3. 求方程y???2y??y?x的一个特解. 4. 求方程y???5y??9y?5xe
高数一复习资料参考答案
一、选择题 1-5: DABAA 6-10:DBCDD 11-15: BCCBD 16-21:ABAAAA
二、求积分
1.求cosxsinxdx.
?3x2的通解.
?解:cosxsinxdx???232sinxd(sinx)?sin2x?C?sin3x?C
332. 求
?34?3lnxdx. x解:
?31114?3lnx3dx??(4?3lnx)d(lnx)??(4?3lnx)3?d(4?3lnx) x341?(4?3lnx)3?C. 43. 求arctanxdx.
解:设u?arctanx,dv?dx,即v?x,则
??arctanxdx?xarctanx??xd(arctanx)
x?1?x2dx 1 ?xarctanx?ln(1?x2)?C.
2 ?xarctanx?4. 求edx 解:edx?3x?3xx?t3t22t2tt2tte3tdt?3tedt?3te?3e?2tdt?3te?6te????dt
?3t2et?6tet?6etdt?3t2et?6tet?6et?C
?
?3ex(3x2?23x?2)?C. 5. 求
3x?3?x2?5x?6dx.
x?3?56解:由上述可知2,所以 ??x?5x?6x?2x?3x?3?5611dx?(?)dx??5dx?6?x2?5x?6?x?2x?3?x?2?x?3dx
??5lnx?2?6lnx?3?C.
6. 求定积分?dx. 301?x8解:令3x?t,即x?t3,则dx?3t2dt,且当x?0时,t?0;当x?8时,t?2,于是
223tdtdx?12?2??3t?t?ln(1?t)?3ln3. ?01?3x?01?t???2?08 7. 计算
??0x2cosxdx.
解:令u?x2,dv?cosxdx,则du?2xdx,v?sinx,于是
??0x2cosxdx??x2dsinx?(x2sinx)0??0??2xsinxdx??2?xsinxdx.
00??再用分部积分公式,得
?
8. 求
?0xcosxdx?2?2?0xdcosx?2?(xcosx)???0?0??cosxdx?
?0?? ?2?(xcosx)???sinx?0???2?.
1?x2?2x?8dx.
解:
1113?(x?1)dx?d(x?1)?ln?C ?x2?2x?8?(x?1)2?963?(x?1)12?xln?C. 64?x ?9. 求
dx?1?3x?2.
3解:令u?x?2,则x?u3?2,dx?3u2du,从而有
dx3u2u2?1?1?du?3?du ?1?u1?3x?2?1?u1u2)du?3(?u?ln1?u)?C ?3?(u?1?1?u211. 求
?212xe?xdx
2解:
?2122xe?x2dx??e12?x2dx?e2?x221?e?4?e?1
12. 求3x?3?x3dx
2解:3x
13. 求
?3?xdx???332323?xd(3?x)??(3?x)?C
333?e1ln2xdx x解:
?e1eln2x111dx??ln2xd(lnx)?lnx?lne?
1x3331e14.求x3?xdx
3311212222223?xd(3?x)???(3?x)?C??(3?x)?C 解:?x3?xdx???22332?2
三、解答题
11. 若lim3x?ax2?x?1?,求a
x??6??解:因为3x?ax?x?1?29x2?ax2?x?13x?ax?x?12,所以a?9
否则极限不存在。
12.讨论函数f(x)?x3?2x2?3x?3的单调性并求其单调区间
3解:f'(x)?x2?4x?3
由f'(x)?x2?4x?3?0得x1?1,x2?3
所以f(x)在区间(??,1)上单调增,在区间(1,3)上单调减,在区间(3,??)上单调增。
x2?x?23. 求函数f(x)?的间断点并确定其类型
x?2解:函数无定义的点为x?2,是唯一的间断点。 因limf(x)?3知x?2是可去间断点。
x?24. 设xy2?sinx?exy,求y?.
解:y2?2xy?y??cosx?exy(y?y?),
y(exy?y)?cosx故 y??
x(2y?exy)
(x?1)3x?25. 求y?的导数.
(x?3)5解:对原式两边取对数得:
1lny?3ln(x?1)?ln(x?2)?5ln(x?3),
2于是
y?3115????, yx?12x?2x?3(x?1)3x?23115[???]. 故 y??5x?12x?2x?3(x?3)
?x?acost6. 求由方程? 确定的导数y?x.
y?bsint?y?(t)bcostb2x???2. 解: y?x?x?(t)?asintay?1x?e,x?0?7. 函数f(x)??1,x?0在x?0处是否连续?
?tanx,x?0??
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