高数同济版第五章答案(2)

6|28

(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

百度搜索“77cn”或“免费范文网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读，免费范文网，提供经典小说综合文库高数同济版第五章答案(2)在线全文阅读。