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：