e?x11xx134.解:?e(1?2)dx=?(e?2)dx=e++C
xxxx135.解:选A 136.解:因为137.解:对138.解:139.解:
2sin2xdx=2sinxcosxdx=2sinxdsinx=sinx+c,故选B ????xf(x)dx=xsinx??sinxdx两边求导得xf(x)=sinx+xcosx?sinx ,故选C
?x?xf'(x)dx=?xdf(x)=xf(x)??f(x)dx=?xe?e?x+c,故选B
?f'(lnx)1dx=f(lnx)+c=+c,故选B xx'140.解:
(?f(x)dx)=f(x),故选A
52141.解:选C 142.解:?xxdx=x2+c,c=1,故选B
5143.解:144.解:
11?dx+c,选B =?x32x2f(x)=(xlnx)'=1+lnx,?xf(x)dx=?(x+xlnx)dx
1112x212121=x+?lnxd=x+xlnx?x2+c=x2(+lnx)+c,选B
4222224145.解:
11=sin?xcosxdxcos2x+c,选A =sin2xdx??24x146.解:选B 147.解:选A
148.解:因为lim?sintdt0xx→0
=lim?xdx0x2sin?tdt0xsinx=1,故选D
x→0x149.解:因为limx→0
2?xdx0sin2x=lim=1,故选D 2x→0x?150.解:limx→0x0sint3dtx4lnx2sinx31=lim=,故选A x→04x342d151.解:因为
dx152.解:因为
lnxt+1eedt=?+102=2ex,故选C xdf(x)=?sintdt=sinx,故选A
dx0=3x3x=?0,所以?(0)为 213x?x+1(x?)2+2425
x153.解:?'(x)
函数?(x)=?3t1]上的最小值 ,故选D dt在区间[0,20t?t+12x212x154.解:
x→+?lim3f'(x)e(3x+1)(3x+1)= =lim=limx→+?cxc?1+2xc2g'(x)x→+?(cxc?1+2xc)e2xx212所以c=1,故选B
d(155.解:
dx?11+x2111+tdt)== +x,故选D
2x2x4x156.解:选C 157.解:a=lim?sintdt0x→0x2=limsinx1=,故选B
x→02x2158.解:由于F'(x)=f(x),故选B
x2f(t)dtxx?2af(t)dt=limxlim=a2f(a),选B 159.解:因为limF(x)=lim?x→ax→ax?aax→ax→ax?a160.解:选C 161.解:选A 162.解:
?+?0e?xdx=?e?x+?=1,选C 0163.解:
??01+cos2xdx=?=??x0?02cos2xdx=??02cosxdx=22,选C
164.解:F(?x)令t=?u,则 f(t)dt,xF(?x)=?f(?u)(?du)=??f(u)du=?F(x),选B
00+?x165.解:因为
?1+??+1+?dx1=x2=2,故选B
31xx?+123166.解:因为
dx1?2+?1=x=,故选A 3?1?22x111?px+?e= e?pa,故选C
Pap+?167.解:
?pxe?dx=?a168.解:
?+?edx1+?=?=1,故选A 2lnxex(lnx)169.解:
?+?0e?kx+??kx1?kx+?dx=?e,所以积分?edx收敛,必须k?0故选A
00k+?lnx0+?=1,选A 171.解:?dx=lnlnx170.解:?edx=e,发散,选B
??e??ex0xx172.解:因为
?+?e11+?dx=?=1,选C 173.解:选B 2lnxex(lnx)26
174.解:若f(x)在区间[a,b]上连续,则f(x)在区间[a,b]上可积。反之不一定成立.因此是充分条件。所以答案为B. 175.解:由于
sinx1+x2在对称区间[-1,1]上为奇函数,因此积分值为0.所以答案为A.
0x4332176.解:
??2x|x|dx=??2(?x)dx+?0xdx=?4101?2x4+411=4+
0117=.所以答案为C. 44e177.解:?(5x+1)edx=?(5x+1)d00545x45x5x1ee(5x+1)??d(5x+1) =
05505x6e5?1e5x?=
55221=e5.所以答案为B.
0111178.解:因为?xf(x)dx= ?f(x2)dx2=?f(t)dt=?f(x)dx,故选A
2202000179.解:因为被积函数为奇函数,故选A 180.解:I'(l)2244=0,I(l)=c,令l=0,得I=?f(x)dx,选B
0T181.解:因为
?0f(x)dx= ?2f(x)dx=2?f(t)dt=2?f(x)dx,故选D x000111f(2x)= ?f(2)?f(0)?,故选C
0222221182.解:
?0f'(2x)dx=183.解:选A 184.解:185.解:
?ba1f'(2x)dx=[f(2b)?f(2a)],选C
2?dx=2,选C
?1a1186.解:
a(arccosx)'dx=arccosx=arccosa?arccos0,选D ?00187.解:选D 188.解:因为
?212x4215x3273=,选A xdx==,?xdx=1414313222xsinxx2sinxdx=0,选D 189.解:因为为奇函数,所以??21+x21+x2190.解:
?11-1xdx=2?xdx=1,选C
0191.解:x+sin2x为奇函数,所以?(x+sinx)dx=0,选D
-2022192.解:
??1xdx=??xdx+?xdx=?105,选D 2193.解:选A
194.解:作出函数的图形知选A
21y2=4?x
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195.解:
13.532.521.510.50.20.40.60.811.2y=ex过原点的切线为y=ex,作出函数的图形知选C
y=ex y=ex 196.解:如图: 曲线
197.解:由
y=x与y=x2所围成平面图形的面积=(x?x2)dx=?01,选A 31.41.210.80.60.40.20.20.40.60.811.2y=x y=x2 y=c?x,y?=?1代入方程x+y?y?=x+(c?x)?(?1)=c+1?1,
所以不是解.所以答案为D.
198.解:将
y=3e2x,y?=6e2x,y??=12e2x,带入微分方程有.y???4y=12e2x?12e2x=0,因此式方程的解.由于
y=3e2x中无任意常数,所以为特解.答案选B.
199.解:由微分方程阶的定义:常微分方程中导数出现的最高阶数知为二阶.
由方程中出现(y??)知,方程为非线性的.所以答案B正确.
200.解:由
2y=C1e?x+C2,y?=?C1e?x,y??=C1e?x代入方程有
y??+y?=?C1e?x+C1e?x=0.且y=C1e?x+C2中有两个独立的任意常数,因此答案为D.
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专升本高等数学必做复习资料200题(含答案)
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