内容讲义说明

【解】应选(B).f(xsinxf(x)cosxC故选fxx0fx在(,上不存在原函数.fxI上连续fxIfxI上不连续fxIfxI上有第一类间断点fxIxt2dt1x31 0x

F(x)

xf(t)dt x 1dtx 1xxxetd x (2)【解】应选(D).F(x)1f(t)dt0 x1edt0(t x ex x1e11x3 x F(xx0处连续F(0)1,F(0)a故应选简单的方法是：若a1,f(xx0F(xx0【解】应选(Ax2t2u，则2tdtdu，当t0时ux2，当tx时u0. 1 x xdx0tf(xt)dt2dx f(u)duxf(xx简单的方法是f(x)1

tf

2t2)dt

tdt

dx

dx【解】应选(D).等式5x340 xf(t)dt两端求导得,f(x)15x2a 5x340 x15t2dt5x3axf(x30x3k

,ac(tanx

c(tanxx)

c3

,故c 3 1 n

n an n1(1xn)2d(1xn) (1xn) 1 12n

n

n1

nn 故limnalim1 1(1e1)2n

n

12 2 n【解】应选(B).limlnn1 1 L1 n n n2lim1[ln(11)ln(12)Lln(1n)]21lnn 【解】应选(A).x

πsinxx,

且cosx单调sin(sinxsinx,cos(sinx)cosx 2sin(sinx)dx2sinxdx 2cos(sinx)dx2cosxdx1.故选 【解】应选(B).x0

4lnsinxlncosxlncotx，故应选（【解】应选(Dysinxyesin2xsinI1I2【解】应选(B).Vsin3xdx22sin3x224 x d(x26x d(x【解】x26x13dx2x26x 2(x3)21ln(x26x13)1arctanx3 x1x【解1】原式arcsinxdxx1x 2】xsint，原式sin2tdsinttdsin dt ln|csctcott|sin 2

sin

【解】原式(x3

)dx2x2xedx

xeex2

【解】x1f(t)dt(x),f(x31)3x21,f(x31) 1..令x317 x2f(7)

13 0f(tdtIf(xx2I[0,1]上积分，0f(xdx0xd2I1dxI12II1f(xx dcos 【解】令xsint,则原式2 2 01cos2 【解】原积分令xt2t2costdt4tsintdt4

et2偶函数，则xet2dt0π π 原式22cosxsinxdx22(1sinx)sinxdx 【解】原式ππcos2xcos4xdxπ|cosx|sinxdx2 2 01】原式2axa2xa20 xaasin 2(1sint)a3cos2tdt22】原式2axa2xa)20

2cos2tdt [(xa)

a2(xa)2

2axx2dxπ2f(xcosxdx两端同乘cosx并从0到积分得,f(xx 0f(x)cosxdx0xcosx0cosxdx0f(x)cos 0f(xcosxdx0xcosxdx2,f(xx 【解】由xf(t)dt3x3x1f(t)dtf(x9x21f f(x是偶函数，在xf(t)dt3x3x1f(t)dtx 1f(t)dt31f(t)dt1f(t)dt2f(x9x2 【解】应填41(1n1 n1(1n1 nn121原式limn

1(n

dx 222 222【解】应填 .原式 1cosxdx0sin2dx

1exsinnxdx

1exd1cosnx 1excosnx111excosnxdx11e1cosn1excosnxdx n n 1e1cosn1excosnxdx1e11exdx2，所lim1exsinnxdx n【解】令txu，则xf(tx)dt0f(u)du0f(u)du1x2x 22x1,x1得1f(u)du2,1f(u)du,1f(x)d22.220【解】由xf(t)dtxexf(x)dxxex0lnxf(ln lnxx则 dx f(lnx)dlnx x1x t，则xt22，dx2tdx d 2 t 原式 t293arctan3 3 0

2312

dx令xsin 1313 2

3sin2td43 3(1cos2t)dt 1sin2t33.3 31 3

6

x (x 2)dxln2 x22 A12e2ad1(e4a2 【解】

tanx，s41(tan0

ddx lnsecxtanx4 d0cos

x

1x(x22x 1x 0 121101f

1(x22x0

5 1ln(1xx

dx

f f(x)0 xxx0xxxx114ln(1 4ln(1x) xdx4ln282x11 01 d dxx2x 1dx

x2x242

112

arcsin12 d d 1 1

31x 12214 2 lnx xx 12214x2 2 2 41 因 ln(2

21 |xx2 2xu

arctan(1t)dtd x【解】原式

12

lim

arctan(132lim2xarctan(1x2)

xtu，则0f(xt)dt0f(u)du xf(x)0f(u)dusin4xx

2[f

0f(u)du]dx

2sin4 01

f

π231231 42 f(x在

]上的平均2x

2f30 302x【解】原式limaf(t)dtxafx

f(x)fxxxa(xa)fa

f f(a)

f(t)dt(xa)fa f(x)f(a) lim x

f(a)

(a)f(a)x xfax

f

f(a)2f 2f2【解】令txuxfx)g(tx)dtfx)g(u)du即fx)g(u)dux2ln(1xxgf(xf(x)

f(x)2ln(1x)x1

2x ，xf 2x 1f(x2x1)ln(1xxCf(0)0得C

1【解】（1）因为|cosx|0且nxn1)，所以0|cosx|dxS(x (n0|cosx|dxn0|cosx|dx |cosx|dx2(n因此当nxn1)时，有2nS(x2(n（2）由（1）知，当nxn1)令n,，由准则得

S(x)2(n1).(n1) S(x)2x 【解】（I）当0t1时，因为ln(1tt，所以|lnt|ln(1t)]ntn|lnt| 10|lnt|[ln(1t)]dt0t|lnt|d1（II）由（I）知0un0|lnt|

t)]ndt1tn|lnt|dt01 1 1 0t|lnt|dt0tlntdtn10tdt(n1)2所以lim1tn|lnt|dt0.从而limun

n1f(1kkxe1xf(xdx00,1

f(1)

kxe1xf(x)dxe11f(

在[,1]上，令(xxe1xf(x).那么，(x)在[,1]上连续，在(,1 (1)f(1)()e1[f()f()f()]2 2【证（I）由积分中值定理可知，存在(0,2)2由题设知0f(xdx2f(0)f()ff(2)f

0f(x)dx2f2

f(2)f定理，存在[2,3]，使f() f(2)f f(0)，故f()f(0).2由于f(0)f()f()，且03，根 定理，存在1(0,)x1】F(x0f(atfx 则F(a0f(atf(t)]dt0f(a atuaf(u)du0，F(0)0由定理知(0,a),使F()0，原题得证x2F(xaxf(t)dtx F(0)af(t)dt0，F(a)0f(t)dt由定理知(0,a),使F()0，原题得证xF(xx1)0f(t)dtxF(0)F(1)0, 定理知(0,1)，使F()即0f(t)dt1f(f(x)0F(x) xf(t)dt1xf(xF(0的01【证法一】（1）F(xxxf(tdtF(0F(1)01F(x) 1f(t)dtxfx对F(x)在区间[0,1]上应 定理知，存在一点x0(0,1)，使F(x0)0，因 0xf(xdxx0f(x00x0f(x00

f(x)dx（2）设(x)1f(tdtxf(xx(0,1)时,有(x)f(xf(xxfx a1内取x.若在区间[x,1]上f(x)0，则(x 2x0，否则可设f(x20为连续函数f(x在区间[x1,11x2[x1,1在区间[0x2上，作辅助函数(x)xf(tdtxf(x)，则(x)连续，且1(0)0.(x) f(t)dtxf(x)(12x)f(x) 2f(x)f(x)f(0)f

f(t)dtx0f(x011c(0,mf(c)M，mxf(x) m0xdx0f(x)dxM0xdx,20f(x)dx1m20f(x)dx1由连续函数介值定理知，至少存在一点[0,1f(21f012F(xf(x2x0f(x)dxF(0)1 0F(x)dx0f(x)dx02xdx0f 0f(x)dx0f(x)dx1c(0,1使得0F(x)dxF(c 定理得，(0,c),使得F()0,即f()21f0【证3】由f(0)0得f(x) xf(t)dt，两端积分0 1f(x)dx1dxxf 1dt1

f

1 11(1t)f0 f()1(1 （积分中值定理bf(x)g(x)dxf()bg(x 1f(2

f()

20f11x f(t)dtxF(xF(0)xF(xxf(xf(

x[f(x)f(

0f(t)dt

f(t)dlimx[f(x)f(x)lim[f(x)f(x)limf(0)limx

lim0f(t)dt0f(t)dtlimf(x)f(x)1f11lim

【证明】连续用两次分部积分公式及一次-公式2f(x)cosxdxf(x)sinx22f(x)sin0

f(x)sinf(x)cosx22f(x)cos f(2)f(0)2f(x)cos0 f(x)(1cosx)dx011

0f(x)dx

fnnkn

k)n kn

kk f(xkn

f(knnk n k fn1f(x)dxk

nknnkk

kf(x)f(

)nk1knnnk

kknn

f(k)x kn kn

k n(nx)dxM2n2k kfx0(x1fxf(x)f(1)

xf(x)dx

x2f2x dx1 dxπ1x2π

1 fx1 fx

f(x)存在且不超过1 4 V22x2dy22(1y2)dy W1032(1y2)(2y)gdy1031[1(1y2)](2y)gd2 103g2(2y2y2)dy1(4y4y2y3)d27 g.即所求的功8

278

g焦耳2（Ⅰ）se1y2dx1e(x1)dxe 2 （Ⅱ）ex(1x21lnx)dx1(e21)(e2 e1

1(e21)(e2

(x lnx)dx

.

1

e3

1】