定積分（與應(yīng)用）習(xí)題及答案

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(A層次)

?2031．?sinxcosxdx；2．?x0a2a?xdx；3．?223dxx211?x2；

4．?7．?1?1xdx5?4xdx；5．?4dxx?11；6．?341dx1?x?1；

e21?dx；8．?2；9．?1?cos2xdx；

?2x?2x?20x1?lnx032xsinxdx；10．?xsinxdx；11．?2?4cosxdx；12．?4

?5x?2x2?1???2??4454lnx1xdx13．??；14．；15．dxxarctgxdx；?2?10sinxx4?3?016．?2e2xcosxdx；17．??3?2?xsinx?dx；18．?sin?lnx?dx；01?e?019．?2?cosx?cosxdx；20．?44?sinxxsinxdx；dx；21．?01?cos2x1?sinx22．?1202??1?x1?xxlndx；23．?dx；24．?2lnsinxdx；40??1?x1?x?25．?(B層次)

??0?dxdx???0?。

1?x21?x????1．求由?edt??costdt?0所決定的隱函數(shù)y對x的導(dǎo)數(shù)

00ytxdy。dx2．當(dāng)x為何值時，函數(shù)I?x???te?tdt有極值？

2x03．

dcosxcos?t2dt。?dxsinx???x?1,x?12?4．設(shè)f?x???12，求?f?x?dx。

0x,x?1??21

5．lim2??arctgtdt?0xx???x?12。

?1x?sinx,0?x??6．設(shè)f?x???2，求??x???f?t?dt。

0?其它?0,?1,當(dāng)x?0時??1?x7．設(shè)f?x????1,當(dāng)x?0時??1?ex，求?f?x?1?dx。

028．lim1n??n2?n?n2n???n2。

kn2kn?9．求lim?n??k?1e。

1n?ne10．設(shè)f?x?是連續(xù)函數(shù)，且f?x??x?2?f?t?dt，求f?x?。

011．若?2ln2xdte?1?12t??61，求x。

12．證明：2e???212e?xdx?2。

2???x?a?2?2x13．已知lim??4xedx，求常數(shù)a。??ax???x?a??2??1?x,14．設(shè)f?x????x??e,xx?0x?0，求?f?x?2?dx。

1315．設(shè)f?x?有一個原函數(shù)為1?sinx，求?2xf??2x?dx。

2?016．設(shè)f?x??ax?b?lnx，在?1,3?上f?x??0，求出常數(shù)a，b使?f?x?dx最

13小。

17．已知f?x??e?x2，求?f??x?f???x?dx。

02100118．設(shè)f?x??x2?x?f?x?dx?2?f?x?dx，求f?x?。19．?

?0?f?cosx?cosx?f??cosx?sinx?dx。

22

20．設(shè)x?0時，F(xiàn)?x???x2?t2f???t?dt的導(dǎo)數(shù)與x2是等價無窮小，試求

0x??f???0?。(C層次)

1．設(shè)f?x?是任意的二次多項式，g?x?是某個二次多項式，已知

?10f?x?dx?b?1??1?，求????f0?4f?f1g?x?dx。?????a6??2??2．設(shè)函數(shù)f?x?在閉區(qū)間?a,b?上具有連續(xù)的二階導(dǎo)數(shù)，則在?a,b?內(nèi)存在?，

b?a?b?13使得?f?x?dx??b?a?f???b?a?f?????。?a?2?243．f?x?在?a,b?上二次可微，且f??x??0，f???x??0。試證

?b?a?f?a???af?x?dx??b?a?f?b??f?a?。

b24．設(shè)函數(shù)f?x?在?a,b?上連續(xù)，f??x?在?a,b?上存在且可積，f?a??f?b??0，試證f?x??1bf??x?dx(a?x?b)。?a211005．設(shè)f?x?在?0,1?上連續(xù)，?f?x?dx?0，?xf?x?dx?1，求證存在一點x，

0?x?1，使f?x??4。

6．設(shè)f?x?可微，f?0??0，f??0??1，F(xiàn)?x???tfx2?t2dt，求lim0x?0x??F?x?。4x7．設(shè)f?x?在

4b?a,b?上連續(xù)可微，若f?a??f?b??0，則

f?x?dx?maxf??x?。??b?a?2aa?x?b8．設(shè)f?x?在?A,B?上連續(xù)，A?a?b?B，求證lim?k?0baf?x?k??f?x?dx

k?f?b??f?a?。

9．設(shè)f?x?為奇函數(shù)，在???,???內(nèi)連續(xù)且單調(diào)增加，F(xiàn)?x????x?3t?f?t?dt，

0x證明：(1)F?x?為奇函數(shù)；(2)F?x?在?0,???上單調(diào)減少。

3

10．設(shè)f?x?可微且積分??f?x??xf?xt??dt的結(jié)果與x無關(guān)，試求f?x?。

0111．若f???x?在?0,??連續(xù)，f?0??2，f????1，證明：

??f?x??f???x??sinxdx?3。

0?12．求曲線y???t?1??t?2?dt在點(0,0)處的切線方程。

0x13．設(shè)f?x?為連續(xù)函數(shù)，對任意實數(shù)a有

????a?asinxf?x?dx?0，求證

f?2??x??f?x?。

14．設(shè)方程2x?tg?x?y???x?y0d2ysectdt，求2。

dx215．設(shè)f?x?在?a,b?上連續(xù)，求證：

h?0lim?1x?f?t?h??f?t??dt?f?x??f?a?(a?x?b)h?ax2?1?x?016．當(dāng)x?0時，f?x?連續(xù)，且滿足?f?t?dt?x，求f?2?。

17．設(shè)f?x?在?0,1?連續(xù)且遞減，證明

??f?x?dx??f?x?dx，其中???0,1?。

001?18．設(shè)f??x?連續(xù)，F(xiàn)?x???f?t?f??2a?t?dt，f?0??0，f?a??1，試證：

0xF?2a??2F?a??1。

19．設(shè)g?x?是?a,b?上的連續(xù)函數(shù)，f?x???g?t?dt，試證在?a,b?內(nèi)方程

axg?x??f?b??0至少有一個根。b?axxab20．設(shè)f?x?在?a,b?連續(xù)，且f?x??0，又F?x???f?t?dt??(1)F??x??2(2)F?x??0在?a,b?內(nèi)有且僅有一個根。21．設(shè)f?x?在?0,2a?上連續(xù)，則?2a01dt，證明：

f?t?f?x?dx???f?x??f?2a?x??dx。

0a22．設(shè)f?x?是以?為周期的連續(xù)函數(shù)，證明：

?0?sinx?x?f?x?dx??0?2x???f?x?dx。

4

2??23．設(shè)f?x?在?a,b?上正值，連續(xù)，則在?a,b?內(nèi)至少存在一點?，使

??af?x?dx??f?x?dx??1b1bf?x?dx。?a2x1f?u?1?24．證明?lnf?x?t?dt??lndu??lnf?u?du。

000f?u?25．設(shè)f?x?在?a,b?上連續(xù)且嚴(yán)格單調(diào)增加，則?a?b??f?x?dx?2?xf?x?dx。

aabb26．設(shè)f?x?在?a,b?上可導(dǎo)，且f??x??M，f?a??0，則?f?x?dx?abM?b?a?2。227．設(shè)f?x?四處二階可導(dǎo)，且f???x??0，又u?t?為任一連續(xù)函數(shù)，則

1af?u?t??dt??0a?1a?f??u?t?dt?，?a?0?。?a0?b?a?b?28．設(shè)f?x?在?a,b?上二階可導(dǎo)，且f???x??0，則?f?x?dx??b?a?f??。a?2?29．設(shè)f?x?在?a,b?上連續(xù)，且f?x??0，?f?x?dx?0，證明在?a,b?上必有

abf?x??0。

30．f?x?在?a,b?上連續(xù)，且對任何區(qū)間??,????a,b?有不等式

???f?x?dx?M???1??(M，?為正常數(shù))，試證在?a,b?上f?x??0。

第五章定積分

(A)

?1．?2sinxcos3xdx

0?解：原式???a0202314cosxdx??cosx?

4403?2．?x2a2?x2dx

解：令x?asint，則dx?acostdt當(dāng)x?0時t?0，當(dāng)x?a時t???2

原式??2a2sin2t?acost?acostdt

05

a4?444a?02sin2tdt?82?st?dt??1?co420??2a?a1???sin4t?a4

828416043．?3dxx211?x2

解：令x?tg?，則dx?sec2?d?當(dāng)x?1，3時?分別為

2sec?原式???32d?

tg?sec?4??，43????3?sin??dsin?

?24??2?4．?1?1233xdx5?4x

1512?u，dx??udu442解：令5?4x?u，則x?當(dāng)x??1，1時，u?3,1原式??5．?4115?u2du?3861??dxx?11

解：令x?t，dx?2tdt

當(dāng)x?1時，t?1；當(dāng)x?4時，t?2原式??212dt?2tdt?2?2??dt??111?t?1?t??222?2t1?ln?1?t?1?2?2ln

3??6．?341dx1?x?1

6

解：令1?x?u，則x?1?u2，dx??2udu當(dāng)x?31,1時u?,04201?2uu?1?1原式??1du?2?2du?1?2ln2

0u?1u?127．?e2dxx1?lnxe211

11?lnxe21e21解：原式??dlnx??11?lnxd?1?lnx?

?21?lnx8．?dx

?2x2?2x?20?23?2

解：原式??0?2dx1??x?1?2?arctg?x?1??2

0?arct??g1???arct1g?4??4??2

9．??01?cos2xdx

?0解：原式??2cos2xdx?2?cosxdx

0??2?coxdxs?2???coxs?dx

202???????2?sin?2?sinx0x???222??10．?x4sinxdx

???解：∵x4sinx為奇函數(shù)

∴?x4sinxdx?0

????11．?2?4cos4xdx

?2??420解：原式?4?2?2cosxdx?2?0?2cosx?dx

227

??2?20?1?cos2x??202dx?2?2?1?2cos2x?cos22x?dx

0?2?2?2x0?cos2xdx????20?1?cos4x?dx

2???2sin2x0?1???2cos4xd4x240???2313???sin4x??

2420x3sin2xdx12．?4?5x?2x2?15x3sin2x解：∵4為奇函數(shù)

x?2x2?1x3sin2xdx?0∴?4?5x?2x2?1513．??3xdx2sinx4??解：原式????3xdctgx

4??3x???3ctgxdx??xctg?44??13?3???lnsin?x????49?4???13?32???ln??ln???49?22???13?13???ln????49?22??14．?4lnxx1dx

4解：原式?2?lnxdx

18

?2?xlnx???11?44xdlnx?

??41???2?4ln2??xdx?

1x??12?8ln2?2?xdx

14??8ln2?415．?xarctgxdx

0111解：原式??arctgxdx2

2023x1?1?2??xarctgx??dx?2023?1?x???8?1111dxdx?2?02?01?x21111??x?arctgx82023????4?1216．?2e2xcosxdx

0?解：原式??2e2xdsinx

0??x?e2xsin?20??2sinx?2e2xdx

0?x?e?2?2e2xdcos0??2x0?x2?2?2cosx?2e2xdx?e?2ecos0?xdx?e?2?4?2e2xcos

0??故?2e2xcosxdx?01?e?25??17．?

2?xsinx?dx0?9

解：原式????xsinx?0?2dx??x20?1?cos2xdx21?21?2xdx?xcos2xdx??0022?1?x36?01?2xdsin2x?04??1?2?xsin2x??sin2x?2xd?x?0?0??64??3??36?1?xdco2sx?04?1??3????xco2sx0??co2sxdx????0??6464??318．?sin?lnx?dx

1e1e解：原式?xsin?lnx?1??xcos?lnx??dx

1xe?esin1??cos?lnx?dx

1e1?e??esin1??xcos?lnx?1??xsin?lnx??dx?

1x??e?esin1?ecos1?1??sin?lnx?dx

1e故?sin?lnx?dx?1ee?sin1?cos1?1?2?19．?2?cosx?cos3xdx

?4??解：原式??2?cosx1?cos2xdx

4????0??4coxs??sinx?dx??2coxssinxdx00??33?2??2?2s?2?????coxs?2????cox?33?????04442??

33

10

?20．?40sinxdx

1?sinx?解：原式??40sinx?1?sinx?dx21?sinxx?sin2???4??tgx?dx20cosx???????40dcoxs2??4secx?1dx20cosx????14?4???tgx?x?02??2?coxs04?21．??0xsinxdx21?cosx解：令x??2?t，則

???????t?sin??t????2??2?原式????2?dt

???21?cos2??t??2?tcots?dt????22221?sint1?sint2???cots????22．?xln120232?cots?2??sgindt??arctt?0241?sint1?xdx1?x120解：原式??1?x?x2lnd?1?x??212????12x1?xx1?x1?x??1?x???1??ln??2??dx2023?x021?x?1?x?221xdx?ln3??2ln208x?1111

1dx?ln3??2dx??22

00x?1811111x?1?ln3??ln

822x?10???1213?ln3281?x2dx23．???1?x4????1?x2dx?2?01?x4解：原式??01?12xdx1?x22x?2???011???x???2x????21??d?x??

x??1x?2xarctg?22?2?

0??24．?2lnsinxdx

0xx??解：原式??2ln?2sin?cos?dx令x?2t2?4?ln2?lnsint?lncost?dt

0022???????ln2?2??4lnsintdt??4lncotdts?

002?????

t??u2??????ln2?2??4lnsintdt???2lnsinudu?

024????2??ln2?2?2lnsintdt

0故?2lnsinxdx??0?2ln2

25．?

??0?dx1?x21?x???????0?

12

11解：令x?，則dx??2dt

tt1dt20??t?dtt原式??????1?t21?t?01?t21?t???2tt?????∴2???0?????dxdxx?dx????2?2?2?001?x1?x1?x1?x1?x1?x?????????????故?

????01???dx?arctgx?2023?x0?dx??2?41?x1?x???(B)

1．求由?etdt??costdt?0所決定的隱函數(shù)y對x的導(dǎo)數(shù)

00yxdy。dx解：將兩邊對x求導(dǎo)得

dy?cosx?0dxdycosx??y∴dxeey2．當(dāng)x為何值時，函數(shù)I?x???te?tdt有極值？

2x0解：I??x??xe?x，令I(lǐng)??x??0得x?0

2當(dāng)x?0時，I??x??0當(dāng)x?0時，I??x??0

∴當(dāng)x?0時，函數(shù)I?x?有微小值。

dcosx2cos?tdt。?sinxdxcostd?a22?cos?tdt?cos?tdt解：原式??a???sinx?dx?cosxd?sinx2??cos?tdt??cos?t2dt???a?a?dx?3．

????????????22s?sinx??sinx??co?scosx??cosx???co?13

2??co?ssincoxs?co?sco2sx??sinx?22??co?ssinxcoxs?sinxco?s??sinx2??sinx?coxs?cos?sinx

???????????x?1,x?12?4．設(shè)f?x???12，求?f?x?dx。

0x,x?1??2解：?f?x?dx???x?1?dx??00212112xdx2218?1???x2?x??x3?

?2?061315．lim2??arctgtdt?0xx???xx?12??arctgtdt?型?0?2。

解：limx???x?12x???lim?arctgx?21?12x?122x2??

?limx???x2?1?arct?gx?limx???x2x1?1arct2gx2x

x??1?22?lim1?2?arctgx??

x???4x?1x?sinx,0?x??6．設(shè)f?x???2，求??x???f?t?dt。

0?0,其它?解：當(dāng)x?0時，??x???f?t?dt??0dt?0

00xx當(dāng)0?x??時，??x???x11?cosxsintdt?022x?x當(dāng)x??時，??x???f?t?dt??f?t?dt??f?t?dt??00?0?x1sintdt??0dt?1?2當(dāng)?0時?0,??1故??x????1?cosx?,當(dāng)0?x??時。

?2當(dāng)x??時??1,

14

?1,當(dāng)x?0時??1?x7．設(shè)f?x????1,當(dāng)x?0時??1?ex?1,當(dāng)x?1時??x解：f?x?1????1,當(dāng)x?1時??1?ex?1，求?f?x?1?dx。

02

?202dx1f?x?1?dx????11??x?1?dx01?ex?112dx1?ex?1?ex?1??dx?1???x?1?01x1?e1?1?ln1?ex?1?ln?1?e?8．lim1n??n2??10?ln2

?n?2n???n2。

??12n?1?解：原式?lim??????n??nn??n?n?lim?n??i?1n1i12???xdx?

0nn39．求lim?n??k?1nekn2kn。

n?nen解：原式?lim?n??k?1ekn2kn1?e1nex?x1dx?arctge?arctg?e??001?e2x4110．設(shè)f?x?是連續(xù)函數(shù)，且f?x??x?2?f?t?dt，求f?x?。

01解：令?f?t?dt?A，則f?x??x?2A，

0115

從而?f?x?dx???x?2A?dx?00111?2A2即A?11?2A，A??22∴f?x??x?111．若?2ln2xdte?1t??6，求x。

解：令et?1?u，則t?ln1?u2，dt?當(dāng)t?2ln2時，u?3當(dāng)t?x時，u?ex?1∴?2ln2x??2udu21?udte?1t??3ex?1?2udu?2arctgu1?u2u?3ex?1

?????2??arctgex?1??

?3?6從而x?ln212．證明：2e?121???212e?xdx?2。

2?11??x2證：考慮??,?上的函數(shù)y?e，則

22??y???2xe?x，令y??0得x?0

2?1?當(dāng)x???,0?時，y??0

2???1?當(dāng)x??0,?時，y??0

2??∴y?e?x2在x?0處取最大值y?1，且y?e?121?212?x21?x2在x??12處取最小值e?12

1故??212edx??edx???2121dx

16

即2e?121???212e?xdx?2。

x2???x?a?2?2x13．已知lim????a4xedx，求常數(shù)a。

x???x?a??2a???2a解：左端?lim?1???e

x????x?a?右端????ax??2xe?d??2x???2?2x??a??a?2x2de?2x

2?2x??2??xe???????a2xe?2xdx??

??2a2e?2a?2?axd?e2x

??a?2x?2a2e?2a?2??xe?????ae?2xdx??

??2a2?2a?1e?2a∴2a2?2a?1e?2a?e?2a解之a(chǎn)?0或a??1。

2??1?x,14．設(shè)f?x????x??e,????x?0x?0，求?f?x?2?dx。

13解：令x?2?t，則

?31f?x?2?dx??f?t?dt??1?tdt??e?tdt?2?1?1010??171?3e15．設(shè)f?x?有一個原函數(shù)為1?sinx，求?2xf??2x?dx。

2?0?解：令2x?t，且f?x???1?sin2x??sin2x

?

?20xf??2x?dx???0t11?f??t?dt??tf??t?dt2240?1?1?????????tdft?tft?ftdt?00???4?04??1?2tsin2t0?1?sint??0??0???4????16．設(shè)f?x??ax?b?lnx，在?1,3?上f?x??0，求出常數(shù)a，b使?f?x?dx最

13小。

17

解：當(dāng)?f?x?dx最小，即??ax?b?lnx?dx最小，由f?x??ax?b?lnx?0知，

1133y?ax?b在y?lnx的上方，其間所夾面積最小，則y?ax?b是y?lnx的切線，

而y??111，設(shè)切點為?x0,lnx0?，則切線y??x?x0??lnx0，故a?，xx0x0b?lnx0?1。

3?a?于是I???ax?b?lnx?dx??x2?bx???lnxdx

11?2?133?4a?2?1?lna???lnxdx

13??4?令I(lǐng)a21?0得a?a2從而x0?2，b?ln2?1

???又Ia32?0，此時?1f?x?dx最小。a2217．已知f?x??e?x，求?f??x?f???x?dx。

01解：f??x???2xe?x

2

?f??x?f???x?dx??011012f??x?df??x???f??x??

2023221??x????2xe??2??2e?2

018．設(shè)f?x??x2?x?f?x?dx?2?f?x?dx，求f?x?。

0021解：設(shè)?f?x?dx?A，?f?x?dx?B，則f?x??x2?Bx?2A

0012∴A??f?x?dx??x2?Bx?2Adx?11?B?2A

0032228∴B??f?x?dx??x2?Bx?2Adx??2B?4A

00314解得：A?，B?，于是

3342f?x??x2?x?

3311????19．?

?0?f?cosx?cosx?f??cosx?sinx?dx。

218

解：原式??f?cosx?cosxdx??sinxf??cosx?dcosx

00????f?coxs?coxdxs?sinxf?coxs?0??f?coxs?coxsdx?00???0

20．設(shè)x?0時，F(xiàn)?x???x2?t2f???t?dt的導(dǎo)數(shù)與x2是等價無窮小，試求

0x??f???0?。

?x?解：limx0x?02?t2f???t?dtx3x0?3??limx?0x02xf???t?dtx2

?limx?02?f???t?dtx?lim2xf?????????0,x??x?0x?2f???0??1故f???0??

(C)

1．設(shè)f?x?是任意的二次多項式，g?x?是某個二次多項式，已知

12?10b?1??1?f?x?dx??f?0??4f???f?1??，求?g?x?dx。

a6??2??解：設(shè)x??b?a?t?a，則

I??g?x?dx??g??b?a?t?a??b?a?dt

a0b1??b?a??g??b?a?t?a?dt

01令g??b?a?t?a??f?t?

?1??b?a?于是f?0??g?a?，f???g??，f?1??g?b?

22????由已知得I??b?a??b?a?????ga?4g?gb???6?2????2．設(shè)函數(shù)f?x?在閉區(qū)間?a,b?上具有連續(xù)的二階導(dǎo)數(shù)，則在?a,b?內(nèi)存在?，

19

b?a?b?13使得?f?x?dx??b?a?f???b?a?f?????。?a?2?24證：由泰勒公式

f?x??f?x0??f??x0??x?x0??f??????x?x0?22!其中x0,x??a,b?，?位于x0與x之間。兩邊積分得：

f?????2?x?x0?dxa2!f?????3b?baf?x?dx??f?x0?dx??f??x0??x?x0?dx??aabb??b?a?f?x0??令x0?

f??x0??b?x0?2??a?x0?2?2??6??b?x?03??a?x0?

?a?b，則2?ba22?a?ba?b1a?ba?b?????????f?x?dx??b?a?f???f?????b????a???

2??2???2??2?2????33f???????a?b??a?b?????a????b?6?22??????????a?b?13??b?a?f????b?a?f?????，???a,b?。

?2?243．f?x?在?a,b?上二次可微，且f??x??0，f???x??0。試證

?b?a?f?a???af?x?dx??b?a?f?b??f?a?。

b2證明：當(dāng)x??a,b?時，由f??x??0，f???x??0知f?x?是嚴(yán)格增及嚴(yán)格凹的，從而f?x??f?a?及f?x??f?a??bbaaf?b??f?a??x?a?b?a故?f?x?dx??f?a?dx??b?a?f?a?

?bab?f?b??f?a???f?x?dx???f?a??x?a??dx

ab?a??f?b??f?a?1?b?a?2

b?a2f?b??f?a???b?a?

2??b?a?f?a??4．設(shè)函數(shù)f?x?在?a,b?上連續(xù)，f??x?在?a,b?上存在且可積，f?a??f?b??0，

20

1b試證f?x???f??x?dx(a?x?b)。

2a證明：由于在?a,b?上f??x?可積，故有

?baf??x?dx??f??t?dt??f??t?dt

axxaxb而f?x???f??t?dt，?f?x???f??t?dt

xbb1?x???ftdt?f??t?dt????x?a?2?b1x1bf?x????f??t?dt??f??t?dt???f??t?dt

?x?a?2a2?于是f?x??5．設(shè)f?x?在?0,1?上連續(xù)，?f?x?dx?0，?xf?x?dx?1，求證存在一點x，

00110?x?1，使f?x??4。

證：假設(shè)f?x??4，x??0,1?

由已知?f?x?dx?0，?xf?x?dx?1，得

00111??xf?x?dx?011?111???fxdx?x???f?x?dx??0022????x?01111f?x?dx?4?x?dx

0221??11???1?2?4????x?dx??1?x??dx??1

02???2???21111f?x?dx?4?x?dx

022故?x?01從而?x?01?f?x??4?dx?02∴f?x??4?0

由于f?x?在?0,1?連續(xù)，則f?x??4或f?x???4。從而?f?x?dx?4或?4，

01這與?f?x?dx?0矛盾。故f?x??4。

016．設(shè)f?x?可微，f?0??0，f??0??1，F(xiàn)?x???tfx2?t2dt，求lim0x?0x??F?x?。x421

1x2解：令x?t?u，則F?x???f?u?du，顯然F??x??xfx2

2023??F?x?F??x?fx2fx2?f?0?11?于是lim4?lim。???lim?lim?f0?2x?0xx?04x3x?04x2x?0444x?0??????7．設(shè)f?x?在

4b?a,b?上連續(xù)可微，若f?a??f?b??0，則

f?x?dx?maxf??x?。??b?a?2aa?x?b?a?b??a?b?證：因f?x?在?a,b?上連續(xù)可微，則f?x?在?a,,b?上均滿足拉?和?22????格朗日定理條件，設(shè)M?maxf??x?，則有

a?x?b?baf?x?dx??a?b2aa?b2af?x?dx??a?bf?x?dx

2bb????f?a??f???1??x?a?dx??a?bf?b??f???2??x?b?dx

2a?b2af???1??x?a?dx??a?bf???2??x?b?dx

2a?b2ab?M?故

4x?adx?M?a?bx?bdx?2bM?b?a?24f?x?dx?M。??b?a?2abak?0b8．設(shè)f?x?在?A,B?上連續(xù)，A?a?b?B，求證lim?f?x?k??f?x?dx

k?f?b??f?a?。

證：?baf?x?k??f?x?1b1bdx??f?x?k?dx??f?x?dx

kkakabb?ka?ka令x?k?u，則?f?x?k?dx??于是?bf?u?du

f?x?k??f?k?1b?k1bdx??f?x?dx??f?x?dx

akka?kka1b?k1a?kf?x?dx??f?x?dx??kbkabf?x?k??f?x?1b?k1a?kdx?lim?f?x?dx?lim?f?x?dx故lim?bak?0ak?0k?0kkk

22

?f?b??f?a?

9．設(shè)f?x?為奇函數(shù)，在???,???內(nèi)連續(xù)且單調(diào)增加，F(xiàn)?x????x?3t?f?t?dt，

0x證明：(1)F?x?為奇函數(shù)；(2)F?x?在?0,???上單調(diào)減少。

證：(1)F??x???

f?x?為奇函數(shù)?x0??x?3t?f?t?dtt??u??0???x?3u?f??u?du

xx?0??x?3u?f?u?du???0?x?3u?f?u?du??F?x?

x∴F?x?為奇函數(shù)。

?xx??(2)F??x??x?f?t?dt?3?tf?t?dt??0?0???f?t?dt?xf?x??3xf?x?

0x??f?t?dt?2xf?x?

0x???f?t??f?x??dt?xf?x?

0x由于f?x?是奇函數(shù)且單調(diào)增加，當(dāng)x?0時，f?x??0，

??f?t??f?x??dt?0??0?t?x?，故F??x??0，x??0,???，即F?x?在?0,???上

0x單調(diào)減少。

10．設(shè)f?x?可微且積分??f?x??xf?xt??dt的結(jié)果與x無關(guān)，試求f?x?。

01解：記??f?x??xf?xt??dt?C，則

01

?0?f?x??xf?xt??dt?f?x???0f?u?du?C

1x由f?x?可微，于是f??x??f?x??0

解之f?x??ke?x（k為任意常數(shù)）

11．若f???x?在?0,??連續(xù)，f?0??2，f????1，證明：

??f?x??f???x??sinxdx?3。

0?解：因?f???x?sinxdx??sinxdf??x?

00??23

?sinxf??x???0??0f??x?coxsdx????0f??x?coxsdx????coxsd?fx???f?x?coxs??00??0f?x?sinxdx?f????f?0????0f?x?sinxdx?1?2????0f?x?sinxdx?3??0f?x?sinxdx所以??0?f?x??f???x??sinxdx?3。

12．求曲線y??x0?t?1??t?2?dt在點(0,0)處的切線方程。

解：y???x?1??x?2?，則y??0??2，故切線方程為：y?0?2?x?0?，即y?2x。

13．設(shè)f?x?為連續(xù)函數(shù)，對任意實數(shù)a有

???a??asinxf?x?dx?0，f?2??x??f?x?。

證：兩邊對a求導(dǎo)

sin???a?f???a????1?sin???a?f???a??0即f???a??f???a?

令a???x，即得f?2??x??f?x?。14．設(shè)方程2x?tg?x?y???x?y2d20sectdt，求ydx2。

解：方程兩邊對x求導(dǎo)，得

2?se2c?x?y??1?y???se2c?x?y??1?y??

從而y??1?cos2?x?y??sin2?x?y?

y???2sin?x?y?co?sx?y??1?y???2sin?x?y?co3s?x?y?15．設(shè)f?x?在?a,b?上連續(xù)，求證：

24

求證

1x?f?t?h??f?t??dt?f?x??f?a?(a?x?b)limh?0?h?a證：設(shè)F?x?為f?x?的原函數(shù)，則左邊?lim?h?01?F?x?h??F?a?h??F?x??F?a??h?F?x?h??F?x?F?a?h??F?a???lim??h?0??hh???f?x??f?a??右邊。16．當(dāng)x?0時，f?x?連續(xù)，且滿足?解：等式兩邊對x求導(dǎo)，得fx2?1?x?2x?3x2?1令x2?1?x??2得x?1將x?1代入得：f?2??5?1故f?2??1。5x2?1?x?0f?t?dt?x，求f?2?。

????17．設(shè)f?x?在?0,1?連續(xù)且遞減，證明

??f?x?dx??f?x?dx，其中???0,1?。

001?證：??f?x?dx?????f?x?dx??f?x?dx??

01?1?0??則??f?x?dx??f?x?dx

001????f?x?dx????1??f?x?dx

?01????1???f??1??????1?f??2?，?1???,1?，?2??0,???????1??f??1??f??2??由于f?x?遞減，f??1??f??2?故??f?x?dx??f?x?dx?0

001?即??f?x?dx??f?x?dx。

001?18．設(shè)f??x?連續(xù)，F(xiàn)?x???f?t?f??2a?t?dt，f?0??0，f?a??1，試證：

0x25

F?2a??2F?a??1。

證：F?2a??2F?a?????2a02a0f?t?f??2a?t?dt?2?f?t?f??2a?t?dt

0a0af?t?f??2a?t?dt??f?t?f??2a?t?dtf?t?f??2a?t?d?2a?t???f?t?f??2a?t?dt

02a2aaa???2aa??f?t?f?2a?t?a??f??t?f??2a?t???f?t?f??2a?t?dt

0a在第一個積分中，令2a?t?u，則

?2aaf??t?f?2a?t?dt??f?u?f??2a?u?du

02a2a而?f?t?f?2a?t?a??f?2a?f?0??f故F?2a??2F?a??1

?a??1

19．設(shè)g?x?是?a,b?上的連續(xù)函數(shù)，f?x???g?t?dt，試證在?a,b?內(nèi)方程

axg?x??f?b??0至少有一個根。b?a證：由積分中值定理，存在???a,b?使f?b???g?t?dt?g????b?a?

ab即g????f?b??0b?a故?是方程g?x??f?b??0的一個根。b?axxab20．設(shè)f?x?在?a,b?連續(xù)，且f?x??0，又F?x???f?t?dt??(1)F??x??2(2)F?x??0在?a,b?內(nèi)有且僅有一個根。證：(1)F??x??f?x??ab1dt，證明：

f?t?1?2??fx(2)F?a???b1dt?0，F(xiàn)?b???f?t?dt?0

af?t?又F?x?在?a,b?連續(xù)，由介值定理知F?x??0在?a,b?內(nèi)至少有一根。又F??x??0，則F?x?單增，從而F?x??0在?a,b?內(nèi)至多有一根。

26

故F?x??0在?a,b?內(nèi)有且僅有一個根。21．設(shè)f?x?在?0,2a?上連續(xù)，則?證：?2a02a0f?x?dx???f?x??f?2a?x??dx。

0af?x?dx??f?x?dx??0a2aaf?x?dx

令x?2a?u，dx??du，則

?2aaf?x?dx??f?2a?u?du??f?2a?x?dx

00aa故?2a0f?x?dx???f?x??f?2a?x??dx

0a22．設(shè)f?x?是以?為周期的連續(xù)函數(shù)，證明：

??sinx?x?f?x?dx???2x???f?x?dx。

002??證：?2?0?sinx?x?f?x?dx

?2?0???sinx?x?f?x?dx??令x???u，則

??sinx?x?f?x?dx

???u????u?f???u?du???sinx?x?f?x?dx??0?sin?02?2?????u???sinu?f?u?du(∵f?x?以?為周期)故?0?sinx?x?f?x?dx??0?2x???f?x?dx

b?23．設(shè)f?x?在?a,b?上正值，連續(xù)，則在?a,b?內(nèi)至少存在一點?，使

??af?x?dx??f?x?dx??xbax1bf?x?dx。?a2證：令F?x???f?t?dt??f?t?dt由于x??a,b?時，f?x??0，故F?a????f?t?dt?0

abF?b???f?t?dt?0

ab故由零點定理知，存在一點???a,b?，使得F????0即?f?t?dt??f?t?dt?0

a?b?27

?f?x?dx???f?x?dx

ab?b又?f?x?dx??f?x?dx??f?x?dx?2?f?x?dx

aa?b??a故?f?x?dx??f?x?dx?a?b?1bf?x?dx。2?a1f?u?1?du??lnf?u?du。0f?u?24．證明?lnf?x?t?dt??ln001x證：設(shè)x?t?u?1，則