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五(00计算订x2+ydx+=y,其中2:(x+D2+(y-D+4=10≥D)取外侧 解设20y=1左侧D(x=03+451则原式=手 f=[ex+y=dv= fcx+ydr=2 ae 2( e sino+rsin0sino+2)r'sinodr 〔wmp+mm+mp)p 原式=兀+2π= 另解设20y=1在侧D(-2+=则原式=手-手 =-d=dx=-2 +y+z)dv,故原式=2(x+y+)dv+2兀 xdv=xdx du 2x-x2,y≥1 ydv=ydx dedx=y2.(2y-y2) 兀,D,:(x-1) 原式 六、(10分)设正项级数∑an收敛,且和为S试求 ()lima+2 a2 t.+ nan;(2)∑4+2n+…+mn 解()4+2a2+…+nan=Sn+Sn-S+S2-S2+…+Sn-Sn1 n S,+s +s S,+s lim a,+2a n B=S-S=0; (2)a1+2+…+mna1+2a2+…+mna1+2a2+…+nap n(n+ n n+ a1+2a +na, a1+2a2 +na+(n+lan+ltan+l n+1( 1), . 4 (10 ) d d d d d d , : ( 1) ( 1) 2 五、 分 计算 2 + 2 + 2 其中 S - 2 + - 2 + = ³ 取外侧 ÚÚ S y z x y z y z x z x y x y π. 3 25 π 2π 3 19 π, 3 19 sin )d 3 2 sin sin 4 1 cos sin 4 1 4 d ( 2 ( )d 2 ( )d 2 d d 2( cos sin sin sin 2) sin d d d 2π, 1, . 4 : 1, , : ( 1) π 0 2 2 π 0 1 0 2 π 0 π 0 2 2 0 0 0 0 0 \ = + = = + + = = + + = + = + + = - = - S = - + £ = - Ú Ú ÚÚ ÚÚÚ ÚÚÚ Ú Ú Ú ÚÚ ÚÚ ÚÚ ÚÚ S+S S S+S S 原式 解 设 左侧 则原式 q q j q j j j x y z v x y v q j r q j r q j r j r z x z y D x V V D π. 3 25 π 2 π 3 11 π 3 8 2 , 4 π, :( 1 ) 6 11 d d d d π 2 ( 2 ) d 2 , 1 , 4 π, :( 1 ) 3 4 d d d d π ( 2 ) d d d 2 π, 2 ( ) d , 2 ( ) d 2 π. 1 , . 4 : 1 , , :( 1 ) 2 2 2 2 0 2 2 1 2 2 2 2 0 2 2 0 2 2 0 0 0 0 0 \ = + + = = = × × - = - + £ - = = - = - + £ - ³ = - = - = + + = + + + S = - + £ = - ÚÚÚ Ú ÚÚ Ú ÚÚÚ Ú ÚÚ Ú ÚÚ ÚÚ ÚÚ ÚÚÚ ÚÚÚ ÚÚ ÚÚ S S + S S + S S 原式 故原式 另解 设 左侧 则原式 y y z y v y x z x y y y y D x x x y z x v x x y z x x x x D y z x x y z v x y z v z y D x y V D x V D D V V y x . ( 1) 2 ; (2) 2 (1) lim (10 ) , . 1 1 2 1 2 1 Â Â • = Æ• • = + + + + + + + n n n n n n n n a a na n a a na a S L L 六、 分 设正项级数 收敛 且和为 试求： . 1 2 2 ( 1 ) 1 2 2 ( 1 ) 2 ( 2 ) 0 ; 2 lim , 1 1 2 ( 1 ) 1 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 1 1 2 1 1 2 1 2 1 + + Æ• - - - + + + + + + + - + + + = + + + + - + + + = + + + + = - = + + + \ - × - + + + = - + + + = - + - + - + + - = + + + n n n n n n n n n n n n n n n n n n n a n a a na n a n a a na n a a na n a a na n n a a na S S n a a na n n n S S S S n S S S S n S S S S S S S n a a na L L L L L L L L L L 解