4?22(??22?sinudu?22(?cosu)040??2?1)?22?2?2(2?1)2
23.解:所围平面图形绕y轴旋转一周所得旋转体体积为:
Vy???(b?a?y)dy???(b?a?y)dy?4?b?
222222?a?aaaa?a1
a2?y2dy?4?b?a2?2?2a2b
2
y
(x?b)2?y2?a2
0
b?a
bb?a
x四、综合题:本大题共3小题,每小题10分,共30分。
x2(x?3)
24.解:定义域为:?xx?1?,f?(x)?,令f?(x)?0,得到:x1?0,x2?3,
(x?1)3由驻点和无定义点划分定义域并列出如下表格:
x
f?(x)f(x)
(??,0)
00
非极值
(0,1)(1,3)
30
极小值
(3,??)
?
增
?
增
?
减
?
增
f??(x)?
6x,令f??(x)?0,得到:x?0,
(x?1)4x
f??(x)f(x)
(??,0)
00
拐点
(0,1)(1,??)
?
凸
?
凹
?
凹
27,4
所以单调递增区间:(??,1),(3,??);单调递减区间:(1,3);极小值:f(3)?凹区间:(0,1),(1,??);凸区间:(??,0),拐点:(0,0)
x3?limf(x)?lim??,故函数f(x)有一条垂直渐近线:x?1x?1x?1(x?1)2x3?lim??,故函数f(x)无水平渐近线x??(x?1)2f?x?x3
?k?lim?lim?1,2x??x??xx(x?1)x3x3?x(x?1)22x2?xb?limf(x)?kx?lim?x?lim?lim2?2
x??x??(x?1)2x??x??x?2x?1(x?1)2故函数f(x)有一条斜渐近线:y?x?2
25.解:(1)
S0??xe?xdx??(2?x)e?xdx???xd(e?x)??(x?2)d(e?x)
01011212??[xe??[e?1
??x1
???edx]?[(x?2)e]??e??e?]?[0?(?e)]??e?1?x
?x
2
2
0
0
1
1
?x1
?1
?x2
1
0
2?x
d(x?2)
??[e?1?e?1?1]?e?1?e?2?e?1?e?2?2e?1?1?
(2)令x?2n?t,dx?dt,S0?
12??12ee
2200?
0f(t)e?(t?2n)dt??f(t)e?te?2ndt?e?2n?f(t)e?tdt
?e?2n?f(x)e?xdx?e?2n(
0212??1)2ee26.解:因为f(x)?sinx?
即:f(x)?sinx?x
x0
?
x
0
(x?t)f(t)dt=sinx?x?f(t)dt??tf(t)dt
0
0
xx
?
x0
f(t)dt?
?
x0
tf(t)dt,两边同时对x求导可得:f?(x)?cosx??f(t)dt,再继续两边同时对x求导可得:f??(x)??sinx?f(x),即:f??(x)?f(x)??sinx,且令上式的变限积分为0可知:f(0)?0,f?(0)?1
下面解微分方程y???y??sinx,满足初值条件:y(0)?0,y?(0)?1的特解??特征方程为:r?1?0,r1?i,r2??i,所以y?y?0的通解为y?C1cosx?C2sinx,2又因为??0,??1,???i?i是特征方程的根,所以k?1,所以设y*?x(acosx?bsinx),?y*???acosx?axsinx?bsinx?bxcosx,
?y?????2asinx?axcosx?2bcosx?bxsinx,所以把y和?y???代入*
**
1
,b?0,所以通解为:2111
y?y?y*?C1cosx?C2sinx?xcosx,y???C1sinx?C2cosx?cosx?xsinx
222
y???y??sinx中可得:a?
又因为y(0)?0,y?(0)?1,所以联立可得:C1?0,C2?
11
f(x)?sinx?xcosx
22
1
,所以2