微積分上習(xí)題5解答

習(xí)題5解答（解（1）當(dāng)0t1時(shí)，因?yàn)閘n(1tt，所以|lnt|ln(1t)]ntn|lnt|，因 1|lnt|[ln(1t)]ndt1tn|lnt|dt0（2）由（1）知0un

1|lnt|[ln(1t)]ndt0

1tn|lnt|dt0因 1tn|lnt|dt1tnlntdt

1tndt

n1 (n所 lim1tn|lnt|dt

0nn從 limu0n

n(n解dyexxetf(t)dtf(xdyexxetf(t)dt

f(x) y

f(xyexxetf(t)dtdyy

f(x

y(x)

e

xetf(t)dt0

xetf ex

xexx

證法1ylnxy1y x

0ylnx在(0,x1x2,xn(xi0,i1,2,n1n

ln

ln(1n

xi 1 1 于是，特別地取tinxi

f(ti

nn

f() f()]， 11n，由定積分的定義有0lnf(x)dxln0f(x)dx112：將[0,1區(qū)間n 0f(x)dxlimf( n f(x1)f(x2)f(xn)n1[lnf(x)lnf(x)lnf(x

nnf(x1)f(x2)f(xnn1lnf(xin1

ei nn1n即f(xin1n

lnf(xi i

i(i1,2,,n)， fn n

i)

lnfnein

innn，得

1 1f(x)dxe0lnf(x)dx 0

f(x)dx1

2

2xx

2xarctan(1x2

x01cosxxsin x0sinxsinxxcos

)x02sinxcosx214 解法２

arctan(1t)dt]dux(1cosx)

2

xxx2xarctan(1

2 2lim

3x

2x x

xxx3 xxxx解：原式x

f(t)dt0tf(t)dtx0f(xt)dt

xxtx

f(t)dt0tf(t)dtx0f(u)duxxlim

f(t)dtxf(x)xf

xf(

f(u)duxf

在0和x之間

xf()xfx0x f 1

f(0)f 解：在fx)f1t)dtxtcostsintdtx 0sintcosf1[f(x)]f(x)xcosxsinx xf(x)xcosxsinx

f(x)cosxsinx

x(0,]4 f(x)sinxcosxdxln(sinxcosx)Cf(0)0，于是C0f(x)1（1）

x ]4

f(x)dx0t0

f(x)dx2t2

tft

f(x)dx002sx2，則有02

f(x)dx0f(s2)ds0f(s)ds

f(x)dxtt

f(x)dx

f(x)dx2t2

f(x)dx020

f(x)dx

f(x)dx（2）由（1）知對(duì)任何的t有

f(s)ds0f(s)ds0f(s)dsa，則G(x20f(t)dtax xG(x2G(x)

f(t)dta(x2)20f(t)dt2所以G(x2

f(t)dt2a20f(t)dt2a0t2（1）

f(x)dx2F(tf(t2f(t0F(tF(tF(022F(02

f(x)dxF(t)2t2

tf(x)dx，即2

f(x)dx

f(x)dx（2）由（1）知對(duì)任意的t2

f(s)dsx

f(s)ds

f(s)dsa，則G(x

f(t)dtax，G(x2)

f(t)dta(x2)xG(x2)2f(x2a2f(xaG(x)2f(xa，所以[G(x2G(x)]0，從而G(x2G(x是常數(shù)，即有G(x2)G(x)G(2)G(0)0，所以G(x)2的周期函數(shù)。x2x解f(x的定義域?yàn)?,)f(x)x21

et

2xdtx

tet

dtx2xf(x)

et

dt

t

t

2xx

et

f(xx0,1x01f—0+0—0+ff(x的單調(diào)增加區(qū)間為(1,0及(1,，單調(diào)減少區(qū)間為(,1及(0,1；極小值為f(1)0f(0)1tet2dt1(1e1。 解f(x)xt(xt)dt1t(tx)dt1x3x1 f(xx210x2

2，x2

（舍去2因f(x)2x (0x1)故x 2為f(x)的極小值點(diǎn)，極小值f

2)1(1 2y

f(x在(0,12

f(x)x21f(x在2

2內(nèi)單調(diào)遞減，在2

2解：當(dāng)1x0F(x)

x(2t3t2dtt21t3|x1x3x21當(dāng)0x1

F(x)1f(t)dt1f(t)dt

f(t)dt1(2t2

)dt0(et1)2(t21t3)

)1

|x x x

et

x et1x

0et1

1

ex

0et(et

ex

lnet12

ex

exex

1x3x21

1xF(x)

ex

exex

。0xxx解x0F(xf(t)dt0dt0xxx當(dāng)0x1時(shí)，F(xiàn)(x) f(t)dt0dttdt1x2x x當(dāng)1x2時(shí)，F(xiàn)(x) f(t)dt0dttdt(2t)dt1x22x1x x2F(xf(t)dt0dt0tdt1(2t)dt20dt111故F(x)2

x21x22x2

x0x1xx解f(xxln(1t)dtf(xln(1x)f(10x x1f(x)dx

xxxxx從而xxx

20f

xf(x)|0

xf(x)dx] 11

xx

xlnx1|040x1dx4ln240x1dxx 1u x令u

x

0x1dx20u21du2(uarctanu|022x所 1f(x)dx824ln2x0

證：令t ，則x

2dt

2du

2dtu

011

x1

1

從而01

dt01

2dt01

2dt

1

2dt

1

20arctanx|02 證In4tannxdx4tann2xsec2x 4tann2xdtanx4tann2 0 0n1

tann1x|

4tann2xdx0

n

In21Inn1In21

0 4tan5xdxI5 I I)0

4tan 4

00

1ln21 證F(t)af(tf(x)]dx3tf(xf(t)]dx。因?yàn)閒(x)0，因而f(x)在[ab上單調(diào)遞增。b所以F(a3af(xf(a)]dx0bbF(baf(bf(x)]dx0bF(t在[ab上顯然連續(xù)，故由零點(diǎn)定理知，至少存在一點(diǎn)abF()0 af(f(x)]dx3f(xf()]dxF(tf(t)[(ta3(bt0tab)，F(xiàn)(t在[ab上單調(diào)遞增，因此僅存在一點(diǎn)ab)證：由于1x2p1，即(1xp)(1xp)p00x1，時(shí)，有(1xp)

1x

1

1(1xp)dx dx1dx 1(1xp)dx[x0

1

x1p

010所

dx11 01xaaa證F(xxf(t)g(t)dt]2xf2t)dtxg2t)dtF(a)0aaaaaa F(x)2f(x)g(x)xf(t)g(t)dtf2(x)xg2(t)dtg2(x)xf2aaax[f2(x)g2(t)2f(x)g(x)f(t)g(t)f2(t)g2ax[f(x)g(t)f(t)g(x)]2dt0aF(x在[ab上單調(diào)遞減，因此，當(dāng)abF(b)F(a)0aaa從而[bf(t)g(t)dt]2bf2t)dtbg2t)dtaaa [bf(x)g(xdx]2bf2(x)dxbg2(x)dxa

x1

cost

1 2證

dt

dcos

2

costtcosx2x1cos(x1)2cosx2x

1x1cost2d x

2 1(11)1x1d2 x 2 1

2x1)1

12x(x 2x(x 證xundxdu2

f(sinx)dx0

))du2

f(sinu)dun

f(cosu)dun

f(cosx)dx0

f(cos(u

))du2

f(cosu)dun

f(sinu)dun 00由于2f(sinx)dx2f(cosx)dx001解：設(shè)切點(diǎn)A的坐標(biāo)為(x,y)，則切線方程為yy (xx)，1x x1將點(diǎn)(0,1xe2y2

S

lnxdx 2

1)2xlnx

dx

12e

1

12

V3

xdx 1)

1)(x5

x2xlnx2x)

3

a解：Vxa

y2dx

x3dx

，a a1 a1

1 a1

7

70Vya307

x2dya305

y6dya3 由

7

10 ，解得a775解（1）y1(x1L 1

1

e21(s1(

dx21(xx)dx2(

1

1

1

lnx)dx

xlnx

)|116

e1

lnx)dx

xlnx2

2x)|112

7)

3(e21)(e2。4(e3解

3(x2x)f(x)dx3(x2x)df(x)(x2x)f(x)|3

3(2x1)f

30020f3(2x1)df(x)(2x1)f(x)|330020f0[7222f(x|3162f(3f(0164200解S(xSy，

x[ex1(1ex)]dx

[lny(y)]dyy yx1

0(2e2)dx1[lny(y)]dyx

1ex1[lny(y)]dy yex

1ex1[x(ex)]ex 于是(ex)x

，從而y)121

y2

12故曲線C3x

y2

12解（1）x0ylnx在點(diǎn)(x0lnx0x0ylnxx00

(xx0由該切線過原點(diǎn)知lnx010x0e1所以該切線的方程為y xeDA1(eyey)dy1e1 1（2）y

xxxexeeV1e2 ylnxxxexe2V1(eey)2dy20因此所求旋轉(zhuǎn)體的體積為VVV1e21(eey)2dy(5e212e3 解ysinx在[0,x

l20

1cos2xdxy

f(x

y

2sin2故s40

1cos2d2

221sin2d20令2

ts

420220

1cos2t(dt)

22221cos2tdt

4l222

2l（1）

V(a)

y2(x)dx0

xaadx

xda

[xaa[xaa