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(3)因为当1?x?2时, 0?ln x?1, ln x?(ln x)2, 所以?1lnxdx??1(lnx)2dx. 又因为当1?x?2时, 0?ln x?1, ln x?(ln x)2, 所以?1lnxdx??1(lnx)2dx. (4)因为当0?x?1时, x?ln(1?x), 所以?0xdx??0ln(1?x)dx. 又因为当0?x?1时, x?ln(1?x), 所以?0xdx??0ln(1?x)dx.
(5)设f(x)?ex?1?x, 则当0?x?1时f ?(x)??ex?1?0, f(x)?ex?1?x是单调增加的. 因此当0?x?1时, f(x)?f(0)?0, 即ex?1?x, 所以?0exdx??0(1?x)dx. 又因为当0?x?1时, ex?1?x, 所以?0exdx??0(1?x)dx. 习题5?2
1. 试求函数y??0sintdt当x?0及x? 解 y??x111111112222?4时的导数.
?2dx?, 当x?0时, y??sin0?0; 当x?时, y??sin?. sintdt?sinx?42dx04tt 2. 求由参数表示式x??0sinudu, y??0cosudu所给定的函数y对x的导数. 解 x?(t)?sin t , y?(t)?cos t ,
yxdyy?(t)??cost. dxx?(t) 3. 求由?0etdt??0costdt?0所决定的隐函数y对x的导数 解 方程两对x求导得 e y y? ?cos x ?0,
dycosx于是 ??y.
dxe 4. 当x为何值时, 函数I(x)??0te?tdt有极值？
x2dy. dx 解 I?(x)?xe?x, 令I ?(x)?0, 得x?0. 因为当x?0时, I ?(x)?0; 当x?0时, I ?(x)?0, 所以x?0是函数I(x)的极小值点. 5. 计算下列各导数:
2dx2 (1)?1?t2dt;
dx0dx31dt; (2)?x24dx1?t7|28
(3)
dcosx2cos(?t)dt. ?dxsinx令x2?ududx2du221?tdt??1?u2?2x?2x1?x4. 解 (1)?01?tdt?0dxdudxdx31d01dx31dt??x2dt??0dt (2)?x2444dxdxdx1?t1?t1?tdx21dx31dt??0dt ???044dxdx1?t1?t ??11?(x2)42x1?x8?(x2)??11?(x3)4?(x3)?
???3x21?x12.
(3)
dcosxdsinxdcosx222cos(?t)dt??cos(?t)dt?cos(?t)dt ???dxsinxdx0dx0 ??cos(?sin 2x)(sin x)?? cos(?cos 2x)( cos x)?
??cos x?cos(?sin 2x)?sin x?cos(?cos 2x) ??cos x?cos(?sin2x)? sin x?cos(???sin2x) ??cos x?cos(?sin2x)? sin x?cos(?sin2x) ?(sin x?cos x)cos(?sin2x)?
6. 计算下列各定积分: (1)?0(3x2?x?1)dx;
a11a?x?1)dx?(x3?x2?x)|0?a3?a2?a.
2221 (2)?(x2?4)dx;
1x211113?35213?3 解 ?(x2?4)dx?(x3?x?3)|1. ?(2?2)?(1?1)?2133338x 解
?0(3xa2 (3)?4x(1?x)dx; 解
9?49x(1?x)dx??491(x2231292312231122?x)dx?(x?x)|4?(9?9)?(42?42)?45. 32323268|28
(4)?13dx231?x;
解
?3dx1231?x?arctanx313?arctan3?arctan1???. 3366??? (5)?12?12dx1?x2;
解
?012?12dx1?x2?arcsinx12?1211????arcsin?arcsin(?)??(?)?.
22663 (6)? 解
3a2dx2?01a?x3adxdx4?x2;
3a01x?arctanaa2?x2a;
11??arctan3?arctan0?. aa3a (7)?0 解
?001dx4?x2?arcsinx211??arcsin?arcsin0?. 0263x4?3x2?1dx; (8)??1x2?14203x?3x?101?2303dx?(3x?)dx?(x?arctanx)|??(?1)?arctan(?1)?1? 解 ??1. ?1??14x2?1x2?1?2dx (9)?;
?e?11?x 解
dx?2??e?11?x?ln|1?x||?e?1?ln1?lne??1.
?2 (10)?04tan2?d?; 解
???40tan?d???24(sec2?0??1)d??(tan???)?40?tan??4?4?1??4.
(11)?0|sinx|dx; 解
2??02?2?|sinx|dx??0sinxdx???sinxdx??cos x|?0?cos x|???cos? ?cos0?cos2??cos??4.
?2?9|28
?x?1 x?1? (12)?0f(x)dx, 其中f(x)??12.
x x?1??221221213281xdx?(x?x)|?(x)|1?. 02?1263 7. 设k为正整数. 试证下列各题: 解
?0f(x)dx??0(x?1)dx??21 (1)???coskxdx?0; (2)???sinkxdx?0; (3)???cos2kxdx??; (4)???sin2kxdx??.
?111 证明 (1)?coskxdx?sinkx|??sink??sink(??)?0?0?0. ????kkk?11111 (2)?sinkxdx??coskx|???cosk??cosk(??)??cosk??cosk??0. ????kkkkk??? (3)?cos2kxdx??? (4)?sin2kxdx?????1?11???(1?cos2kx)dx?(x?sin2kx)|????. ?????222k221?11???(1?cos2kx)dx?(x?sin2kx)|????. ???2??22k22 8. 设k及l为正整数, 且k?l . 试证下列各题: (1)???coskxsinlxdx?0; (2)???coskxcoslxdx?0; (3)???sinkxsinlxdx?0.
???1?[sin(k?l)x?sin(k?l)x]dx 2???11cos(k?l)x]??[?cos(k?l)x]? ?[??????0. 2(k?l)2(k?l) 证明 (1)?coskxsinlxdx????1?[cos(k?l)x?cos(k?l)x]dx 2???11sin(k?l)x]?sin(k?l)x]? ?[???[???0. 2(k?l)2(k?l) (2)?coskxcoslxdx???? (3)?sinkxsinlxdx?????1?[cos(k?l)x?cos(k?l)x]dx. 2???10|28
?[?11sin(k?l)x]??[sin(k?l)x]??????0. 2(k?l)2(k?l) 9. 求下列极限:
cost?0 (1)limx?0x2dtx2;
(2)limx?0(?0etdt)2x?0tex?0xx2t2.
dt2cost?0 解 (1)limx(?0etdt)22xdtcosx2?lim?1. x?012?0etdt?(?0etdt)?xe2x2x2x2 (2)limx?0?0tex?0x2t2?limx?0?limx?02?0etdt?exxe2x2x22?limx?02?0etdtxex2x2
dt ?lim2ex222ex?2x2ex2?lim?2. x?01?2x2?x2 x?[0, 1]x 10. 设f(x)??. 求?(x)??0f(t)dt在[0, 2]上的表达式, 并讨论?(x)在(0, 2)内的
?x x?[1, 2]连续性.
xx1 解 当0?x?1时, ?(x)??f(t)dt??t2dt?x3;
003x1x11111 当1?x?2时, ?(x)??f(t)dt??t2dt??tdt??x2??x2?.
00132226?1x3 0?x?1?因此 ?(x)??3.
211?x? 1?x?26?211111111 因为?(1)?, lim?(x)?limx3?, lim?(x)?lim(x2?)???,
x?1?03x?1?023x?1?062633x?1?0所以?(x)在x?1处连续, 从而在(0, 2)内连续.
?1sinx 0?x??x? 11. 设f(x)??2. 求?(x)??0f(t)dt在(??, ??)内的表达式.
??0 x?0或x?? 解 当x?0时, ?(x)??0f(t)dt??00dt?0;
xx