ’ o Sau dˆy l` dinh ngh˜ mo rˆng vˆ nguyˆn h`m - ’ Dinh ngh˜ 11.2.1 H`m F (x) du.o.c goi l` nguyˆn h`m cua h`m ıa a e a a a ´ f (x) trˆn doan [a, b] nˆu e e 1) F (x) liˆn tuc trˆn [a, b] e e ’ e e ’ 2) F (x) = f (x) tai c´c diˆm liˆn tuc cua f (x) a ’ o e a Nhˆn x´t H`m liˆn tuc trˆn doan [a, b] l` tru.`.ng ho.p riˆng cua a e a e e ng doan Do d´ dˆi v´.i h`m liˆn tuc dinh ngh˜a 11.2.1 ´ h`m liˆn tuc t` a e u e o o o a ı ` vˆ nguyˆn h`m l` tr`ng v´.i dinh ngh˜a c˜ tru.´.c dˆy v` F (x) = f (x) e e a a u o o a ı ı u o.c suy t` t´nh kha vi ’ ∀ x ∈ [a, b] v` t´ liˆn tuc cua F (x) du a ınh e ’ u ı - Dinh l´ 11.2.2 H`m f (x) liˆn tuc t`.ng doan trˆn [a, b] l` c´ nguyˆn y a e u e a o e h`m trˆn [a, b] theo ngh˜ cua dinh ngh˜ mo rˆng Mˆt c´c a e ıa ’ o a ıa ’ o ıch a a Chu.o.ng 11 T´ phˆn x´c dinh Riemann 62 nguyˆn h`m l` e a a x F (x) = f (t)dt a - ´ Dinh l´ 11.2.3 (Newton-Leibniz) Dˆi v´.i h`m liˆn tuc t`.ng doan y e u o o a c Newton-Leibniz: trˆn [a, b] ta c´ cˆng th´ e o o u b f (x)dx = F (b) − F (a) (11.3) a ’ ıa ’ o d´ F (x) l` nguyˆn h`m cua f (x) trˆn [a, b] v´.i ngh˜ mo rˆng o a e a e o - ’ ´ ’ ’ a o e Dinh l´ 11.2.4 (Phu.o.ng ph´p dˆi biˆn) Gia su.: y (i) f (x) x´c dinh v` liˆn tuc trˆn [a, b], a a e e ’ (ii) x = g(t) x´c dinh v` liˆn tuc c`ng v´.i dao h`m cua n´ trˆn a a e u o a o e doan [α, β], d´ g(α) = a, g(β) = b v` a g(t) b o a Khi d´ o β b f(x)dx = a f (g(t))g (t)dt (11.4) α - ´ ` Dinh l´ 11.2.5 (Phu.o.ng ph´p t´ch phˆn t`.ng phˆn) Nˆu f (x) v` y a ı a u a e a g(x) c´ dao h`m liˆn tuc trˆn [a, b] th` o a e e ı b b b a f(x)g (x)dx = f (x)g(x) − a f (x)g(x)dx a ´ CAC V´ DU I ` ’ a e V´ du Ch´.ng to r˘ng trˆn doan [−1, 1] h`m ı u a  1  v´.i x > 0, o   f(x) = signx = v´.i x = 0, x ∈ [−1, 1] o    −1 v´.i x < o (11.5) a ınh ıch a a i 11.2 Phu.o.ng ph´p t´ t´ phˆn x´c d nh 63 ’ o ’ ıch, a) kha t´ b) khˆng c´ nguyˆn h`m, c) c´ nguyˆn h`m mo rˆng o o e a o e a ng doan ’ ’ ıch ı o a a Giai a) H`m f (x) kha t´ v` n´ l` h`m liˆn tuc t` a e u ng minh h`m f (x) khˆng c´ nguyˆn h`m theo ngh˜ c˜ b) Ta ch´ u a o o e a ıa u Thˆt vˆy moi h`m dang a a a  −x + C F (x) = x < x + C2 x ` ` o a ´ dˆu c´ dao h`m b˘ng signx ∀ x = 0, d´ C1 v` C2 l` c´c sˆ t`y e a o a a a o u ´ ´ ´ y Tuy nhiˆn, thˆm ch´ h`m “tˆt nhˆt” sˆ c´c h`m n`y ´ e a ı a o a o a a a F (x) = |x| + C ’ ´ (nˆu C1 = C2 = C) c˜ng khˆng c´ dao h`m tai diˆm x = Do d´ e u o o a e o ng doa n) khˆng c´ dao h`m h`m signx (v` d´ moi h`m liˆn tuc t` a a o a e u o o a a diˆm gi´n doan ’ ´ ’ trˆn khoang bˆt k` ch´ e a y u e a ’ o c) Trˆn doan [−1, 1] h`m signx c´ nguyˆn h`m mo rˆng l` h`m e a a a o e a F (x) = |x| v` n´ liˆn tuc trˆn doan [−1, 1] v` F (x) = f(x) x = ı o e e a a √ a2 − x2 dx, a > V´ du T´ ı ınh π ´ ´ ´ ’ th` x chay hˆt ı e a e e Giai D˘t x = a sin t Nˆu t chay hˆt doan 0, doan [0, a] Do d´ o a √ a2 − x2 dx = π/2 π/2 + cos 2t dt a2 cos2 tdt = a2 0 π/2 a2 = π/2 a2 dt + cos 2tdt = V´ du T´ t´ phˆn ı ınh ıch a √ 2/2 1+x dx 1−x I= a2 π · ıch a a Chu.o.ng 11 T´ phˆn x´c dinh Riemann 64 ’ ’ ´ ´ a ’ e e o e Giai Ta thu.c hiˆn ph´p dˆi biˆn x = cos t Ph´p dˆi biˆn n`y e o e ’ thoa m˜n c´c diˆu kiˆn sau: a a ` e e (1) x = ϕ(t) = cos t liˆn tuc ∀ t ∈ R e √ π π ´ ´ (2) Khi t biˆn thiˆn trˆn doan e e e , th` x chay hˆt doan 0, ı e 4 √ π π = , ϕ = (3) ϕ 2 π π (4) ϕ (t) = − sin t liˆn tuc ∀ t ∈ , e ’ ´ ’ Nhu vˆy ph´p dˆi biˆn thoa m˜n dinh l´ 11.2.4 v` d´ a e o e a y a o x = cos t, ϕ π dx = − sin tdt, √ π = 0, ϕ = · Nhu vˆy a π π t cotg (− sin t)dt = I= −π (1 + cos t)dt π = t + sin t π/2 π/4 √ π · = +1− V´ du T´ t´ phˆn ı ınh ıch a √ 3/2 dx · x − x2 √ I= 1/2 ’ ´ ’ e e o e Giai Ta thu.c hiˆn ph´p dˆi biˆn x = sin t ⇒ dx = cos tdt ’ ´ v` biˆu th´.c du.´.i dˆu t´ phˆn c´ dang a e u o a ıch a o   dt ´  nˆu cos t > 0, e cos tdt sin t √ = sin t cos2 t − dt  ´ nˆu cos t < e sin t a ınh ıch a a i 11.2 Phu.o.ng ph´p t´ t´ phˆn x´c d nh ’ ’ ıch a C´c cˆn α v` β cua t´ phˆn theo t du.o.c x´c dinh bo.i a a a a = sin t ⇒ α = √ = sin t ⇒ β = π , π · 5π 2π ’ v` β1 = a ) Trong ca hai tru.`.ng ho.p o √ ´ ´ ` ´ ´ ’ Ta s˜ thˆy kˆt qua e a e biˆn x = sin t dˆu chay hˆt doan [a, b] = , e e e 2 ´ t´ phˆn l` nhu Thˆt vˆy tru.`.ng ho.p th´ nhˆt ta c´ ıch a a a a o u a o cos t > v` a ’ ´ (Ta c˜ng c´ thˆ lˆy α1 = u o e a π/3 π/3 t dt = ln tg sin t I= π/6 √ 2+ · = ln π/6 5π 2π Trong tru.`.ng ho.p th´ hai t ∈ , ta c´ cos t < v` o a o u 2π/3 I =− dt t = − ln tg sin t 2π/3 5π/6 √ 2+ · = ln 5π/6 V´ du T´ t´ phˆn ı ınh ıch a π/3 x sin x dx cos2 x I= ` ` ’ a ıch a u a Giai Ta t´ b˘ng phu.o.ng ph´p t´ phˆn t`.ng phˆn ınh a a D˘t u = x ⇒ du = dx, sin xdx ⇒v= · dv = 2x cos cos x 65 ... 2x + 2) 2( x2 35 ´ ıch a a Chu.o.ng 10 T´ phˆn bˆt dinh 36 12 x4 − 2x2 + dx (x2 − 2x + 2) 2 (DS x + 13 x2 + 2x + dx (x − 2) (x2 + 1 )3 (DS 14 3? ??x + ln(x2 − 2x + 2) + arctg(x − 1)) x2 − 2x + 11 3. .. a o a x 26 1+ 27 28 29 30 31 32 √ dx x2 x3(1 + 2x2)− dx dx √ x4 + x dx (1 + x3 )5 /3 x dx x6 (DS ? ?2 3 (x− + 1 )2 ) (DS − √ ) 2( x + 1 )2 u7 − u + u3 − u2 , u2 = √ u5 2u3 − + u, u= √ − x? ?2 ) dx... x? ?2 = (t3 − 1) x3 = Do vˆy a −5 /3 t3 t2 (t3 − 1)− dt = 3? ??1 t + 2t3 t? ?2 ? ?3 −t+C = C − = t dt − dt = ? ?2 2t3 + 3x =C− · 2x (1 + x3 )2 I3 = − (t3 − 1 )2 /3 − t3 dt t3 ´ ıch a a Chu.o.ng 10 T´ phˆn