数学分析讲义第四版刘玉琏傅沛仁著高等教育出版社课后答案第十二单元.pdf

1?2?C? ?SK12.1 56e?1265? 1.e?; (2) Z + 2 dx x2+ x 2 ) Z + 2 dx x2+ x 2 =lim p+ Z p 2 dx x2+ x 2 =lim p+ 1 3 Z p 2 ? 1 x 1 1 x + 2 ? dx = 1 3 lim p+ln(x 1) ln(x + 2) ? ? ? p 2 = 1 3 lim p+ ? ln p 1 p + 2 + ln4 ? = 2 3 ln2 . (2) Z + 1 e x xdx. ) Z + 1 e x xdx=2 Z + 1 e x d(x) =2e x? ? ? + 1 = 2 e . (5) Z + ea|x|dx(a 0) . ) Z + ea|x|dx= Z 0 ea|x|dx + Z + 0 ea|x|dx =2 Z + 0 eaxdx = 2 ae ax? ? + 0 = 2 a . (6) Z + 0 eaxsinbxdx .(a 0) . 1 ) fi ?Reaxsinbxdx = eax a2+ b2(asinbx bcosbx) + C(?56?7.26),k Z + 0 eaxsinbxdx = eax a2+ b2(asinbx + bcosbx) ? ? ? + 0 = b a2+ b2 . 2.?Oe?5: (2) Z + 1 dx 4 x3 + 1 . )lim x+ x 3 4 1 4 x3 + 1 . = 3 4 0.u,u?. (4) Z + 1 xm 1 + xndx ,(n 0,m 0). )lim x+ xnm xm 1 + xn = 1 d = 1,? = n m 1,;? = n m 6 1,u? . (6) Z + 0 argtanx x dx . )lim x0+ argtanx x = lim x0+ 1 1 + x2 = 1. l?, argtanx x ?3:0Ym?. lim x+ x argtanx x =lim x+ argtanx = 2 . = 1,d = 2 0,u?. (8) Z + x ex+ exdx . ) Z + x ex+ exdx = Z 0 x ex+ exdx + Z + 0 x ex+ exdx . ?x = t,dx = dt,k Z 0 x ex+ exdx = Z 0 + t et+ etdt = Z + 0 x ex+ exdx lim x+ x2 x ex+ ex =lim x+ x2 x3 ex+ ex = 0 . = 2,d = 0.?.l?R + x ex+ exdx . 3.y:ef(x)31,+)?N?,?x +?,f(x) 0KR + 1 f(x)dx? 2 P n=1 f(n)?u?. yyy fi ?f(x)31,+)?N?,? lim x+ f(x) = 0,Kx 1,+),kf(x) 0,?k N,k f(k + 1) = Z k+1 k f(k + 1)dx 6 Z k+1 k f(x)dx 6 Z k+1 k f(k)dx = f(k) . l?, n P k=1 f(k + 1) 6 n P k=1 Rk+1 k f(x)dx 6 n P k=1 f(k) n P k=1 f(k + 1) 6 Rn+1 1 f(x)dx 6 n P k=1 f(k) ? nR n+1 1 f(x)dx o ?NO?,Sn= n P k=1 f(k)? P n=1 f(n)?. e P n=1 f(n),K?Snk?.,l? nR n+1 1 f(x)dx o ?k?u, nR + 1 f(x)dx ; e P n=1 f(n)u?,K?Sn?.,l? nR n+1 1 f(x)dx o ?u, nR + 1 f(x)dxu ?, =R + 1 f(x)dx? P n=1 f(n)?u?. 5.y:eR + a f(x)dx?,?(x)3a,+)?Nk.,KR + a f(x)(x)dx . yyy fi ?R + a f(x)dx?,d?OK, 0,A a,p1,p2 A,k ? ? ? Z p2 p1 |f(x)|dx ? ? ? 0,x a,+),k|(x)| 6 M. qfi ?f(x)(x)?.k ? ? ? Z p2 p1 f(x)(x)dx ? ? ? 6 ? ? ? Z p2 p1 |f(x)|(x)|dx ? ? ? 6 M ? ? ? Z p2 p1 |f(x)|dx ? ? ? 0.?n8,R + 1 sinx x . x 1,k|sinx| sin2x,l? ? ? ? sinx x ? ? ? sin2x x = 1 cos2x 2x = 1 2x cos2x 2x . 3 fi ?0 0,A 0,p1,p2 A,k ? ? ? Z p2 p1 |f(x)|dx ? ? ? 2A? x 2 ?p 1= x 2,p2 = x?, df(x)3a,+)?K,?N?,k Z x x 2 f(t)dt f(x) Z x x 2 dt = x 2f(x) xf(x) 0,n N,xn n,k|f(xn)| 0. l?,?3?xn,? lim n+ xn= +,k|f(xn)| 0. fi ?f(x)30,+)?Y,= 0 2 0, 0,x0,x00 a,+) : |x0 x00| 6 ,k |f(x0) f(x00)| 0 2 . l?,x xn,xn+ : |x xn| 6 ,k|f(x) f(xn)| |f(xn)| |f(x) f(xn)| 0 0 2 = 0 2 .(1) 4 x xn,xn+ ,f(x)?f(xn)7?.Kef(x)?f(xn)?, Kk|f(x) f(xn)| = |f(x)| + |f(xn)| 0,g. ef(xn) 0,Kf(x) 0,d(1)“,kf(x) 0 2 .l? Z xn+ xn f(x)dx 0 2 Z xn+ xn dx = 0 2 (?); ef(xn) 0,n N,xn n,k ? ? ? Rxn+ xn f(x)dx ? ? ? 0 2 . ?OK?Q,R + 0 f(x)dxu?,?fi ?g.u lim x+ f(x) = 0 . 11.y:ef(x)3a,+)kY?f0(x),?R + a f(x)dx ?R + a f0(x)dx? ,K lim x+ f(x) = 0 . yyy fi ?R + a f0(x)dx,=4? Z + a f0(x)dx =lim x+ Z x a f0(t)dt =lim x+ f(t) ? ? ? x a =lim x+ f(x) f(a) ?3,?4? lim x+ f(x)?3.? lim x+ f(x) = . ey: = 0.?y.b? 6= 0.? 0.= lim x+ f(x) = 0.dY? 5,A 0,x A,k f(x) 2 . l?,p A(?,p + 1 A),k Z p+1 p f(x)dx 2 (?) . ?OK?Q,R + a f(x)dxu?,?fi ?g.u,lim x+ f(x) = 0 . 12.?f(x)3a,+)?,?N?, lim x+ f(x) = 0,y: R+ a f(x)dx R+ a xf0(x)dx . y: Z + a xf0(x)dx =lim x+ xf(x) af(a) Z + a f(x)dx.(1) eR + a f(x)dx,qfi ?f(x)3a,+)?N?,d19K,K lim x+ xf(x) = 0.u, R + a xf0(x)dx. eR + a xf0 (x)dx,qfi ?f(x)3a,+)?N?,? lim x+ f(x) = 0, Kx 5 a,+),kf(x) 0,?f0(x) 6 0. ?a 0.x a,b x. ?n, x,b,k Z b x tf0(t)dt = Z b x f0(t)dt = f(b) f(x) = f(x) f(b) xf(x) f(b) 0 fi ? lim b+ f(b) = 0.l?,?b +?,k Z + x tf0(t)dt xf(x) 0 . ?n2?1, lim x+ h R+ x tf0(t)dt i = 0.l?,k lim x+ xf(x) = 0. d(1)“,R + a f(x)dx. ?SK12.2 56e?1275? 1.e?: (2)R 1 0 dx (2 x)1 x . )1?:. : 0 0,x : a 0,x : a 0,x (a,b,k |F(x)| = ? ? ? Z x a f(t)dt ? ? ? 6 C, K? 0?,?R b a(x a) f(x)dx. yyy fi ?dF(x) = f(x)dx.F(b) = 0. 0( 0,=1 0,?f(x)3(0,1?N?. 8 fi ?R 1 0 f(x)dx,d?OK,= 0, 0( c) =lim p+ Z c p f(x)dx +lim p+ Z p c f(x)dx 9 = Z + f(x)dx = A . ?.X, V.P. R+ sinxdx = 0 . ?,R + sinxdx%u?. ?SK12.3 56e?1305? 1.?k?f(x,y) = sgn(x y),(x,y) R2.y:? F(y) = Z 1 0 f(x,y)dx 3RY,F(y)? . yyy0 6 x 6 1.y R,k f(x,y) = sgn(x y) = 1,y 1. ?y 1?,z = F(y) = R1 0 f(x,y)dx = R1 0 (1)dx = 1. u, z = F(y) = 1,y 1. ?X12.a w,z = F(y)3(,0),(0,1),(1,+)?Y,? lim y0 F(y) = lim y0+ F(y) = 1 = F(0), lim y1 F(y) = lim y1+ F(y) = 1 = F(1) , =z = F(y)30?1?Y.u,?F(y)3RY. 2.e?4?: (1)lim y0 R1 1 px2 + y2dx . )px2+ y23?/?D(1 6 x 6 1,a 6 y 6 a)Y(a0).?n1,k lim y0 Z 1 1 p x2+ y2dx = Z 1 1 ? lim y0 p x2+ y2 ? dx = Z 1 1 |x|dx = 1, 10 (3) lim n R1 0 dx 1 + ? 1 + x n ?n. )? f(x,y) = 1 1 + (1 + xy) 1 y ,0 6 x 6 1,0 0?y0 (1,1),y y0 ,y0+ (1,1).?f(x,y) = 1 (1 + y sinx)2? yf(x,y) = 2sinx (1 + y sinx)3 3?D( 6 x 6 ,y0 6 y 6 y0+ )Y.?n2,y (1,1),k F0(y) = Z h y dx (1 + y sinx)2 i dx = 2 Z sinx (1 + y sinx)3dx. (3)F(y) = Rb+y a+y sinxy x dx. )lim x0 sinxy x = lim x0 sinxy xy y = y. (0,y)?m?:.?Ym?.d,? sinxy x ? y ? sinxy x ? = cosxy3R2 Y,a + y?b + y?y?Y?.?n4,k F0(y) = Z b+y a+y y ? sinxy x ? dx + siny(b + y) b + y siny(a + y) a + y = Z b+y a+y cosxydx + siny(b + y) b + y siny(a + y) a + y 11 1 ysiny(b + y) siny(a + y) + siny(b + y) b + y siny(a + y) a + y = ?1 y + 1 b + y ? siny(b + y) ?1 y + 1 a + y ? siny(a + y). 5.?F(x) = Rh 0 nR h 0 f(x + + )d o d(h 0),?f(x)Y,F00(x). )?t = x + + ,Kdt = d.k F(x) = Z h 0 nZx+h x+ f(t)dt o d. ?n2?n4,k F0(x) = Z h 0 n x Z x+h x+ f(t)dt o d = Z h 0 n x Z x+h x+ f(t)dt o d = Z h 0 nZx+h x+ xf(t)dt + f(x + + h) f(x + ) o d = Z h 0 f(x + + h) f(x + )d. ?u = x + + h,Kdu = d;v = x + ,Kdv = d,k F0(x) = Z x+2h x+h f(u)du Z x+h x f(v)dv. 2?n4,k F00(x) = f(x + 2h) f(x + h) f(x + h) + f(x) = f(x + 2h) 2f(x + h) + f(x). 7.y;ef(x)3ma,bY,Kx a,b,k Z x a nZy a f(t)dt o dy = Z x a f(t)(x t)dt. yyy?“?m?Ox?,x a,b,k ?Zx a nZy a f(t)dt o dy ?0 x = Z x a f(t)dt ? ?R x a f(t)(x t)dt ?0 x = Rx a ? xf(t)(x t) ? dt + f(x)(x x) = Rx a f(t)dt. u,x a,b,k ?Zx a nZy a f(t)dt o dy Z x a f(t)(x t)dt ?0 x = 0. 12 d Rx a nR y a f(t)dt o dy Rx a f(t)(x t)dt = C(). -x = a,w,C = 0,u,x a,b,k Z x a nZy a f(t)dt o dy = Z x a f(t)(x t)dt. 555?K?,?SK8.4117K. 8.y:ef(x)3a,AY,Kx a,A),k lim h0 1 h Z x a f(t + h) f(t)dt = f(x) f(a). yyy fi ?f(x)3a,AY,x a,A),t a,x,t + h a,A). ?t + h = y,dt = dy.k Z x a f(t + h)dt = Z x+h a+h f(y)dy = Z x+h a+h f(t)dt u, lim h0 1 h Z x a f(t + h) f(t)dt =lim h0 1 h ?Zx a f(t + h)dt Z x a f(t)dt ? =lim h0 Rx+h a+h f(t)dt Rx a f(t)dt h ?0 0,?K ? =lim h0 ?R x+h a+h f(t)dt Rx a f(t)dt ? h0 =lim h0f(x + h) f(a + h) = f(x) f(a). 555?K,?y?SK8.4111K. 9.?e?e?: I(a) = Z 2 0 ln(sin2x + a2cos2x)dx,a 0. ) : 0 0).u,R + 0 etxsinxdx3a,+)?. 11.ye?3?m?: (1)R + 0 yeyxdx,0 6 y 6 1. ?A0N?,o?3?y0 0,1, ? ? ? R+ A0 y0ey0xdx ? ? ? = ey0A0?u?u,?.?L?y0= 1 A0 0,1=?. yyy0= 1 3 0,A 0,A0 A,y0= 1 A0 0,1,k ? ? ? Z + A0 y0ey0xdx ? ? ? = ? ? ?e y0x|+ A0 ? ? ? = ey0A0 =e 1 A0A0 = e1= 1 e 1 3 = 0, =R + 0 yeyxdx30,1?. (2)R + 1 y (x+y)2dx, 0 0,A0 A,y0= A0 (0,+),k ? ? ? Z + A0 y0 (x + y0)2dx ? ? ? = ? ? ? ? y0 x + y0 ? ? ? + A0 ? ? ? = y0 A0+ y0 = A0 A0+ A0 = 1 2 1 3 = 0, =R + 1 y (x+y)2dx3(0,+)?. 12.?u ,:(b,u)?f(x,u)?:.?R b a f(x,u)dx 3m,?, 14 Q?,?y?R 1 0 (1 x)u1dx 3ma,+)(a 0)?,3 m(0,+)?. ?R b a f(x,u)dx 3m,?(:(b,u)?:) 0, 0( 0( 0,u a,+)(a 0),?“(0 0, : 0 0, 0( 0. )?f(x,a) = ex 2 eax 2 x .f0 a(x,a) = xeax 2. ?a 0,k lim x0+ f(x,a) = lim x0+ ex 2 eax 2 x = 0, 15 ,f(x,a)3(x,0)?Ym?.f(x,a)?f0 a(x,a)3?D(0 6 x 0( 1,a , Z + 0 a ?ex2 eax 2 x ? dx = Z + 0 xeax 2dx 3,?.fl ?,a ,k |xeax 2| 6 xex2 fi ?R + 0 xex 2dx,KR+ 0 xeax 2dx3,?. ?n10,a ,k I0(a) = Z + 0 a ?ex2 eax 2 x ? dx = Z + 0 xeax 2dx = 1 2ae ax2? ? + 0 = 1 2a. l?,I(a) = R da 2a = 1 2 lna + C -a = 1,fi ?I(1) = 0,k I(1) = 1 2 ln1 + CC = 0. u,a 0,k I(a) = 1 2 lna. 15.A?e?,: I = Z + 0 eax ebx x sinxdx,a 0,b 0. yyy?L? eax ebx x sinx = sinx Z b a exydy = Z b a exysinxdy. l?, I = R+ 0 eax ebx x sinxdx = R+ 0 nR b a exysinxdy o dx. fi ?f(x,y) = exysinx3?D(0 6 x 0,m 1. yyy?t = xn,dx = 1 nt 1 n 1dt.k Z + 0 xmex ndx = 1 n Z + 0 t m nt 1 n 1etdt = 1 n Z + 0 t m + 1 n 1etdt =1 n ?m + 1 n 1 + 1 ? = 1 n ?m + 1 n ? . 17.?Be?: (2)R + 0 x2 1 + x4dx. )?t = 1 1 + x4,x 2 = t 1 2(1 t) 1 2,dx = 1 4t 5 4(1 t) 3 4dt.k 17 Z + 0 x2 1 + x4dx = 1 4 Z 1 0 t 3 4(1 t) 1 4dt = 1 4B ? 1 3 4,1 1 4 ? = 1 4B ?1 4, 3 4 ? = 1 4 ?1 4 ? ?3 4 ? ?1 4 + 3 4 ? = 1 4 ?1 4 ? ?1 4 + 1 2 ? = 1 4 2 2 41 ?2 4 ? (A18?“) = 22 ?1 2 ? = 22. (4)R 1 1(1 x 2)ndx,n N. )?t = x2,dx = 1 2t 1 2dt.k Z 1 1 (1 x2)ndx = 2 Z 1 0 (1 x2)ndx(1 x2)n? =2 1 2 Z 1 0 (1 t)nt 1 2dt = B ? 1 1 2,n + 1 ? = B ?1 2,n + 1 ? = ?1 2 ? (n + 1) ?1 2 + n + 1 ? = n! ?1 2 ? ? n + 1 2 ? n 1 2 ? . 3 2 1 2 ?1 2 ? = 2n+1n! (2n + 1)! = 2(2n)! (2n + 1)!. 18.y:(2) In= R 2 0 sinnd = (2m 1)! (2m)! 2, n = 2m, (2m)! (2m + 1)!, n = 2m + 1. yyyd“(13),k In= Z 2 0 sinnd = ?n + 1 2 ? ?1 2 ? 2 ?n 2 + 1 ? = 2 ?n + 1 2 ? ?n 2 + 1 ?. ?n = 2m?,k(d(1) I2m= 2 ? m + 1 2 ? (m + 1) = 2 (2m 1)! 2m m! = (2m 1)! (2m)! 2 ?n = 2m + 1?,k(A(1)?(J) I2m+1= 2 (m + 1) ? m + 1 2 + 1 ? = 2 m! (2m + 1)! 2m+1 = (2m)! (2m + 1)!. 18 555dK8.4?7.p?A?B,?. 19.y;?E(k) = R 2 0 p 1 k2sin2d?v? E00(k) + 1 kE 0(k) + E(k) 1 k2 = 0,0 0,(x,u) R,k|f(x,u)| 6 M. u ,u+ M u ,k(u+ M u) = Rb(u+Mu) a(u+Mu) f(x,u+ M u)dx = Ra(u) a(u+Mu)f(x,u+ M u)dx + Rb(u) a(u) f(x,u+ M u)dx + Rb(u+Mu) b(u) f(x,u+ M u)dx. ? ? ? ? Ra(u) a(u+Mu)f(x,u+ M u)dx ? ? ? 6 M|a(u+ M u) a(u)|, ? ? ? Rb(u+Mu) b(u) f(x,u+ M u)dx ? ? ? 6 M|b(u+ M u) b(u)|. fi ?a(u)?b(u)3uY,d?“,klimMu0 Ra(u) a(u+Mu)f(x,u+ M u)dx = 0 ?limMu0 Rb(u) b(u+Mu)f(x,u+ M u)dx = 0. 20 ?n1,qklimMu0 Rb(u) a(u) f(x,u+ M u)dx = Rb(u) a(u) f(x,u)dx. u,limMu0(u+ M u) = Rb(u) a(u) f(x,u)dx = (u), =(u) = Rb(u) a(u) f(x,u)dx3uY,l?3m,Y. 22.yn7. n7.ef(x,u)3?D(a 6 x 0)Y,?F(x,u) = Rx a f(t,u)dt