2009級(jí)一元微積分期末考題答案

答一．填空題（每空3分，共15空（請(qǐng)將答案直接填寫(xiě)在橫線上lnln

dx

1x

ln|1x|lnx 1cos2

1arctan2222

tanxarctanxdxx xarctan

dxx

1

xf(x)dxarctanxC，則1dx f2

x4C4C 2

(1x)cos1sin2

dx 2dx2t dx

edt 2xex4tf(xf(0)0F(x)xt

f

x0F(xxk0無(wú)窮小，則k 將(x3)2y21繞y軸轉(zhuǎn)一圈，則所得圖形圍成的體積為 答案：62設(shè)m0，且廣義積分 收斂，則m的范圍為x xmxm n冪級(jí)數(shù)2n(5)n的收斂域n(5,(1)nsin級(jí)數(shù)n

n條件收斂，則參數(shù)p的范圍 np1p

0

xet2dt0

x，xy'exexy，的通解

1

exxdy(x2y)dx0滿(mǎn)足y(1)0的解 yxxy2xy2初值問(wèn)題y(0)1,y(0)0的解 y二．計(jì)算題（每題10分，共40分 p的范圍，使得1sinxlnpx

sinxlnpx1sinxlnpx

sinxlnpxx1附近，sin 1~11 ，所以?xún)H當(dāng)2p0時(shí)

pxlnpp

x(x

xlnp 5 x,sinxlnpx~xlnpxp成立,所以只需要考慮廣義積分

xlnp

lna 的收斂性。因?yàn)?/p>

xlnpxln2upp1時(shí)廣義積分

xlnpxdx收斂

5 xlnp計(jì)算擺線xtsint,y1 t[0,2]，繞x軸旋轉(zhuǎn)體的體積和表面積V2y2dx2(1cost)3dt2(13cost3cos2tcos3t)dt5

5 t S2 yx2y2dt2(1cost)22costdt8 dt 5(1)n求級(jí)數(shù)n

2 (1)n

1S(x)n

，則

S 2n

Sn2xn

xSdxnxn 3 n1xSdx

0

nxSdxdxxn xx0

n

0x0

n

1

3 1xSdx xSdx x0

0

(1S(x)x

1x,(1x)3

S(x)x(1x)(1)nn2 1

S

4n 設(shè)f(xC(0,)，且對(duì)任意x0xf(tx)dt2f(t)dtxf(xx4 f(1)0f(x00解：令utx，則原方程可化為xf(u)du2xf(t)dtxf(x)00f(x)2f(xf(xxf(x4x3從而得一階線性O(shè)DE，f(x)2f(x)4x2 5xf(xx2C4x)f(1)0，得C所以f(x)x2(44x) 5三．證明 1.（8）y

fx)xlnx(n1n1fxyf(xxyyxexlimy(00xlnx的定義域?yàn)?/p>

,

n0(n1)(n所以函數(shù)f(x)的定義域?yàn)?2fxy1lnx

(nn0(n1)(nxyxxlnx

(n2)xn2n0(n1)(nxyyx

(n2)xn2

n0(n1)(n n0(n1)(n xn xx(n1)!xx(1)1n!x

4 limyxlimxlnxlim

2

x0n0(n1)(n2（71f(x)dx21f(x)3 證明：記(x0f(t)dt0f(t)

(x)2f

0f

f(x)f

f(t)dtx0x

2記(x2xf(t)dtf2 (x)2f(x)1f

2f(x在[0,1f(000f(x