??a?af(x)g(x)dx??a0f(?x)g(x)dx??a0f(x)g(x)dx???f(x)?0af(?x)?g(x)dx?A?g(x)dx0a
9．设f(x)连续，证明：?lnf(x?t)dt?01?x0lnf(1?t)f(t)dt??10lnf(t)dt
证明：依题意?lnf(x?t)dt令x?t?u?01x?1xlnf(u)du
＝?lnf(u)du?x0?10lnf(u)du?x?x?11lnf(u)du
又?? 所以
x?11lnf(u)duu?t?1?lnf(t?1)dt
0?10lnf(x?t)dt??x0lnf(1?t)f(t)dt??10lnf(t)dt
?10．设n为正整数，证明?2cosxsinxdx?nn012n??20cosnxdx
证明：令t=2x,有
?
?20cosxsinxdx?nn12n?1??20(sin2x)d2x??n12n?1??0sintdt
n?1?2nsintdt? ?n?1???02??0?n?, sintdt???2?? 又，??sintdtt???u?2n??sin2n(??u)du???20sinnudu，
??所以，?2cos0nxsinnxdx?120n?1?(?2sin0ntdt???20sinntdt)?12n?20sinntdt?12n??2?sinnxdx
又，??sin2?nxdxx??2n?t???cos2ntdt??20cosnxdx
?因此，?2cosxsin0nxdx?12n??20cosnxdx
11．设函数f(x)可导，且f(0)=0, F(x)??x0tn?1f(x?t)dt证明:
nn
lim证明：F(x)?F(x)xn2nx?0?12nnnf?(0)
f(u)du
?x0tn?1f(x?t)dtn令u?x?t?1?nxn0于是，F?(x)?xn?1f(xn),
limF(x)x2nx?0?limF?(x)2nx2n?1x?0?12nlimf(x)xnnx?0?12nlimf(x)?f(0)x?0nnx?0?12nf?(0)
12．设?(t)是正值连续函数，f(x)?在??a,a?上是凹的。 证明：f(x)??a?ax?t?(t)dt,?a?x?a(a?0),则曲线y?f(x)?xx?a(x?t)?(t)dt?x?axx(t?x)?(t)dt
a ?x??(t)dt??t?(t)dt??t?(t)dt?x??(t)dt
?a?a?ax f?(x)??x?a?(t)dt??ax?(t)dt??x?a?(t)dt??xa?(t)dt
f?(x)??(x)??(x)?2?(x)?0 故，曲线y?f(x)在??a,a?上是凹的。
13．设g(t)是?a,b?上的连续函数，f(x)?使
f(b)b?a?g(?).
xb?xag(t)dt,证明：在?a,b?上至少存在一点?，
证明：由已知条件f(x)??ag(t)dt,有f(b)??ag(t)dt,
又，由于g(t)在?a,b?上连续，由积分中值定理，有
?bag(t)dt?g(?)(b?a),a???b,
f(b)b?a?g(?) 故 f(b)?g(?)(b?a)?14.证明：?1
dx1?x21x??xdx1?x21
证明：?1dx1?x2令x?1u1x??11?1u21x?(?1u21du)??xdu1?u211??xdx1?x21
15．设f(x)是定义在全数轴上，且以T为周期的连续函数，a为任意常数，则 证明：??a?Taf(x)dx???T0f(x)dx
?a?T令x?u?TTf(x)dx?a0f(u?T)du??a?f(x)以T为周期0f(x?T)dxf(x?T)?f(x)??a0f(x)dx
??f(x)dx?0T0a?a?TTf(x)dx?0
a?T在等式两端各加?f(x)dx，于是得?af(x)dx??T0f(x)dx
xu16．若f(x)是连续函数，则???f(t)dt?du??0?0???x0(x?u)f(u)du
xuux证明：???f(t)dt?du?u?f(t)dt?0??0??00?x0uf(u)du
?x?f(t)dt??uf(u)du
00xx ??bx0(x?u)f(u)du
17．设f(x)，g(x)在?a,b?上连续，证明至少存在一个??(a,b)使得 f(?)?g(x)dx?g(?)?f(x)dx
?a?证明：作辅助函数F(x)??x?xa由于f(x),g(x)在?a,b?上连续，所以F(x)f(t)dt?g(t)dt，
xxb在?a,b?上连续，在（a,b）内可导，并有F(a)?F(b)?0 由洛尔定理F?(?)?0,??(a,b)
xb?即?f(t)dt?g(t)dt???x?a?x??b??f(x)?g(t)dt??x??af(t)dt?g(x)????ax??
?f(?)?g(x)dx?g(?)?f(x)dx
?b ＝0 亦即，f(?)?g(x)dx?g(?)?f(x)dx
?ab?
18．设f(x)，g(x)在?a,b?上连续，且g(x)?0,x??a,b?，试证：至少存在一个??(a,b)使得
?babf(x)dx?g(x)dxf(?)g(?)
?
a
证明：令F(x)? F(b)??baxaf(t)dt,G(x)??xabg(t)dt,于是 g(x)dx,
x?f(x)dx, G(b)?x?a作辅助函数W(x)?F(b)?g(t)dt?G(b)?f(t)dt
za由题设条件，显然W(x)在?a,b?上连续，由变上限积分定理W(x)在?a,b?上可导 又 W(a)?0，W(b)?F(b)G(b)?G(b)F(b)?0
由洛尔定理，在?a,b?内至少存在一点??(a,b)，使W?(?)?0 即 F(b)g(?)?G(b)F(?)?0，亦即
F(b)G(b)?f(?)g(?)，故
?babf(x)dx?g(x)dxf(?)g(?)
?abb??19．设f(x)在?a,b?上连续，证明：??f(x)dx??(b?a)?f2(x)dx
a?a?xx2? 证明：令F(x)????f(t)dt??(x?a)?f(t)dt a?a?22 ?F?(x)????f(t)?a2xf(x)?dt?0
2 故f(x)是 ?a,b?上的减函数，又F(a)?0，F(b)?F(a)?0
bb2?故 ???f(x)dx??(b?a)?f(x)dx a?a?
20．设f(x)在?a,b?上可导，且f?(x)?M，f(a)?0证明：
?baf(x)dx?M2(b?a)2
证明：由题设对?x??a,b?,可知f(x)在?a,b?上满足拉氏微分中值定理，于是有
f(x)?f(x)?f(a)?f?(?)(x?a),???a,x? 又f?(x)?M，因而，f(x)?M(x?a) 由定积分比较定理，有
?baf(x)dx??baM(x?a)dx?M2(b?a)2
《高等数学Ⅰ》答疑题
11．limx?0(arcsinxx(1?)x2
1原式=limx?0arcsinx?xx)x2
1因为limx?0arcsinx?xx?1x22=limx?01?x3x1622?1=limx?01?3x21?1?x2x2
=limx?0x3x2=
1?x2(1?1?x2)1所以 极限=e6
12．求limx?0(xxx??????a1a2anxnnn)
1原式=limx?0(1?xxx??????a1a2an?nx)
xa1xlna1?????anlnan?
limx?0xxx??????a1a2an?n?1x2=limx?0
n=
lna1?a2???annaa12=lnnaa1???an
?极限=n3???an
3.求limx???x2(x?2?2x?1?3x)
x)] ] x原式=limx???x2[(x?2?32x?1)?(x?1??1x?1?
=limx???x[[[1x?2?x?13=limx???xx2x?(x?2?(x?2?x?2x)?2]
]
x)(x?x?2)x?1)(x?1?x?1)(x?1?143=limx???23=limx???x24．求limn??tan2x?2x?2x1n?(?) 4n[?2]=-