高数同济第五版 第四章答案

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习题4?1

1. 求下列不定积分:

1 (1)?2dx;

x111 解 ?2dx??x?2dx?x?2?1?C???C.

?2?1xx (2)?xxdx; 解 ?xxdx??1x3x2dx?12?12x?C?x2x?C. 35?123 (3)? 解

dx;

?1xdx??x?12dx???11x2?C?2x?C. 1??121 (4)?x23xdx; 解 ?x23xdx??1x27x3dx?17?137?13x?C?333xx?C. 10 (5)? 解

xdx;

?x21xdx??x?52dx???1131x2?C????C. 52xx??125 (6)?mxndx; 解

?mxdx??nnxmdx??11mxm?C?xnn?m?1mnm?nm?C.

(7)?5x3dx;

5 解 ?5x3dx?5?x3dx?x4?C.

4 (8)?(x2?3x?2)dx;

13 解 ?(x2?3x?2)dx??x2dx?3?xdx?2?dx?x3?x2?2x?C.

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(9)?dh2gh(g是常数);

解

?dh2gh?12g?h?12dh?12g1?2h2?C?2h?C. g (10)?(x?2)2dx;

1 解 ?(x?2)2dx??(x2?4x?4)dx??x2dx?4?xdx?4?dx?x3?2x2?4x?C.

3 (11)?(x2?1)2dx;

12 解 ?(x2?1)2dx??(x4?2x2?1)dx??x4dx?2?x2dx??dx?x5?x3?x?C.

53 (12)?(x?1)(x3?1)dx;

解 ?(x?1)(x?1)dx??(x?x?x?1)dx??xdx??122 ?x3?x2?x2?x?C.

3353532321x2dx??3x2dx??dx

(13)?(1?x)2xdx;

121?2x23?x21)dx?2x235 解

?(1?x)2xdx??1?2x?x2xdx??(x?42?x2?x2?C. 353x4?3x2?1dx; (14)?x2?1 解

?3x4?3x2?1x2?1dx??(3x2?1x2?1)dx?x3?arctanx?C.

x2dx; (15)?1?x2 解

?1?x2dx??x2x2?1?11?x2dx??(1?11?x2)dx?x?arctanx?C.

3 (16)?(2ex?)dx;

x31 解 ?(2ex?)dx?2?exdx?3?dx?2ex?3ln|x|?C.

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(17)?(31?x31?x22?21?x2)dx;

解 ?(?21?x2)dx?3?11?x2dx?2?11?x2dx?3arctanx?2arcsinx?C.

(18)?e(1?xe?xx)dx;

121?2x2 解 ?e(1?xe?xx)dx??(e?xx?)dx?ex?C.

(19)?3xexdx;

(3e)x3xex?C??C. 解 ?3edx??(3e)dx?ln(3e)ln3?1xxx

2?3x?5?2xdx; (20)?x32x()2?3x?5?2x2x52x3 解 ?dx?[2?5()]dx?2x?5?C?2x?()?C. ?23ln2?ln333xln3 (21)?secx(secx?tanx)dx;

解 ?secx(secx?tanx)dx??(sec2x?secxtanx)dx?tanx?secx?C.

x (22)?cos2dx;

2x1?cosx11 解 ?cos2dx??dx??(1?cosx)dx?(x?sinx)?C.

2222 (23)? 解

1dx;

1?cos2x111dx?dx??1?cos2x?2cos2x2tanx?C.

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(24)?cos2xdx;

cosx?sinxcos2xcos2x?sin2xdx??dx??(cosx?sinx)dx?sinx?cosx?C. 解 ?cosx?sinxcosx?sinx

(25)?cos2x22cosxsinxcos2xcos2x?sin2x11dx?dx?(?)dx??cotx?tanx?C. 解 ???cos2xsin2xcos2xsin2xsin2xcos2x

1 (26)?(1?2)xxdx;

x??1?4? 解 ??1?2?xxdx??(x4?x4)dx?x4?4x4?C.

7?x?3571dx;

2. 一曲线通过点(e2, 3), 且在任一点处的切线的斜率等于该点横坐标的倒数, 求该曲

线的方程.

解 设该曲线的方程为y?f(x), 则由题意得

1 y??f?(x)?,

x1所以 y??dx?ln|x|?C.

x 又因为曲线通过点(e2, 3), 所以有?3?2?1

3?f(e 2)?ln|e 2|?C?2?C, C?3?2?1. 于是所求曲线的方程为 y?ln|x|?1.

3. 一物体由静止开始运动, 经t秒后的速度是3t2(m/s), 问 (1)在3秒后物体离开出发点的距离是多少？ (2)物体走完360m需要多少时间？

解 设位移函数为s?s(t), 则s??v?3 t2, s??3t2dt?t3?C. 因为当t?0时, s?0, 所以C?0. 因此位移函数为s?t 3. (1)在3秒后物体离开出发点的距离是s?s(3)?33?27. (2)由t 3?360, 得物体走完360m所需的时间t?3360?7.11s. ex12xxx

4. 证明函数e, eshx和echx都是的原函数.

chx?shx25|34

exexex 证明 ?x?x??x?e2x. x?xchx?shxe?ee?ee?22 因为

1 (e2x)??e2x,

2ex12x所以e是的原函数.

chx?shx2 因为

ex?e?xex?e?x?)?e2x, (eshx)??eshx?echx?e(sh x?ch x)?e(22x

x

x

x

xex所以eshx是的原函数.

chx?shx 因为

x

ex?e?xex?e?x?)?e2x, (echx)??echx?eshx?e(ch x?sh x)?e(22x

x

x

x

xex所以echx是的原函数.

chx?shx习题4?2

x

1 1. 在下列各式等号右端的空白处填入适当的系数??使等式成立(例如??dx?d(4x?7):

4 (1) dx???d(ax); ?

1 解dx???d(ax).

a (2) dx? d(7x?3);?

1 解dx? d(7x?3).

7 (3) xdx? d(x2); 解xdx? 1 d(x2).

2 (4) xdx? d(5x2);?

1 解xdx? d(5x2).

10 (5)xdx? d(1?x2);

1 解 xdx? ? d(1?x2).

2 (6)x3dx? d(3x4?2);?

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