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1、NO.*补充习题2（微积分基本定理）.f ( x)1 ( x2ex )(g(x) dx)g( f ( x)f ( x) ，0xf (x)g ( f (x)f (x) ( x22 x)ex，f (x)( x2) exf ( 0) 01)ex，f ( x)( x1。xx3 t 2 ) dt2 当1x0时， F ( x)f (t ) dt(2t112当0x1x0x03 t 2 )dtF ( x)f (t ) dtf (t ) dtf (t ) dt(2t11012lnexxln 21；x1x12eex3x21,1x0所以F (x)22exx1,0x1lnln 2x1ex12earctanxt2arc
2、tan2 x13 ky x(edt) x xe1，0001x2 x0y( 0) 0yx ；x3x21；22时，xtet2 dtt1)0 (e.f (2f (2f ( 0)lim n f ( 2 )limn2 limn2 f (0)2 12。nnn1n2nn4xuxx(f (t )dt)duxf (t )dttf (t )dtaaaa（ Just think it over : what aboutax；bub? It means to prove( f (t)dt)du(b t) f (t)dt ）xxxxb5 设 F ( x)f (x) dx g( x)dx ；F (a) F (b) 0 ；
3、ax6；1N0.*NO.*；117f (x)dxf (x)dx (1)f (x)dxf (x)dx000(1)f ( )(1) f ( ) 0 , 01 。另证 利用积分区间可加性，把结论不等式等价变形为aa1aaf (x) dxf ( x) dxf ( x) dx，即：0a08limf ( 2x a)存在xaxa 0limf (2xa)0f (a)0xa0f ( x)f ( a)0x取F ( x)x2 ， G( x)f ( x)dx ，a由柯西定理F (b)F (a)F ()G(b)G(a)G()910 单增区间：( 2,1)；单减区间： (0,2 )， 上凹区间(0,1) ； f极小f ( 2 )21 。222632N0.*
