北大版高等数学第三章 积分的计算及应用答案 第三章总练习题

第三章总练习题

1.为什么用Newton-Leibniz公式于下列积分会得到不正确结果?(1)?1?1??1d?x?d?x?x?e?dx.?e????e?2[?1,1]无界,从而不可积.dx?dx?????xdtanx2?tanx2111(2)?2?0dx.u?tanx在(0,2?)的一些点不可导.2.证明奇连续函数的原函数为偶函数,而偶连续函数的原函数之一为奇函数.证设奇连续函数f的原函数为F, 现在证明F是偶函数.F?(x)?f(x).(F(?x)?F(x))???F?(?x)?F?(x)??f(?x)?f(x)?0,F(?x)?F(x)?C,C?F(?0)?F(0)?0.F(?x)?F(x)?0.设偶连续函数f的原函数为F,现在证明F是奇函数.F?(x)?f(x).(F(?x)?F(x))???F?(?x)?F?(x)??f(?x)?f(x)?0,F(?x)?F(x)?C.设F(0)?0,则C?F(?0)?F(0)?0.F(?x)?F(x)?0.?sinx,x?0,3.f(x)f(x)??3求定积分?x, x?0,解?baf(x)dx??其中a?0,b?0.0a3b0?xba4f(x)dx?a?b0af(x)dx?a4?b0f(x)dx??xdx??sinxdx?4?cosx|0?1?04?cosb.4.求微商解ddxddx?10sin(x?t)dx.ddx?10sin(x?t)dx??x?1xsin(u)du?sin(x?1)?sin(x).5.试证明limh?0?10f(x?ht)dx?f(x),其中f(x)是实轴上的连续函数.证limh?01?hx?hxf(x?ht)du?10??u0f(t)dt??u?x?f(x).6.求极限limn???(1?x)dx.2n解?10(1?x)dx?22n??/20cos2n?1tdt?I2n?1?(2n)!!(2n?1)!!.?I2n?1??(2n?1)!!1??,(2n?1)!!(2n?2)!!n?11n?1dx.?0(n??),limn??2(2n)!!0?I2n?1?7.??10(1?x)dx?0.2n

sinx?cosx2sinx?3cosx解令sinx?cosx?A(2sinx?3cosx)?B(2sinx?3cosx)??A(2sinx?3cosx)?B(2cosx?3sinx)?(2A?3B)sinx?(?3A?2B)cosx,?2A?3B?115,A??,B?.?1313??3A?2B?1?2sinx?3cosxdx??sinx?cosx?A(2sinx?3cosx)?B(2sinx?3cosx)?2sinx?3cosx113513dx

?Ax?Bln|2sinx?3cosx|?C??x?ln|2sinx?3cosx|?C.8.通过适当的有理化或变量替换求下列积分:(1)?e?2dx.e?2?u,x?ln(2?u),dx?xxx22udu2?u2.?e?2dx?2?udu2?u22du???2?u?2?2?2?u??e?2?x??2?u??(2)?xexx?u?2arctan??C?2??2??dx?2arctanxe?2???C.?2?e?2x?xxe?2xd(e?2)?2?xdxe?2x?2xe?2?2???2xe?2?4???xxe?2dxe?2?x2arctanxxe?2???C.?2??2e?2(x?2)?42arctanx1?x431?xdx1?x?x?21?xx2dx1?xxxe?22?C.(3)???dx??3??23?21?xx?Cx?C.(1?(1??12x?x?1?x)dxx?1?x)x?1?x)?C.

??1?x)(1?(4)???(1?1?x)dxx?2x?x?x(1?x)?ln(?9.?dxsinx?cosx2444??secxdtanx1?tanx2242???(1?u)du1?u42.?2u11?u?21?u1?u?1?u(1?u?22u)(1?u?2u)1?1?2?1?u2???2u?????1?11?.??222?12?1?12?1???)??(u?)??????(u??22?2??2????sindx4x?cosx4?1arctan(?22u?1)?arctan(2u?1)?C.1TT0?10.设函数f(x)在(??,??)上连续,以T为周期,令g(x)?f(x)?函数h(x)??f(x)dx,证明:?x0g(t)dt也以T为周期.证(此即习题3.4第24题)11.设函数f(x)在区间[a,b]上连续,且使f(c)?0.证若不然,f(x)在(a,b)没有 零点,由f的连续性和连续函数的中间值定理,f在(a,b)不变号.不妨设f(x)?0,x?(a,b).取c,d满足,a?c?d?b,则f在[c,d]取最小值m?0.于是?baf(x)dx?0.证明:在(a,b)内至少存在一点c,?baf(x)dx??caf(x)dx??dcf(x)dx??babdf(x)dx?m(d?c)?0.矛盾.12.设函数f在区间[a,b]上连续,且?f(x)dx?0,证明:f(x)?0,x?[a,b].2证若不然,存在c?[a,b],f(c)?0.由f在c的连续性,存在区间 [d,e]?[a,b],|f(x)|?2|f(c)|22,x?[d,e].2

2?baf(x)dx?2?edf(x)dx?|f(c)|2(d?e)?0.矛盾.13.设f(x)在(-?,??)上可积,证明(1)对于任意实数a,有?(2)?(3)??0a0f(x)dx??a0f(a?x)dx;xsinx1?cosx22dx??24;120a

ln(1?2).?/20sinxcosx?sinxa0dx?证(1)?f(x)dx(x?a?t)???f(a?t)dt??a0f(a?t)dt??a0f(a?x)dx.(2)I?I???0?0xsinx1?cosxsinxdx2dx??????0(x??)sinx1?cosxdcosx1?cosx22dx?10??0(??x)sinx1?cosx2dx?1??02?sinx1?cosx.2dx?I,?2??21?cosx?/222??0???du1?u22??arctanu|0?dx2?4(3)I??/20?sinxcosx?sinx0dx??0?/20sin(?/2?x)cos(?/2?x)?sin(?/2x)cosx2????cosxcosx?sinxdxcosx?sinx2dx,2I??dx??/2cosx?sinxdxdx???1?/20sinxcosx?sinxdx?/2?/20??/202sin(x??/4)2ln|csc(x??/4)?cot(x??/4)||0?1???ln2???I?12??????1???1??1?ln?1?????????cos???sin??4???4??2ln(2?1),ln(2?1).

214.一质点作直线运动,其加速度a(t)?(2t?3)m/s.若t?0时x?0且v??4m/s,求(1)质点改变动方向的时刻;(2)头5秒钟内质点所走的总路程.解(1)x??(t)?2t?3,x??t?3t?C1,?4?C1,x??t?3t?4,x?0?C2.x(t)?t322t33?32t?4t?C2,2

3?32t?4t.x??t?3t?4?(t?4)(t?1)?0,t0?4.322?t?32s?x(5)?x(4)?|x(4)|???t?4t?2?3?t?5?t?32?2??t?4t?2?3?3?t?4432m.15.一运动员跑完100m,共用了10.2s,在跑头25m时以等加速度进行,然后保持等速运动跑完了剩余路程.求跑头25m时的加速度.?at, 0?t?t0;解v(t)???at0, t0?t?10.2.?at, 0?t?t0;?s(t)??2?att?C t?t?10.2.0?0?at0/2?at0?C?22a?3m/s.?at0/2?25?100?10.2at0?C2?222

16.(1)利用积分的几何意义证明:

1n?1?lnn?1n12?1n,n?1,2,?1n?1?1n?lnn,(2)令xn?1?yn?1?12???1n?1????lnn,证明序列xn单调上升,而序列yn单调下降.111??(3)证明极限lim?1??????lnn?存在(此极限称为Euler常数).n??2n?1n??证 (1)1n?1??n?1ndxn?1??n?1ndxxn+1?lnx|ndxn?1n.n?1?ln(n?1)?lnn?lnn?1n??n1111????(2)xn?1?xn??1?????ln(n?1)???1?????lnn?2n2n?1?????11???ln?1???0(由(1)).nn??11??1?????lnn??2n??111??yn?1?yn??1??????ln(n?1)??2nn?1???1???ln?1???0(由(1)).n?1n??1(3)yn?xn?x2?1?ln2?0(n?2).yn单调下降有下界,故有极限limyn.n??17.证明:当x?0时,?1x11?tdt?21?1/x11?t21dt.证?1x1?tdt(x?1/u)?2?1/x11?1/u21?1u2dx??1/x11?t21dt.18.设f(x)在(??,??)上连续(书上为可积，欠妥),且对一切实数x,均有f(2?x)??f(x).求实数a?2,使?2af(x)dx?0.解(条件f(2?x)??f(x)相当f关于x?1为奇函数f(1?1?x)??f(1?x?1))?20f(x)dx??120f(2?u)du???20f(u)du,?20f(x)dx?0.取a?0即可.19.利用定积分的性质，证明不等式ln(1?x)?arctanx,0?x?1.证11?t?21?t,t?[0,1],在[0,x]上积分得?x0dt1?t??x0dt1?t2,

ln(1?x)?arctanx,0?x?1.20.(1)设f(x)在[0,a]上可积，证明?a0f(x)dxf(x)?f(a?x)dx?a2;(2)利用(1)中的公式求下列积分的值：

?202x2x?2x?2dx;??/20sinxsinx?cosxdx?a0dxa0证(1)I?a0?a0f(x)f(x)?f(a?x)f(x)?f(a?u)f(u)?f(a?u)f(a?u)dua0du2I?a0?f(x)?f(a?x)f(x)dx?a0?f(u)?f(a?u)f(a-x)??f(x)?f(a?x)202dx??f(x)?f(a?x)202dx??1dx?a,I?a2

.解(2)?x2x?2x?2sinxdx?2?x2x?(2?x)dx?2?222?2.??/20sinx?sin(?/2?x)dx??/22??4.21.设f(x)?解f(x)?df(x)dx?tanxsinx(1?xt)dtdx求22df(x)dx.tanxsinx2?tanxsinx2(1?xt)dt?tanx?sinx?x?22tdt,2?secx?cosx?xtanxsecx?xsinxcosx?tanx?tanxsinxtdt23t?2222?secx?cosx?xtanxsecx?xsinxcosx???3?sinx

??secx?cosx?xtanxsecx?xsinxcosx??secx(1?xtanx)?cosx(1?xsinx)?22.求定积分I?解I?222222213?tan3x?sinx?313?tan3x?sinx?.3??/20cos3?d?的值.12??/20cos3?d??2?2?/20(1?cos6?)d???4?112sin6?|0?/2??4.23.求定积分I???2?0|sinx?cosx|dx的值.解I=2?|sinx?cosx|dx0?2???/20|sinx?cosx|dx??????/2|sinx?cosx|dx|sin(????t)|dx?2???/2=2??|sinx?cosx|dx?0??/20?2?t)?cos(???2???2?/20|sinx?cosx|dx?(cosx?sinx)dx??/4?/20|cost?sint|dx???/20?/40???/2/4(sinx?cosx)dx??/2(cost?sint)dx?/2?2.?2??sinx?cosx?|0?(?cosx?sinx)|?/4?(sinx?cosx)|0??4

24.设0?x0?x1,求定积分I?解I???x1x0(x?x0)(x1?x)dx的值.?x1x02(x?x0)(x1?x)dx????x1x0?x?(x1?x0)x?x0x1dxx?x0?(x1?x0)???x?1??x0x1dx?24??x?x0?(x1?x0)x1?x0?????x?1?dxu?x????242??????u??222222?x1x0?x1x0?(x1?x0)/2?(x1?x0)/2a0(x1?x0)4x1?x022du?2???u???a?udx(a?2)aa?u222u?2?aarcsina??20?a2??(x1?x0)8

.

25.求下列曲线所围图形的面积:2(1)y?x?6x?8与y?2x?7.?y?x?6x?82解?2x?7?x?6x?8,?y?2x?7x?8x?15?0,(x?3)(x?5)?0.122 x

?3,x2?5.S??53(2x?7?(x?6x?8))dx?352?53(?x?8x?15)dx2?x?2????4x?15x? ?3?43?343.(2)y?x?x?16x?4与y?x?6x?8x?4.42?y?x?x?16x?43?232解?x?16x?4?6x?8x?4,x?6x?8x?0,42??y?x?6x?8x?4x?0,x?6x?8?0,(x?2)(x?4)?0,x?2,4.S?4243

?20{(x?x?16x?4)?(x?6x?8x?4)]dx42434342??[(x?6x?8x?4)?(x?x?16x?4)]dx2

2??(x3?6x24?8x)dx?3x2?8x)dx0?(?x?62 42??x?44??2x3?4x2??x ?4??042x3?4x2?????8.???2

(3)y2?x?1与y?x?3.

解(x?3)2?x?1,x2?7x?10?0,

(x?2)(x?5)?0,

x?2,5.y??1,2. 2S? ?2?1[(y?3)?(1?y)]dx 32 ??yy2?9????32?2y??.??12

(4)y?sinx,y?cosx与x??/2.解S??/2??(sinx-cosx)dx?(?cosx?sinx)|?/2/4?/4?2?1; S??5/4?5?/4?(sinx?cosx)dx?(?cosx?sinx)|/2?/2?2?1. 5? 4 ?? 4226.设区域?由曲线y?cosx,y?1及x??/2所围成,将?绕x轴旋转一周,得一旋转体V.试用两种不同的积分表示体积V,并且求V的值. 1解V=??/2?(1-cos21x)dx?20??y???2?arccosy?y?20?d?????yarcsinydy0/22V???2?0(1?cos2x)dx??4.11V?2??0yarcsinydy????0arcsinydy2??arcsiny(y)2210???10y?211?y2dx??2???y?2?1?y?212arcsiny???1

??202??24??24.27.求下列定积分的值:(1)?(2)?22duuu?1(91x212??22duu21?1/u972??1/2dx1?x21/2?arcsinx|1/2?1/2?4??6??12.200?200?80x33?5580x?1)dx?400.2028.设f(x)在[0,7]上可积,且一直已知?(1)求(2)求f(x)dx?5,?f(x)dx?6,?2570f(x)dx?3.?5075f(x)dx的值;f(x)dx的值.?

(3)证明:在(5,7)内至少存在一点,使f(x)?0.解(1)?f(x)dx?05?720f(x)dx??552f(x)dx?5?6?11.(2)?75f(x)dx??0f(x)dx??0f(x)dx?3?11??8.证(3)若不然,f(x)?0,x?(5,7),?75f(x)dx?0,但是?75f(x)dx??8?0,矛盾.29.设f(x)?sinx,h(x)?(1)??/2??/23?1, ???x?2,,g(x)?试求下列定积分的值或表达式:?2x?2, 2?x??.1xf(x)g(x)dx;(2)?g(x)h(x)dx;(3)?1?/2f(t)g(t)dx.解(1)?31?/2??/2f(x)g(x)dx???/2??/2sinxdx?0.(2)?g(x)h(x)dx???21g(x)h(x)dx?1x2?5632g(x)h(x)dx

?21xx1dx?2?322xdx??2x?12x3?2.(3)??/2?sintdt??cosx,t???x?2???/2f(t)g(t)dx??2x?sintdt??22sintdt?cos2?2cosx,2?x??.???/230设函数f(x)在区间[a,b]上连续，严格单调递增(a?0),g(y)是f(x)的反函数,利用定积分的几何意义证明下列公式?baf(x)dx?bf(b)?af(a)??f(b)f(a)g(y)dx.

并作图解释这一公式.解

b31.(1)设函数?(x)在[0,??)上连续且严格单调递增,又设当x?+?时?(x)???且?(0)=0.证明:对于任意实数a?0,B?0,下列不等式成立:aB??a0?(x)dx?(x)是a?B0??1(x)dx其中??1?a0?(x)的反函数.证由30题,??(x)dx?0??(a)0??1(x)dx?a?(a)(*).B?0时不等式显然成立.设B?0??(0),由于x?+?时?(x)???,存在a??0,?(a?)?B,?在[0,a?]连续,根据连续函数的中间值定理,存在a1?0,?(a1)?B.若a1?a,则由(*)得aB?若a1?a,则??(x)dx?0a?Ba0?(x)dx??1?B0??1(x)dx.?0?(x)dx??a0?(x)dx??B?(a)0??1(x)dx???B(a)??1(x)dx?a?(a)????a?(a)???(a)??1(x)dx?1a(?(a))(B??(a))?aB.若a1?a,则??(x)dx?0?B0??1(x)dx??a0?(x)dx???(a)0??1(x)dx???(a)B??1(x)dx?a?(a)???(a)B?1??1(x)dx?a?(a)??(?(a))(?(a)?B)?aB.1p?1q?1,证明下列Minkowski

(2)利用(1)中的不等式,对于任意实数a,b?0,p,q?1ap不等式ab?p?bqq.p?1证不妨设p?1.在(1)中取?(x)?xab?,则??1(x)?xap1/(p?1).?ap?a0xdx?p?b0x1/pdx?app?b1/(p?1)?11/(p?1)?1xex2?p?bp/(p?1)p/(p?1)p?bqq.32.设a?0,求a的值,使由曲线y?1?旋转所得之旋转体的体积等于2?.,y?1及x?a所围成的区域绕直线y?1

解??a0(y-1)dx?2?.?(022axe2x2)dx?2,12?a0xe2xdx?2,1?4a0e2x2d2x?2,?42a02edu?2,eu2a2

?1?8,2a2?ln9,a?ln3.33.作由极坐标方程r?1?sin2?所确定的函数的图形,并求它所围区域的面积.解S???0(1?sin2?)d??2??0(1?2sin2??1?cos4?2)d??3?2.

解??a0(y-1)dx?2?.?(022axe2x2)dx?2,12?a0xe2xdx?2,1?4a0e2x2d2x?2,?42a02edu?2,eu2a2

?1?8,2a2?ln9,a?ln3.33.作由极坐标方程r?1?sin2?所确定的函数的图形,并求它所围区域的面积.解S???0(1?sin2?)d??2??0(1?2sin2??1?cos4?2)d??3?2.

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