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7. 计算下列积分: (1)?20?x?sinxdx;
1?cosxx?sinx? 解
?2dx?0201?cosx??xx2cos22?dx??20?d(1?cosx)?x??02xd(tan)?ln(1?cosx)1?cosx2?2020?
x ?(xtan)2??20x?x??02tandx?ln2??2lncos222?ln2??2.
(2)?04ln(1?tanx)dx;
2sin(?x)4dx 解 ?04ln(1?tanx)dx??04lncosx????? ??04ln2dx??04lnsin(?x)dx??04lncosxdx.
4 令
???4?x?u, 则
所以 (3)?0a?4lnsin(?x)dx?4lnsin(?0044??????4?u)du??4lncosudu?4lncosxdx,
00???
???04ln(1?tanx)dx??04lndxx?a?x222dx?ln2??4??82.
;
解 令x?a sin t, 则
?0adxx?a2?x2??02?costdt.
sint?cost 又令t??u, 则
2 所以
????20a?costdtsinudu, ??02sint?costsinu?cosu?01?sint?cost1??2??0dt??02dt?.
24x?a2?x22sint?costdx (4)?021?sin2xdx;
?? 解
?021?sin2xdx??02|cosx?sinx|dx
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??4(cosx?sinx)dx?2(cosx?sinx)dx?04???
?(sinx?cosx)??40??(sinx?cosx)?24?2(2?1).
(5)?20dx. 21?cosxdx1?cosx2 解
?20???02?dxcosx(secx?1)?2022??02?dtanx2?tanx2
?12arctantanx2?2?. 4xxt 8. 设f(x)为连续函数, 证明?0f(t)(x?t)dt??0[?0f(u)du]dt. 证明
?0[?0f(u)du]dt?t?0f(u)duxxxttx0??0td[?0f(u)du]
xt ?x?0f(u)du??0tf(t)dt
?x?0f(t)dt??0tf(t)dt??0f(t)(x?t)dt.
9. 设f(x)在区间[a, b]上连续, 且f(x)>0, F(x)??af(t)dt??b (1)F ?(x)?2;
(2)方程F(x)?0在区间(a, b)内有且仅有一个根.
1?2. 证明 (1)F?(x)?f(x)?f(x) (2)因为f(x)?0, a?b, 所以
adtb?0, F(b)??f(t)dt?0, F(a)??baf(t)由介值定理知F(x)?0在(a, b)内有根. 又F??(x)?2, 所以在(a, b)内仅有一个根.
xxxxxdt, x?[a, b]. 证明: f(t)?1 x?02?1?x 10. 设f(x)?? , 求?0f(x?1)dx.
?1x x?0?1?e 解
?02011令x?1?t11f(x?1)dxf(t)dt?dt???1??11?et?01?tdt
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???10e?t?te?1 11. 设f(x)在区间[a, b]上连续, g(x)在区间[a, b]上连续且不变号. 证明至少存在一点
dt??011dt??ln(e?t?1)1?t0?1?ln(1?t)10?ln(1?e).
x?[a, b], 使下式成立
?af(x)g(x)dx?f(?)?ag(x)dx (积分第值定理) .
bb 证明 若g(x)?0, 则结论题然成立.
若g(x)?0, 因为g(x)不变号, 不妨设g(x)>0.
因f(x)在[a, b]上连续, 所以f(x)在[a, b]上有最大值M和最小值m即 m?f(x)?M,
因此有 mg(x)?f(x)g(x)?Mg(x). 根据定积分的性质, 有
m?ag(x)dx??af(x)g(x)dx?M?ag(x)dx, 或
f(x)g(x)dx?am??M. b?ag(x)dxbbbbb 因为f(x)在[a, b]上连续, 根据介值定理, 至少存在一点x?(a, b), 使
?af(x)g(x)dx?f(?), b?ag(x)dx?af(x)g(x)dx?f(?)?ag(x)dx.
bb即
*12.(1)证明: 证明
?02??xne?xdx?2n?1??n?2?x2xedx,(n?1) ?02?0??xne?xdx?1??n?1?x2xd(e) 2?022??1?? =?[(xn?1e?x)|0??0e?xd(xn?1)]
2?n?1??n?2?x2xedx(n?1) ?02??2n?1?x2xe0(2)证明?1dx??(n?1)(n?N)
2