数学分析(Ⅱ)期末考试题(内附答案)

一、选择题(每题3分,共15分)

1、下列函数在[1,e]上满足拉格朗日中值定理条件的是( )

1 A、ln(lnx) B、lnx C、 D、ln(2x)

lnx2、设f(x)在区间I上二阶可导,则下述命题正确的是( ) A、若xI,f(x)0,则f(x)在区间I上严格单调递减. B、若xI,f(x)0,则f(x)在区间I上严格单调递增. C、若xI,f(x)0,则f(x)在区间I上为严上凸函数. D、若xI,f(x)0,则f(x)在区间I上为严下凸函数. 3、( ) 是函数sin2x的一个原函数.

11 A、2cos2x B、cos2x C、cos2x D、sin2x

224、积分中值定理f(x)dxf()(ba)中( )

abA、是[a,b]内任意一点

C、是[a,b]内唯一一点

B、是[a,b]内必定存在的某一点 D、是[a,b]内中点

15、若an收敛,则级数an( )

nn1n1 A、一定收敛 B、一定发散 C、可能收敛,也可能发散 D、收敛且其和为0 二、填空题(每题3分,共15分)

11、曲线yxln1的渐近线为 ．

x2、若F(x)f(x)，则f(2x3)dx ．

3、曲线yex,yex及x1所围面积可用定积分表示为 4、sinxdx11x21．

5、若级数an收敛，则级数n11 （收敛、发散或不能判断）． an1n三、求函数f(x)(t21)etdt的极值(8分).

0x四、求解下列各题(每题8分,共32分)

edt1、求极限limxt20x2x0edt2t2.

2、求不定积分3、求定积分2x1x3dx.

2cosxcos3xdx.

4、计算由抛物线yx2，直线x2及x轴所围成的平面图形面积及该图形绕x 轴旋转一周

得到旋转体的体积。

五、(8分) 确定下列幂级数(n1)xn的收敛域,并求其和函数.

n0六、(7分)证明函数项级数n12sinnx3nx44,x(,)一致收敛.

七、(7分) 证明:若级数an和bn都收敛,则级数|anbn|,2n1n1n1n1an也收敛. n八、(8分) 证明不等式:0

111,其中x0.

ln(1x)x参 考 答 案 及 得 分 要 点 评分标准（得分） 一、选择题（每题3分，共15分） 1、B 2、C 3、C 4、B 5、B 二、填空题（每空3分，共15分）

111、x1,y1. 2、F(2x3)c 3、exexdx 4、0 5、发散

02三、(8分) 求函数f(x)(t21)etdt的极值.

0x解 函数f(x)的定义域为(,)

························································ 2分 f(x)(x21)ex,f(x)(x22x1)ex ·

令f(x)0,求得驻点x1,x1. ··············································································· 4分

由于

f(1)2e0,f(1)2e0 ········································································· 6分 所以函数在x1处取得极小值

················································································· 7分 f1(t21)etdt1 ·

01在x1处取得极大值

41. ·········································································· 8分

0e四、求解下列各题(每题8分,共32分)

f1(t21)etdtxt22x1edt1、limx0x0edt2t2lim2ex2xex002x2etdt2 ············································································ 2分

lim2etdte22exx22x ················································································· 3分

2xe1lim ···························································································· 7分 xxxlimx2 ······················································································ 5分

0 ···································································································· 8分

2、令t6x,则xt6,dx6t5dt,于是 ········································································· 2分

t35····················································································· 4分 dx6t13x1t2dt ·

x1·························································· 5分 6t6t4t21dt ·21t1·························································· 6分 6t6t4t21dt ·21t111t11··········································· 7分 6t7t5t3tlnc ·532t17175116111x116x6x6x2x6ln1·································· 8分 c ·7532x613、22cosxcosxdx23202cosxcos3xdx ························································ 1分

202coxs12······························································· 2分 sxind x 20cosxd(cox ·s·)························································· 4分

3224········································································ 6分 cosx ·

304 ·························································································· 8分 34、Sx2dx8 ········································································································· 4分

03232Vx4dx ································································································· 8分

052五、(8分) 确定下列幂级数(n1)xn的收敛域,并求其和函数.

n0n11,故级数的收敛半径为R1.当x1时, 解 由于limnnnn(n1)x1nn0n0n(n1),而

lim1(n1)0,故当x1时,幂级数发散,于是级数的收敛域为(1,1). ············ 4分

设f(x)(n1)xn,x(1,1),则

n0

x0f(t)dt(n1)tdtxn1nn00n0xx······························· 6分 ,x(1,1) ·

1x 1xf(x), x (1,·1)· ····························································· 8分 21x(1x)六、(7分)证明函数项级数n1sinnx3nx44,x(,)一致收敛.

解 因x(,)有

sinnx3¥11£4/3,而å4/3收敛 ·········································································· 6分 44nn+xn=1n由M判别法知n1sinnx3nx44在(,)上一致收敛. ···················································· 7分

七、(7分) 证明:若级数an和bn都收敛,则级数|anbn|,22n1n1n1n1an也收敛. n证明 由于

|anbn|122anbn2（,anbn)222a2na112····················· 5分 bn2n,an22 ·

n2n而级数级数an,bn,n1n11均收敛,由比较判别法知,级数 2n1n（anbn),|anbn|,2n1n1n1ann也收敛. ············································································· 7分

111,其中x0.

ln(1x)x八、(8分) 证明不等式:0证明 设f(x)ln(1x),则f(x)在[0,x]上满足拉格朗日中值定理的条件,于是0,1使得

ln(1x)f(x)f(0)x,01 ··························································· 4分 1x 11x1 ················································································ 6分

ln(1x)xx01

0111 ························································································· 8分

ln(1x)x