微積分課后答案復旦第6章

第六章習題6-解y=x2+1在[a,b]上連續(xù),x2+1在[a,b]上可積,從而可特殊地將[a,b]n等分,abai,Δxba,f()(abai)21 b bf(i)Δxi[(a

i) (b

[a2(ba)2 2a(ba)

(ba)[na2(ba)211n(n1)(2n1)2(ba)a1n(n1)n]n2

S

(x1)dxlimf(i)Δxi(bn

a(ba)102xdx 1

aa2x2dx01解(1)根據(jù)定然積分的幾何意義知,02xdxy=2x,x=0,x=1x1的面積,1,所以02xdx1根據(jù)定積分的幾何意義知, a2x2dx表示由曲線y0

a2x2x0xa軸所圍成的1圓的面積,而此1圓面積為 a2,所以aa2x2dx

a2 1x2dx與1x3dx；(2)1exdx與1(1x)dx 解(1)∵x[0,1]時x2x3x21x0,x2x3 x3,所以1x2dx1x3dx f(xex1x,f(xex1,因0x1,f(x0 從而f(x)f(0)0,說明ex1x,又 1+x.所以1exdx1(1x)dx 34(x21)dx 3xarctanxdx3

aex2dx

0ex2xdx2解(1)在區(qū)間[1,4]上,f(xx21是增函數(shù),故在[1,4]Mf(417,即

f(1)2,所以2(41)4(x21)dx17(41)164(x21)dx511令f(x)xarctanx,則f(x)arctanx 1

,當x[,3]時,f(x)0, 而f(x)在 3]上是增函數(shù),從而f(x)在 ,3]上的最大值Mf(3) ,最 值mf ) ,所ππ 9

)1xarctanxdx 3

)3 3 3

3f(x)ex2,f(x2xex2,f(x0x=0,f(0f(af(aea,a>0時,ea21,f(x在[-a,a]M=1,mea2 2aea exdx2af(xex2x,f(x2x1)ex2x,f(x0x1,f(02f(1)e1,f(2)

f

M

me 2,2

1

4,

2e4 ex2xdxe2 0ex2xdx2ex2xdx 2e2

2exxdx2e42習題6- 5

dx

1t2dt

dxln

t dtcos

cos(t

d2x

dttd解

x

1t2dt (x2)dxx111

t5e3tdtdxln cosxcos(πt2)dtcosxcos(πt2)dtsinxcos(πt2)dt sin

cosxcos(πt2)dtsinxcos(πt2)dt x x πsintdt

d

dsinxtdx2x t0

xcosxsinxsinxxcosx x (1)limxarctantdt (2)lim0sin3tdt (3)lim0e xt0

xte2t200arctan

arctan

1解(1)lim lim 1x

x2

x0x x0

3

x3e 0tedt

lim6sin3x2ex 2xet2dt 2

2

2xet2dt 2xet2 x

0x0te2t2 x0xte2t2dt xe2 0 x 2et lim lim

x2

x0ex2xex2

x01 求由方程yetdtxcostdt0y=y(x) cos解xcoseyycosx0 y 又由已知方程有etysintx0,即ey1sinxsin0 即ey1sinx,ycosx

cos.sinxx

xtet2dt有極值0解I(xxex2,I(x0x0,I(xex2(12x2I(0)1042(1) xdx (2)1xxdx422x,0x (3)

f(x)dx,f(xsinxx

max1,xdx23

解3

xdx

x2

2(4232)

(83(2)

201111xxdx0(x2x)dx1(xx201111 x3 x2 x2 x3 x3 2 xf(x)dx2xdxπsinxdx

cosxπ1 2 f(xmax{1,x2

2x1x

2

11dx xdx

x2f(u)duf(x)f(2)=3,求

2 (x x2f x2f(u)dudt

xf(u)

xf

解lim

lim

lim (x x22(x 2(xf

limf(x) f(2) 2 習題6-計算下列積分： 3sin(x)dx 31

2(115x)3(3)

54xdx 2sincos3d0

2cos2udu

xx1ln

12 (8)12

x2dx

(9)ln2exex

2x2x22 32

x(x3x)dx cosxcos3xdx2 解(1)sin(x

)dxsin(x

)d(x

)[cos(x 3

1d(5x11)

2(11 52(5x

10(5x

dx1 d(54x)

1555555 2sincos3d2cos3dcos1cos42 1cos 1 1

2cos2udu

du

2cos

4 3π1ππ1sin2u2π3π 11

2 6 d(lnxd(lnx1lnx1x1ln1

63321lnx1 3令x=tant,則dx=sec2tdt,當x=1時,tπ;當x 時,tπ3 33于

dt111πsincos4

ππ2 3234π令x 2sint,則dx

costdt,當x0時,t=0;當x 時,t 222于

2x2dx22cos2dt2(1cos2t)dt(t1sin2t)2π 令ext,xlntdx1dt,xln2時t2;xln3時t3t 3 13 1 t1 于 ln2ex

x2t2122

dt2

t t1 t1

13(

1)dx

1

x12

x

2(x

1)2

3

x x

x2 1(ln2ln1)ln21ln 6xt,xt6dx6t5dt,x=1時,t=1;x=2,t623x(x3x 63x(x3x于 dx1tt2dt61(tt66(lntln(t 7ln26ln(166(12)

cosxcos3xdx

2xsin2x 22

cosx(sinx)dx0

cosxsin 22

cosxdcosx0

cos cos23

2

cos2x2 aln(x1x2)dx(a為正常數(shù)

x3sin2 dx 24cos4d5x42x2

2解(1)f(xlnx1x2是奇函數(shù)alnxf(x)

1x2dx0x3sin2是奇函數(shù)

x42x2x3sin2dx5x42x2f(cos4是偶函數(shù) 24cos4d2

24cos4d0π

2(1cos2022(12cos2cos2220π π22(2cos2 cos40 π2(34cos2cos400

π2sin

π

sin

π a 1

0xf(x)dx21xf(x)dx(a為正整數(shù)證明1dx2

2x1 112設f(x)是定義在(-∞，+∞)上的周期為T的連續(xù)函數(shù)，則對任意a∈［－∞，＋∞ f(x)dx0f(x)dx12證(1)令x2=t,則x t,dx dt12x=0時,t=0;x=a時a

1

t于 0xf(x)dx0 f(t)2tdt20tf(t)dt20xfta 1 0xf(2x1則dx1dt

)dx20xf(x)dx t21dx2

11 1 12dt1 1

dt

11 dt x1 11 dt t

t3 t1 11 111 1 2 x1 11 (3)由于 f(x)dx0f(x)dx f(x)dx, f(x)dx令xtT0f(tT)dt0f 0f(x)dx0f(x)dxaf 故 f(x)dxx

f(x)dxf(t)是連續(xù)函數(shù)且為奇函數(shù)，證明x證明0f(t)dt是奇函數(shù)x證F(x0f(t)dt)=-

F(x)x

f(t)dttu

f(u)du

f(u)duF(x)F(x0f(t)dt是偶函數(shù) F(x) f(t)dttu0f(u)du0f(u)duF(x)xF(x0f(t)dt是奇函數(shù)x調(diào)不增

x0(x2tf(t)dt,f(x)F(x) F(x)xxf(t)dtx2tf(t)dxxf(t)dtxf(x)2xf x0f(t)dtxf(x)f()xxf(x)x[f()f(x)]x其中x0之間.x>0時,x,f(x)f(f(x0,F(x0;時x,f(x)f(f(x0,F(x0;F(x)單調(diào)不增x0x0

習題6-f00 f00

f(x)dxdu證F(u)uf(x)dxF(uF(u0 x(uf(x)dx)du xf(u)duuF(u)x

xuF0 x

即等式成立

xF(x)0uf(u)dux0f(x)dx0uf x0f(u)du0uf(u)du0xf(u)du0ufx0(xu)f 1xexdx (2)exlnxdx xsin24lnxsin2

dx 42

dx 20

1xlog2xdxe

π(xsinx)2dx 0

1(9) x3exdx (10)2xln dx 1解 1xexdx1xdexxex11 0e1ex1e1e1e012 1

1 1

12 x11xlnxdx21lnxdx2xlnx121xdx2e4x14(ex1 4lnxdx ln 2xlnx4

1dx8ln2 48ln2xx xx

xx (4)sin2(4)

dx3xdcotxxcotx

3cot ππl(wèi)nsin

π1ln3 9

2 cosxdx2

dsinx sinx222

eπ

eπ2e2xcosx

20

42

coseπ242e2xcos0 故2 cosxdx

2)xlog2xdx lnxdx x2lnx2 1

2ln2 2ln (4ln23)2 4ln 1π 1120(xsinx)dx20x(1cos2x)dx (2

1π020xππ31(x2sin2xπ2xsin2xdx)π31ππ

4 (xcos2x

0 sin2xπ 1sin(lnx)dxxsin(lnx)11cos(ln1eesin1xcos(lnx)esin(ln1e1eesin1ecos111sin(ln 1sin(lnx)dx2(esin1ecos11)2 x3ex2dx1 x2dex21 2 2 2ln21ln2dex2ln21 ln222 1111

1 1

2xln dx 2 dx2 x2 22 1 2 1 0x 1 ln32(1 )dx

ln3x2

2(

1x1x11x1x1

2 x

x1ln311

13ln83.

1,f′(2)=0,2f(x)dx1,求2x2f(x)dx 2x2f(x)dx2x2df(x)x2f(x)222xf 4f(2) xdf(x)2xf(x)2 f 04f(2)214122習題6-y=ex與直線x=0及 (2)y=x3與(3) 1(5)

,x軸與直線y=x及 (6)y=(x-1)(x-2)與x軸x(7)y=ex,y=e-x與直線 (8)y=lnx,y軸與直線y=lna,y=lnb,(0ab)解(1)y=ey=e的交點坐標(1,ey=exx=0的交點為(0,1),6-1中陰影部分,其面Sexdyelnydy(ylnyy)e 圖6-22y x0x 22 解方程組y2x得y0y22y yx3y2x的交點坐標分別為(0,0(2,22),(2,22),它們S0(x32x)dx2(2xx3)dx(1x4x2) (x21x4 2

解方程

3S

(x2

x3)dx

x3

x4)4164 4 yx2yx的交點為(0,0),(11);yx2y2x的交點為 S1(2xx)dx2(2xx2)dx1xdx1(2x 1x21(x21x3)2 y 與yx的交點為(1,1),y ,x軸與直線x=1,及x=2所圍成的圖形如圖6-x

2 22Sxdx dx

11lnx2 1 圖6-yx1)(x2)x321,頂點坐標為31),與x ( 14)6-6中陰影所示,由y(x321得x314)

S 3 1y3 1y 14014440 4402

4 321y2 2

3

1 yexyex的交點(0,1),yexyexx=16-7陰影所示,其面積S 1

x 0(ee)dx(ee)0ee0 Slnb lnbeydy

lnbbln

(2)y=x3,x=2,x軸，分別繞x軸與y軸(3)y=x2,x=y2，繞y軸 (5)x-2)2+y2≤1,y軸(1)6-9OABD,CBDy軸旋轉(zhuǎn)所成的幾何體體積之差,y=exx=1的交點為(1,ey=exy軸的交點為(0,1),所以,eVπ12eπe(lny)2dyπeπy(lny)2e e

(ln πeπe2π(ylny)eedy2π(ee1)2 2Vy2x

x dπxx

7 8 Vπ228

x2dy

y3dy32ππy

8

解方程組

2Vπ1xdxπ1x4dxπx2x513 2 50 Vπy2dxπ2pxdxπp

apa1. 1.1所求旋轉(zhuǎn)體的體積是由右半圓x1

繞x軸旋xx1Vy121y222 1y221 1y2dy16π11y2dy - (a＞0)與

6-xx兩曲線與x軸圍成的平面圖形的面積解(1)由題意有點(x0y0在已知曲線上,且在點(x0y0處兩函數(shù)的導數(shù)相等.y0y0ax12xxx

1

xy0 xy

0y解 即 ax2 x2

x(2)由(1)知兩曲線的交點為(e2,1),又在區(qū)間(0,1)上,曲線yx

在曲線y 的S1e2y(ey)2dy1e2y1e2y311e210 x(由y x得x(ey)2,由yxe

xe2yx設f(x) xn1x2

=2 f(x)lim

xn1xy1

x

得交點為11且易知當 2

6-y 1x(1,0)時,y

xy1x2的上方.所圍圖形如陰影部分所示,20 S x dx x2 ln(1x2 ln21 1x2 1 與x軸所圍圖形的面積最小.解由拋物線過(0,0),(1,2)點,c=0,a+b=2,yax2bxx為(0,0),b,0,x軸所圍圖形的面積a a aSb(ax2bx)dxa

b22 (2由ab2得b2a,代入上式有S S(2a)2(aS ,

S

0

2

a4a0a4,從而b2a6所以a4,b6,c0解設產(chǎn)品產(chǎn)量為Q(t,則Q(tf(t,5Q15f(t)dt5(2t5)dt(t5t)5 025Q

f(t)dt

t(2t5)dt(t25t)5 R(萬元)的變化率為產(chǎn)量Q(百臺)的函數(shù)R′(Q)=7-2Q.問：生產(chǎn)量為多少時，總利潤最大?最大利潤為多少臺，總利潤減少了多少?解(1)L(QR(QL(Q0R(QC(Q0即72Q20Q=2.5百臺時,總利潤最大,25時的總成本2 C C(Q)dQ 2dQ LRC11.2556.25(萬元(2)在利潤最大的基礎(chǔ)上又生產(chǎn)了50臺,此時產(chǎn)量為3百臺0總成本C3C(Q)dQ32dQ60 2總收入R3R(Q)dQ (72Q)dQ(7QQ

12 LRC1266(萬元解投資后Tya(1ert,ra25,r5%0.05,Ty25(1e00510196.73習題6- x(1) x4 xe0

(4)0cosxdx(5)exsinxdx 0

x22x(7)

xdx (8)lnxdxx1x1x1x1ln2(9) (10)0(1x)21(11)1

11

1解(1)d1

x x1

xx1

eaxdx1eax1,此廣義積分收斂

cosxdxsin0

limsinxsin0limsinx不存在,所以,此廣義積分發(fā)散 exsinxdxexdcosxexcosxexcosxdxexcosxexcosxexsinxexsinxdx1(exsinxexcosx)1ex(sinxcos1 1 exsinxdxlimexsinxdx

e(sinxcos b lim1eb(sinbcosb)1 2

d(x arctan(x1)πx22x

(x1)2 22 2x 2 lim dxlim(x1)22x 0 2

0

1lim 2 ,收斂0 11lnxdxxlnx1 dxln1 1lnxdxlim1lnxdxlim(ln1)1,收斂e

arcsin(lnx)e

π,收斂

x1(ln1(lnx1(ln1(ln

dx0(1

(1

(1 lim

2lim

2

111x 1 011

1 1-1

121

11

0

1

k為何值時，廣義積分

x(lnx)k收斂?k為何值時，這廣義積分發(fā)散?k解k=1時

+

+dlnxln(lnx) xln ln 1當k1時 (lnx)1

k(k1)(ln 2 k所以,k>1時,此廣義積分