一、填空題（10220分1f(xy,y)x2y2x

f(x,y)

x2(1y2。(12、

ln(1x2y2

(x,y）(0,0)arcsin(x2d

(t2ex)f(t)dt=sinxcos2xf(cosx)

dxcos41

exe2

=

5y''2y'15y0yCe3xCe 6zxe2y,P(1,0)P(1,0)Q(2,-1)22

22

(zxcoszycos

(2y

2xe2

1 12222272x2y4z50,xy2a=–2,b=2/3,則此平面方程為x3yz 8D

x||y|1，則xD

y)dxdy=43(xy)dxdy81dx1xxdy81x(1x)(x D

2n

n9、若級(jí)數(shù)(unSnSn1

的部分和數(shù)列為

，則

,n2, 10、若級(jí)數(shù)un條件收斂，則級(jí)數(shù)

的斂散性 發(fā) 二、選擇題（5210分1、若xf(t)dt1f(x)1，且f(0)=1，則f(x)=（ 1e

1e2

e2

2、設(shè)u(x,y)在平面有界閉區(qū)域D

0，則u(x,y)的（ DDDDDD3、設(shè)f(x,y)在(x0,y0)處偏導(dǎo)數(shù)存在，則f(x,y)在該點(diǎn)（ 極限存 (B)連(C)可 (D)以上結(jié)論均不成4f(x,y)D：x2y2a2a0a2 f(x,y)dxdy的極限為 a2D(A)不存 (B)f (C)f (D)f(1,5、下列結(jié)論正確的是（ nn若u2收斂，則nn

nn若u2收斂，則nn

若

收斂，則收斂，則u 不一定收n若

收斂，則收斂，則u 必發(fā)n三、計(jì)算題（8756分11、計(jì)算01

1 1 0解：exdxt x2tetdt2et(t1) 02、求方程(x1)dy2yex(x1)3yy2xy

yex(xx1[(x1)2

x1dxC](x1)2[exzf(xy,y2

2z3

，求x2yzx

f1

zxx

f1",zy

f1'2yf2x24y=f(x2

arctanydy

[lnx x2y2]' 2y2]'2x2yy'x[lnx 2x2y2 x2y )'yx

()x(1y(x

xy'yx2xy'yx2

xyyx2

解得dyx x15、計(jì)算01

sinxdxyxy解： dx

1dxxsinxdy

1sinx(x2ysin1 ysin1

0(xcosxsinxcosx)1106f(x,y)Dxy|x2y2yx0}f(x,y)

8f(x,1x1x2f(x,解：Af(xD

A

1x2y2dxdy8A

2d

1r2rdr 1 3

1

2A3

2(1r200

d3

2(cos3 f(x,y)

A1

1x21x2

，其中zxy，yx,x1z08、求冪級(jí)數(shù)(2n1)xnRlim

2n n

n2(n1)x1時(shí)(2n1)(1)n故收斂域?yàn)?1,1

x(2n1)xnt x

t2n＋1)'( )1t1t(1t2

1(1

四、應(yīng)用題（8分（萬(wàn)元）之間的關(guān)系有如下經(jīng)驗(yàn)LR(x1x2) 8x

318x1 1514x132x2L138x

318x1

x1x2五、證明題（6分 20f(x)dxxf(y)dy

f解: 0f(x)dxxf(y)dy0

f(x)f( 0f(y)dy0f(x)dx0f

f( 20dxxf(x)f(y)dy