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xx1111x 当0?x??时, ?(x)??f(t)dt??sintdt??cost|0; ??cosx?002222 当x??时, ?(x)??f(t)dt??00x?x1111sintdt???0dt??cost|???cos??cos0?1. 022220 x?0???因此 ?(x)??1(1?cosx) 0?x??.
?2?1 x??? 12. 设f(x)在[a, b]上连续, 在(a, b)内可导且f ?(x)?0,
1xf(t)dt. x?a?a证明在(a, b)内有F ?(x)?0. F(x)? 证明 根据积分中值定理, 存在??[a, x], 使?af(t)dt?f(?)(x?a). 于是有 F?(x)?? ?f(t)dt?f(x)?f(x)?2?ax?ax?a(x?a)(x?a)21xx111f(?)(x?a)
1[f(x)?f(?)]. x?a 由f ?(x)?0可知f(x)在[a, b]上是单调减少的, 而a???x, 所以f(x)?f(?)?0. 又在(a, b)内, x?a?0, 所以在(a, b)内 F?(x)?1[f(x)?f(?)]?0. x?a习题5?3
1. 计算下列定积分:
(1)??sin(x?)dx;
32??? 解 ??sin(x?)dx??cos(x?)332 (2)??2 解
1????2??cos4?2?11?cos???0. 3322dx311?2???2(11?5x)35??2(11?5x)?0(11?5x)1dx;
1?2??1151?16?2??1?2?. 1010512 (3)?2sin?cos3?d?;
1 解 ?02sin?cos?d????02scos?dsin???cos3?433???201?11??cos3?cos30?. 4244 (4)?0(1?sin3?)d?;
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解
32?0(1?sin?)d???0d???0sin?dcos???????0??0(1?cos2?)dcos?
?1 ???(cos??cos3?)3 (5)??2cos2udu;
6?04???.
3? 解
???2cos2udu?61?12(1?cos2u)du?u2??26??261?sin2u4??2
61??1??3 ?(?)?(sin??sin)??.
2264368 (6)?0 解
22?x2dx;
?令x?2sint?22?xdx?02cost?2costdt??02(1?cos2t)dt
2?021 ?(t?sin2t)2?20??2.
(7)?? 解
228?2y2dy; 8?2ydy?2???222??22?令y?2sinx4?ydy2?4?2cosx?2cosxdx
2?41 ?22?4?(1?cos2x)dx?22(x?sin2y)?24112?4??4?2(??2).
(8)?1?x2x2dx;
解
?1121?x2x2?令x?sint?cost122(dx?costdt???sin2t??sin2t?1)dt?(?cott?t)44??24?1?.
4? (9)?0x2a2?x2dx; 解
a?0ax2令x?asint?a4222a?xdx?0asint?acost?acostdt?422?02sin?22tdt
a4 ?8?2(1?cos4t)dt?0?a4t8?20a4?sin4t32?20a4??. 1613|28
(10)?13dxx21?x2;
解
?13dxx?2令x?tant21?x??tan2t?sect?sec34?12tdt
1 ???dt??2sint4sint3cost??34?2?23. 3 (11)??1 解
1xdx5?4x;
1??141xdx令5?4x?u111132(5?u)du??(5u?u)?38835?4xdxx1?. 36 (12)
?11?4;
211?2udu?2(1??11?u?11?u)du?2(u?ln|1?u|)221 解
?11?14dx令x?uxdx1?x?12?2(1?ln).
3 (13)?3 解
;
110112?(?2u)du?20(1?)du?2(u?ln|u?1|)210u?12u?1?134dx令1?x?u1?x?12a???1?2ln2.
(14)?0xdx3a?x22;
解
?012axdx3a?x?t2222??12a12222d(3a?x)??3a?x?203a2?x22a0?a(3?1).
(15)?0te 解
dt;
1?t22?0tee21?t22dt???0edxtd(?)??e22?t2210?1?1?e2.
(16)?1 解
?1e2x1?lnxdxx1?lnx; ??1e211?lnxdlnx?21?lnxe21?2(3?1).
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(17)?02dxx?2x?2 00dx1???2dx?arctan(x?1) 解 ??222x?2x?21?(x?1) (18)?2?cosxcos2xdx;
?2?2;
0?2?arctan1?arctan(?1)??2.
? 解
?2cosxcos2xdx?2(1?2sin2????22???2x)dsinx?(sinx?sin3x)3?2??22?. 3 (19)?2?cosx?cos3xdx;
?2? 解
??2??2cosx?cosxdx??2?cosx1?cos2xdx
?23? ????cosx(?sinx)dx??20?2032cosxsinxdx?cos2x30??232?cos2x3?204? 3 (20)?01?cos2xdx. 解
??0??1?cos2xdx?2?0sinxdx??2cosx??0?22.
2. 利用函数的奇偶性计算下列积分: (1)???x4sinxdx;
解 因为x 4sin x在区间[??, ?]上是奇函数, 所以???x4sinxdx?0. (2)?2?4cos4?d?;
?2?? 解
?24cos4?d???2??2?24cos4?d?0??8?02(?1?cos2x2)d? 2??31 ?2?2(1?2cos2x?cos22x)d??2?2(?2cos2x?cos4x)d?
00221 ?(3??2sin2x?sin4x)4?20?3?. 215|28
(3)?12?12(arcsinx)21?x2dx;
解
?12?12(arcsinx)21?x2dx?2?120120(arcsinx)21?x2dx?2?02(arcsinx)2d(arcsinx)
12 ?(arcsinx)335??3324.
x3sin2xdx. (4)??54x?2x2?1 解 因为函数 3. 证明:
ax3sin2xx4?2x2?1是奇函数, 所以??5a5x3sin2xx4?2x2?1dx?0.
22??a?(x)dx?2?0?(x)dx, 其中?(u)为连续函数.
证明 因为被积函数?(x2)是x的偶函数, 且积分区间[?a, a]关于原点对称, 所以有
22??a?(x)dx?2?0?(x)dx.
bbaa 4. 设f(x)在[?b, b]上连续, 证明??bf(x)dx???bf(?x)dx. 证明 令x??t, 则dx??dt, 当x??b时t?b, 当x?b时t??b, 于是 而 所以
??bf(x)dx??bbb?bf(?t)(?1)dt???bf(?t)dt,
b??bf(?t)dt???bf(?x)dx, ??bf(x)dx???bf(?x)dx.
bbbbb 5. 设f(x)在[a, b]上连续., 证明?af(x)dx??af(a?b?x)dx. 证明 令x?a?b?t, 则dx?d t, 当x?a时t?b, 当x?b时t?a, 于是 而 所以
?af(x)dx??bf(a?b?t)(?1)dt??af(a?b?t)dt, ?abbabf(a?b?t)dt??af(a?b?x)dx,
bb?af(x)dx??af(a?b?x)dx.
?x1?x21b 6. 证明:
dx??1x1dx1?x2(x?0).