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1、练习十二二课堂自测题11二(1)若f(x)1f(x)dxxf(2x)dx,则f(x)______42x.0011分析令af(x)dx,bf(2x)dx,00则f(x)1abx,111b2f(x)dx(1abx)dx(xaxx)0020b1aa(1)21112f(2x)dx(1a2bx)dx(xaxbx)0001abb(2)解方程组(1),(2)得a3,b2.所以f(x)42x.1二(2)设f(x)t(tx)dt,则f(x)__________.0分析当x0,1112f(x)t(tx)dt
2、tdtxtdt00011f(x)tdt;02x1当0x1,f(x)t(xt)dtt(tx)dt0xxx1122xtdttdttdtxtdt00xxx121f(x)tdttdtx;0x21112当x1,f(x)t(xt)dtxtdttdt00011f(x)tdt.023x4t六、求函数I(x)2(t1)edt之极值。112解:3x2I(x)(x1)e3x令I(x)0,解得x=0或1。1212124x3x133xI(x)9xe6x(x1)e36x(x1)e1
3、I(1)9e0I(1)2为极大值。11x(,),I(x)0.22x0不是极值点。九、求数列极限n1n2n3nnlim()22222nn1n4n9nn解:inn1nin1原极限lim22limni1ninini11()2n11x111xdxdxdx01x201x201x21121arctanx
4、ln(1x)
5、0021ln242十设函数f(x)在[a,b]连续,且f(x)0,证明1b在(a,b)内有点,使f(x)dxf(x)dxa3ax1b证明:令g(x)f(
6、t)dtf(t)dt(x[a,b])a3a显然g(x)在[a,b]上连续,f(t)0(x[a,b])1b2bg(a)f(t)dt0,g(b)f(t)dt03a3a由零值定理知:(a,b)s.t.g()0原式得证。十一求函数f(x),使f(x)对任意正数a在[0,a]可积,x且当x0时f(x)0,又满足f(x)f(t)dt0解:f(0)0且x[0,a],f(x)0,xxf(t)dt连续f(t)dtf(x)在[0,a]连续(a0)aaxxf(t)连续f(t)dt可导f(x)af(t)dt可导af(x)f(x)1f
7、(x)(x0)x2f(x)22f(t)dtax则f(x)C,又f(0)021f(x)x(x0)2十二设函数f(x)在(,)连续,且xF(x)(x2t)f(t)dt0试证:若f(x)单调不增,则F(x)单调不减。xx证明:F(x)xf(t)dt2tf(t)dt00xF(x)f(t)dtxf(x)2xf(x)0xxf(t)dtxf(x)[f(t)f(x)]dt00f(x)单调不增,且tx,f(t)f(x)F(x)0即F(x)单调不减。十三若函数f(x)在[a,b]连续单增,证明babbxf(x)dxf(x
8、)dxa2atatt证明:令g(t)xf(x)dxf(x)dx,g(a)0,a2a只要证g(b)g(a)0即可。at1tg(t)tf(t)f(t)f(x)dx22a11t1t(ta)f(t)f(x)dx[f(t)f(x)]dx022a2ag(t)g(b)g(a)0得证。txe十四若F(x)dt(x0),问x在什么范围取值时,有1tF(x)lnx解:令g(x)F(x)lnx,(x0)xe1g(x)0,g(x)xx而g(1)0,0x1时,g(x)0,x1时,g(x)0,F(x)lnx0x1十五
9、若函数f(x)在[0,3]上可导,且有13xxef(x)dxef(x)dx02证明:存在(1,3),使f()f().x证明:g(x)ef(x)f(x)在[0,1]连续,由积分中值定理,x1[0,1],x2[2,3]s.t.13g(x)dxg(x1),g(x)dxg(x2)02g(x1)g(x2)(0x1x23)又g(x)在[x1,x2][0,3]可导,由罗