高數(shù)答案第五章練習(xí)二

二．1．0；2．1，1；3．1， 2 3

3 nlim n1xn 2

1xdx n11xn2

lim1x 解：由分部積 ，原式π

1ln2sin sin解：原式

cosxdx 0

sinxdsinx 2

sinxdsin232 (sinx)3

π230 (sinx)2π2302

43解：原式 4

dx

2ln(1x32)4

4ln3 1x 2解：所求積分

2f2

22

costcos3tdt 0

1

dt πcost2

0 20

2|sint|dt2

costsin2

20 20

costdcost

π33π332π201cos

dt01cos

d(1cost)ln(1cost)2ln2所求積分2ln23

πsec2π tan2tsec

πcosπsin2

π12 12解：令xtant，原式 4

dt 4

dt=

34

230解：原式x2ex|10

1xexdxe12(xex|11ex 0e12(e1ex|1)25e10 1cos2x2cos2 π 1 01cos

02cos

d(tanx)0

[xtanx]4

4tan2π1[lnsec

et et2ln

du2arctan 2π2 etet

e1(u2 3ex3exex故 ,所以xex411解：因?yàn)樵赱e21]上lnx0,在[1,e2上lnx0,11e2|lnxx是x

dx

xx

e2lnxx

dx

lnxd

x) lnxd(2x1x e2x

xln

1

dxxx

xln1

x xe

4e

8(1e111.

0(xx)exdx=0

(xx)exdx(xx)ex xexdx2(xex

exdx)2ex02 xx

t，則原式＝

(t2

t2

t，則原式＝0d(1t211(t21x0x1d(x1

x 1

2)10arcsin(2x1)x(1x(1

(1)2

(x

12

02πππ

12或 x(1x)2

x1x

(1uu2

201

2(π0)π2

則x u 1u1exsin

exsinnx1n1excos exsinnx1nexcosnx 21

sin1 e1sinnne1cosn exsinnxdx0進(jìn)而lim1exsinnxdx0n

n2f''(30（二“直線l1l2C在(0,0),(32處的切線，其交點(diǎn)為(24)明三個問題：①f(0)0,f(3)24(0,0處切線的斜率為202，即f'(0)2(3,2)處切線的斜率為422，即f'(322（三）f(0)0，這樣一來，問題就轉(zhuǎn)化為：已知f(0)0，f(3)2，f'(0)2，f'(32f''(3)0，求3x2xfx)dx200四．1．證：由已知有f(x)f(xF(x)0(x2tf(t)dt，令ut F(x)0(x2u)f(u)du0(x2u)f(u)du0(x2t)f(t)dtF(x)ππ

I

u

e(uπ)2

(uπ)du

e(xπ)2cos2xdx[0,π上有ex2cos2x、e(xπ)2cos2x連續(xù)，ex2cos2 ex2cos2x不恒等于e(xπ)2cos2x，II

exπ)2cos2x 1 xt

1x

1 t

1t4dt 1x

x dx 1 設(shè)xπtdxdt,x0tπ,xπt 0 2f(sinx)dxπfsintdt2f(cost)dt2f(cos (2)xπtdxdtx0tπxπt 0xf(sinx)dxπ(πt)f[sin(πt)]dt0(πt)f(sin π0f(sint)dt0tf(sint)dtπ0f(sinx)dx0xf(sinπxf(sinx)dxππf(sin0πxsin01cos2

sin dxπ 01cos2 201cos2 π π πarctan(cosx)0 2 4 易 易 I0 2dx

,I1 2sinxdx1,當(dāng)n20In

(n1)2sinn2x(1sin2x)dx(n1)2sinn2xdx(n1)2sinn (n1)In2(n Inn1