《微积分》习题6.docx
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《微积分》习题6
《微积分》习题6
习题六 1.根据定积分的几何意义说明下列各式的正确性 解:
该定积分的几何意义如右图所示阴影部分面积的代数和,对称性可知正确.该定积分的几何意义如右图所示阴影部分面积的代数和,且在(?
2,2)范围内对称,所以是正确的. 该定积分的几何意义如右图所示阴影部分面积的代数和,且关于原点对称,所以正确.原式?
2?
2?
0cosxdx?
0 ?
2?
2(x2?
1)dx?
21?
(x022?
1)dx ?
xdx?
0 ?
3?
111?
12xdx?
4?
xdx 0?
1?
1xdx 等式左边的定积分的几何意义是右边图形阴影部分面积的代数和的2倍,且又因为阴影部分在(?
1,1)范围内关于轴对称,所以等式两边相等. 2.不计算积分,比较下列积分值的大小 解:
定积分的比较性可知在(0,1)范围内x2?
x3,所以前者大于后者.定积分的比较性可知在(1,3)范围内x2?
x3,所以前者小于后者.定积分的比较性可知在(3,4)范围内lnx?
(lnx)2,所以前者小于后者.a?
1定积分的比较性可知在(0,)范围sinx?
x,所以前者小于后者. 2?
xdx与?
xdx ?
xdx与?
xdx 232300111133?
43lnxdx与 ?
(lnx)dx ?
234?
20sinxdx与 ?
?
20xdx ?
3.用定积分性质估计下列积分值 解:
因为e?
x在[0,1]范围内的最大值为1,最小值为e?
1所以定积分的估值定理可知:
2?
10e-x2dx ?
?
5?
4(1?
sin2x)dx 4?
1x51?
x?
0dx ?
20sinxdxx?
10e?
1dx?
?
110e?
xdx?
22?
ldx 01?
e?
1?
?
0e?
xdx?
1 因为1?
sin2x在[?
5?
]2的最大值为2,最小值为1。
442所以定积分的估值定理可知:
?
?
5?
4ldx?
?
5?
44?
5?
4(1?
sin2x)dx?
?
4?
5?
42dx 4 ?
?
?
?
?
(1?
sin2x)dx?
2?
4设f(x)?
4x51?
x x5则f‘(x)?
令 5x1?
x?
421?
x?
x(10?
9x)1?
x21?
x(1?
x) f‘(x)?
0 则x4(10?
9x)?
0,1?
x?
0解得:
x?
0,x?
?
所以所以 109f(x)在(0,?
?
)上单调递增 f(x)在[0,1]的最小值为 0,最大值是2 2所以定积分的估值定理可知:
?
100dx?
?
1x51?
xx51?
x0dx?
?
102dx2?
0?
?
10dx?
22 图中易知:
AB?
?
?
ADAB 其中AB?
sinx,AD?
tanx,AC?
x即:
sinx?
x?
tanx亦得到:
1?
0?
x?
x1?
sinxcosx ?
2,从中cosx?
sinx?
1x定积分性质有:
?
?
?
20cosxdx?
?
?
20sinxdx?
x?
?
201?
dx ?
1?
?
2?
sinx?
dx?
x2 1 A yx4.利用定积分的几何意义计算下列积分x0BCD 解:
该定积分的几何意义是以原点为圆心2为半径的一个圆面积的一半,且在x轴的上方. ?
2?
22?
x2dx ?
21(1?
2x?
x2)dx 所以原式?
?
R2 ?
?
12 该定积分的几何意义是以(1,1)为圆心,以1为半径的一个圆面积的一半且在x轴 的上方.所以原式?
?
R2 ?
?
5.求下列函数的导数f(x)12 12?
x2?
1te?
t2dt f(x)?
?
0exxln(1?
t2)dt f(x)?
?
x3xedt f(x)?
t2?
(tx3?
x3)sintdt 解:
设te?
tdt?
g(t) ?
22x则f(x)?
g(e)?
g(x2)?
g(?
1)令x2?
m ?
1f‘(x)?
g‘(m)m‘?
x2e?
x2x?
2x3e?
x44 设ln(1?
t2)dt?
g(t) ?
则 f(x)?
g(ex)?
g(x)令ex?
m f‘(x)?
g‘(m)m‘?
g‘(x)?
exln(x?
e2x)?
ln(x?
x2) 设etdt?
g(t) ?
2则 f(x)?
g(x3)?
g(x)令x3?
m,x?
n 6f‘(x)?
g‘(m)m‘?
g‘(n)n‘?
ex3x2?
ex12x 设(t3?
x3)sint?
g(t) ?
则f(x)?
g(x)?
g(0) f‘(x)?
g‘(x)?
g‘(0)?
0 6.求下列极限lim 1x?
0x3?
x1xsint2dt lim2arctantdt00x?
0x?
x?
0lim1x3?
0(x1?
t2?
1?
t)dt2 limx?
0?
1?
x2?
1xln(1?
t)dt0 lim1?
1x(1?
sin2t)tdtlntdt1?
t1 lim?
x x?
0x01?
xt2xlim?
?
?
?
?
edt?
?
?
?
x20?
?
t2x?
dtxlim?
xe?
0?
0x?
sinx?
arctanx 解:
?
lim1x?
0x3?
x0t2dt ?
lim113x?
0x3?
3tdt?
13?
lim1xx?
0x2?
0tdt ?
xlim11?
0x22x2x0 ?
12 ?
1xxlim?
0x3?
(1?
tan2t?
1?
sin20t)dt ?
lim1xx?
0x3?
0(1cosx?
cosx)dt ?
lim1?
xx?
0x30tanxsinxdt ?
lim1x?
0x3?
x0x2dt ?
xlim1?
0x313t3x0x?
1(x?
1)21xxlim?
?
?
x?
(t?
t2)et2?
x2dt0
?
?
lim1?
x2?
113x-?
0?
?
x 0ln(1?
t)dt1?
x2?
1 ?
limx-?
0?
x 0ln(1?
t)d(1?
t)1?
x2?
1?
0 ?
limx?
0?
?
(ln?
?
1)1lim?
x?
0xx?
x01(1?
sin2t)tdt ?
lim?
01(1?
sin2t)tx?
0dt 1x洛必达法则lim(1?
sinx) x?
0?
lim(1?
2x)x?
01?
22x?
e2 lntdt1?
t1limx?
1(x?
1)2?
x lnx?
lim1?
xx?
12x?
2 11lnx1x?
1?
lim?
limx?
12x2?
12x?
12x41 1?
lim?
x?
?
?
?
?
lim1x2?
lim?
x2x?
?
?
ln(edt?
?
e?
0?
t22x?
x0edt)xt22 ?
ex?
?
?
xt?
ln?
0edt?
e1x2?
x2?
e limx?
?
?
?
(arctant)dt 20x1?
x2?
?
x0x?
?
?
x?
?
?
(limarctan)2dtlim1?
x2 ?
2?
x?
?
?
4lim1?
x2?
0 ?
x?
?
?
lim1x?
x0(t?
t2)et2?
x2dt ?
lim?
x0(t?
t2)etx?
ex22x?
?
?
dt 洛必达法则lim (x?
x2)ex22x?
?
?
ex(1?
2x)2?
127.设F(x)在[a,b]上连续,且f(x)?
0 F(x)?
?
xaf(t)dt?
?
xb1dtf(t)求证:
F‘(x)?
2; F‘(x)在[a,b]内有且仅有一个实根. 解:
证明:
设f(t)dt?
g(t)?
?
1dt?
h(t)f(t)F(x)?
g(x)?
g(a)?
h(x)?
h(b)?
F‘(x)?
g‘(x)-g‘(a)?
h‘(x)-h‘(b)?
f(t)?
1f(t) 又因为f(t)?
0,?
F‘(x)?
2 因为F(x)在[a,b]上单调增加,又因为F(a)?
?
ab1de?
?
f(t)?
ba1de?
0f(t)F(b)?
?
baf(t)dt?
0 又因为F(x)在区间[a,b]上连续.所以在区间[a,b]内紧有一个实根.8.设 f(x)为连续函数,且存在常数a满足 ex-1?
x?
?
axf(t)dt 求 f(x)及常数a. 解:
设f(e)de?
g(t) ?
则ex?
1?
x?
g(a)?
g(x)对等式两边求导,得:
ex?
1?
x?
g‘(a)?
g‘(x)?
?
f(x) 所以所以 f(x)?
1?
ex?
1 ?
axf(t)dt?
?
axex?
1dx?
x?
ex?
1a?
a?
ea?
1?
ex?
1?
xx?
1所以a?
19.设解:
Q‘(t)?
f(t),P‘(t)?
Q(t)x?
(x?
t)f(t)dt?
1?
cosx,说明?
0x?
20f(x)dx?
1. ?
(x?
t)f(t)dt0?
tf(t)dt?
x[Q(x)?
Q(0)]?
td(Q(t)?
?
x[Q(x)?
Q(0)]?
tdQ(t)?
x?
x[Q(x)?
Q(0)]?
tQ(t)?
Q(t)dt0?
?
x[Q(x)?
Q(0)]?
0x0x0x0x?
?
Q(x)dt?
1?
cosx0x即 P(x)-P(0)?
1-cosx?
Q(x)?
sinx?
f(x)?
cosx?
?
2?
f(x)dx?
sinx02?
10 10.用牛顿-莱布尼茨公式计算下列积分 ?
8dx 1x?
e?
x)dx 13x?
(e?
12?
12arcsinx2dx 21?
x ?
?
0cosxdx ?
2xdx ?
1?
?
ex2?
lnx2d31xx?
?
tanxdx 62?
4?
?
x?
1?
?
?
d1?
x?
x?
?
?
21?
2?
sindx?
1?
sin2xdx00?
3max{1,x2}dx ?
1 解:
?
8dx2x?
32x381?
9132 ?
1(ex?
e?
x)dx?
(ex?
e?
x)1?
1?
1?
0 ?
e2(lnx)2e2(lnx)2d(lnx)?
1e21xdx?
?
(lnx)313122 ?
12arcsinxdx?
21?
x2?
12arcsinxdarcsinx 2?
122arcsin2x21?
2?
21?
2(16?
36) s3) ?
e2(lnx)21xdx 7) ?
1dx ?
14?
x2?
11)?
?
3tan2xdx 6?
15)?
xcosx?
sinx?
2dx4(xsinx)28) ?
2dx04?
x212) ?
x0cosxdx?
2cosxdx?
?
?
cosxdx?
sinx2?
sinxx?
2 x0022?
?
?
?
2?
11xdx?
?
20?
1?
xdx?
?
20xdx?
1221205x?
x?
202?
12?
dx4?
x?
1?
arcsinx1?
?
2?
13 ?
?
2dx4?
x20e?
1x2?
arctg?
2208 1x2?
lnx2dx?
x?
xdx?
?
1ee121lnx1edx?
x2?
2x21?
e1lnxdlnx ?
e1121(e?
1)?
2(lnx)2?
(e2?
3) 1222?
?
tanxdx?
?
lncosx3 ?
?
3?
?
61ln326?
?
?
?
3tanxdx?
?
3626?
1cosx2?
?
ldx?
?
3(sec2x?
1)dx 6?
?
2ln2?
1 ?
(3?
?
3)?
(3?
?
)36?
442?
3?
36 1dxx ?
1(x?
1x)2dx?
?
1x?
2x?
1dx?
x?
411?
2x?
4?
(x?
44?
lnx 1?
2ln2?
1 ?
?
20?
111?
sinxdx?
6(?
sinx)dx?
?
2(sinx?
)dx2202?
?
?
6
11?
(?
cosx?
x)26?
(x?
cosx)2?
026?
3?
1?
?
?
?
12 ?
x ?
x01?
sin2xdx?
?
x0sinx?
cosxdx?
?
40(cosx?
sinx)dx?
?
x4(sinx?
cosx)dx ?
?
(sinx?
cosx)4?
(?
cosx?
sinx)x?
22x04 ?
?
?
2xcosx?
sinx(xsinx)2?
?
1(xsinx)2d(xsinx)?
?
12?
42?
2xsinx?
?
4dx?
?
244?
?
3?
1max{1,x2}dx?
?
1?
1ldx?
?
31x2dx?
x1?
1?
x232?
10313 11.设f(x)?
解:
设 ?
t(l?
t)e0x?
2tdt,问x取何值时,f(x)取极大值或极小值. ?
t(t?
l)e-2tde?
g(t) f(x)?
g(x)?
g(e) 则 所以 因为 所以 f‘(x)?
g‘(x)?
x(x?
1)e?
2x?
?
x(x?
1)e?
2x f‘(x)在(?
?
0)f(x)在 ,(?
1,?
?
),(-1,?
)上大于0,在(0,1)内小于0 (?
?
0),(1,?
?
)上单调递增,在(0,1)内单调递增. 所以当x?
0时,f(x)取极大值,x?
1时,f(x)取极小值。
12.设 ?
I1?
?
?
2?
sinx1?
x2cos2xdx2?
I2?
?
?
2?
(sinx?
cosx)dx 2?
I3?
?
?
2(sin5x?
cosx)dx?
2比较I1,I2,I3的大小.解:
?
I1?
?
2sinx?
?
21?
x2cos2xdx ?
0 ?
I2?
?
2?
?
(sinx?
cosx)dx2 ?
2 ?
I3?
?
2?
?
(sin5x?
cosx)dx2 ?
?
2 ?
?
?
?
cosxdx2 ?
?
?
2?
?
cosxdx?
02 ?
?
?
2?
?
?
cosxdx?
02 ?
I3?
I1?
I2 13.用换元积分法计算下列各定积分 ?
?
sinx1?
cos2xd0x ?
ln3dx 01?
ex ?
1dx 0(1?
x2)3?
2x2?
11xdx 3) ?
e2dx1?
ln 1x?
ax2a2?
x2dx0 ?
3dxx1?
x21 ?
1xexe?
eee?
?
x0dx ?
40tan(lncosx)dx ?
?
e6e223lnx?
2dx xdxx2?
dxx(1?
lnx)lnx ?
2?
sin9xdx 0x?
12 ?
1x?
3?
2x?
5?
1x2dx 解:
?
?
x0?
11?
cos2xdcosx x?
?
arctan(cosx) 0?
?
2令1?
ex?
t 则x?
ln(t2?
1) x?
ln3,时t?
2;x?
0,时t?
2 ?
2?
221t2?
1dt ?
lnx?
12x?
12?
?
[ln3?
2ln(2?
1)] = ?
e2dxx1?
lnx1 2?
?
?
?
0xsinxdx?
?
?
xsinxdx?
21?
lnxe21?
23?
2 ?
x31?
322x12?
03?
dx(1?
322x) ?
x3(1?
x2)1302?
12dx 3(1?
x2)102 ?
2 2?
?
21x2?
1dxx212x1 ?
x2?
1?
arccos1 ?
3?
x8?
3 aa4xaarcsin8a0?
(2x2?
a2)a2?
x2?
0 ?
0?
?
16a4?
?
16a4 令x?
tant,则积分区域为 3?
?
到.43?
dx1?
x2?
1?
?
?
?
34(sect)2?
dttant?
sect ?
?
?
34?
dt?
dcost?
?
3sint1?
cos2t?
41d(?
cost)13dcost?
?
?
3?
?
?
21?
cost21?
cost?
?
?
4?
4?
?
?
1313ln(1?
cost)?
ln(1?
cost)?
2?
24411?
ln(2?
2)?
ln3(2?
2)221?
ln(2?
2)?
ln62 令ex?
t ?
?
1exex?
e?
xe0dx?
?
e11dt1tt?
tt?
1?
?
edt?
?
lnx?
x2?
1?
?
?
1t2?
11e?
e2?
11?
2?
ln 令t?
lnx则x?
et ?
e6e3lnx?
2dx?
x3?
613t?
2dt62?
(3t?
2)219?
14 令lnx?
t,则积分上下限变为与1. 12?
eedxx(1?
lnx)lnx?
?
112dt(1?
t)t 令t?
sin2x积分上下限为:
42?
?
?
?
?
22sinxdsinx1?
sin2x?
sinx?
?
?
244?
2dsinx1?
sin2x?
?
ln(sinx?
sin2x?
12?
4?
ln1?
22?
2322?
?
ln2?
21?
3 ?
?
2?
?
0sin9xdx ?
?
?
?
?
(cosx?
4cos?
6cos?
4cos02(1?
cos2x)4dcosx86420x?
1)dcosx ?
0令x2?
sec3a则dx?
seca?
tana ?
22dxx2x2?
1?
?
?
?
34?
seca?
tana?
dasec2?
tana?
3?
23?
?
?
3cosada?
?
sina?
?
244?
令x?
1?
a
?
1x?
32?
1x2?
2x?
5dx?
?
2a?
0a2?
4da?
1d(a2?
4)0a2?
4?
?
222?
20a2?
4da?
122ln2?
?
20a2?
4da ?
12ln2?
arctana220?
12ln2?
(?
4?
0)?
12ln2?
14?
14.用分部积分法计算下列各定积分?
10x2e?
xdx ?
1(x?
x)e?
xdx ?
1 ?
e(lnx)3dx elnxdx 1?
1e ?
10xarctanxdx ?
e2lnx(x?
1)x e2d?
ln2?
x2dxsinxdx 0x3ex ?
2?
0?
12x1?
x2arcsinxdx ?
e2lnx0e?
1xdx 解:
?
?
1x2de?
x 0?
?
(x2e?
x11?
x0?
0e2xdx ?
?
(e?
1?
?
12xdx?
x)0 ?
?
(e?
1?
2e?
1?
2e?
x10) ?
?
(3e?
1?
2e?
1?
2) ?
2?
5e?
1?
?
0()?
1x?
xexdx?
?
1(x?
x)e?
x0dx ?
?
12xde?
x 03) ?
e?
1xln(1?
x)dx 0?
6) ?
2xsinxdx 09) ?
1e2x(4x?
3)dx 0?
12) ?
e2sin(lnx)xdx12= 1210?
e?
10ln(x?
1)dx2 ?
(x2ln(x?
1)?
x(lnx)3?
?
e?
e12e?
1?
0?
e?
10x21dxx?
1e1?
e11x3lnxdx x?
3lnxdx 1 ?
e?
3(xlnx?
e1?
e11xdxx ?
e?
3[e?
(e?
1)] ?
e?
3 1?
?
(xlnx1?
e?
1e11ee1xdx)?
xlnx?
1x?
e11xdxx ?
?
[?
?
(1?
)]?
e?
(e?
1) ?
1?
1 ?
2?
?
?
201e?
xdcosx ?
?
(xcosx?
20?
?
?
20cosxdx) ?
sinx?
x0 ?
1 1?
2?
10arctanxdx2 111?
(x2arctanx?
0211?
?
[?
(x?
024?
0x211?
x2dx)] dx) ?
11?
x211?
?
[?
1?
arctanx] 024?
?
4?
12e2 1(x?
1)?
?
?
elnxd ?
?
[lnx1e2x?
1e2?
ln] (x?
1)exee1?
e?
ln(1?
e)?
2x?
12?
(4x?
3)de01 ?
[(4x?
3)e3x?
12121210?
e012x?
4dx] ?
(7e2?
3?
4e2x) ?
[7e2?
3?
2(e2?
1)] ?
(5e2?
1)?
?
11212102?
ln20x2de?
x2 ?
?
[x2?
e?
x122ln20?
?
ln202xe?
xdx] 2 ?
1[ln2e?
ln2?
2?
ln202de?
x] 2 ?
?
[ln2?
e?
x1122ln20] ?
?
[ln2?
e?
ln2?
1] ?
?
[ln2?
] ?
?
(ln2?
1)?
141122112212?
xsinxdx?
?
?
0?
2?
xsinxdx ?
?
(xcosx?
0?
?
?
0cosxdx)?
(xcosx2?
?
?
?
?
2?
cosxdx) ?
4?
?
?
e2sinlnxx2?
1dx?
?
?
?
20sinlnxd1x1?
?
?
sinlnxe2?
x11?
?
sinlnxe2?
x1?
?
?
?
12111?
coslnx?
dxxx1x?
?
?
2coslnx?
d?
11sinlnx?
dxxx11?
?
sinlnxe2?
coslnxe2?
xx11?
111?
?
(sinlnx?
coslnx)e22xx1?
11(1?
?
)2e2?
?
21?
2x011?
x2arcsinxdx3?
?
102?
(?
1)?
?
arcsinx?
d(1?
x2)233?
12?
?
?
(1?
x2)2?
arcsinx?
03?
?
?
103(1?
x2)21?
(1?
x2)?
2?
?
dx?
?
?
?
?
23?
10(1?
x2)dx?
49 ?
e2lnxxdxe?
1 ?
2?
e2e?
1lnx1?
dx2?
2?
lnx1?
x2e2e?
0e0e?
112?
2?
e2e?
01?
x21x2121?
dxx1?
?
2lnxx2e2?
2?
e0e?
11?
dxx?
2 x?
12dx?
4e?
2e01?
?
8?
6e2?
x?
?
dx?
?
2e?
?
e0e?
1 15.利用函数奇偶性计算下列积分 解:
设f(x)?
sin2xln(x?
1?
x2)则f(?
x)?
sin2xln(?
x?
1?
x2) f(?
x)?
f(x)?
sin2x[ln(?
x?
1?
x2)?
ln(x?
x?
2)]?
sin2xln(1?
x2?
x2)?
0 ?
?
3?
22 ?
sinxln(x?
1?
x)dx3?
112?
x2?
1(11?
)dxcosxarccosxdx ?
11?
ex21?
所以 因为 f(x)为奇函数 f(x))在上连续且为奇函数,所以原式等于 0; 设f(x)?
12?
x2(1?
)1?
ex21(11?
)21 ?
f(?
x)?
?
x2?
x21?
e?
f(x)?
f(?
x)?
所以 12?
x2(11?
e?
x?
11?
ex?
1)?
0 f(x)为奇函数且f(x)在上连续。
所以原式等于0;设因为 ?
?
4?
5(?
5)dx?
?
?
3?
4(?
4)dx?
?
?
4dx 45f(?
x)?
cosxarccosx?
f(x)
?
?
xdx?
?
xdx?
?
xlnxdx 0111ee ?
x2?
x2?
x2(lnx?
) ?
?
(e2?
1)?
[e2?
?
(?
)] ?
?
e2?
?
e2?
?
e2?
?
14141212121414121212121212121102e1121e21 ?
10(x3?
3x2?
2x)dx?
?
21(x3?
3x2?
2x)dx ?
(x4?
x3?
x2)?
(x4?
x3?
x2 ?
?
1?
1?
[(4?
8?
4)?
(?
1?
1)] ?
12141414101421 ?
5?
?
(sinx?
cos)4?
(sinx?
cosx)4?
04 ?
42 ?
?
(y3?
36y) ?
48 25.求抛物线y?
?
x2?
4x?
3及其在点(0,3)和点(3,0)处两条切线所围成图形的面积. 解:
抛物线g?
?
x2?
4x?
3及两点处的切线所围的图形如后图:
113360y 10 2 3 x-3 ?
t(0,3)处切线的斜率k1?
?
2x?
4x?
0?
4切线方程为:
y?
4x?
3 ?
t(3,0)处切线的斜率k2?
?
2x?
412x?
3?
2切线方程为:
y?
?
2x?
6 则阴影部分面积S?
?
04x?
3?
(?
x2?
4x?
3)dx?
?
312?
2x?
6?
(?
x2?
4x?
3)?
dx ?
?
120x?
6dx?
2?
312(x2?
6x?
9)dx ?
334 26.过原点作曲线y?
lnx的切线,求切线,x轴及曲线y?
lnx所围平面图形的面积. 解:
阴影部分如右图所示可求解直线方程为y?
x二曲线的交点为(e,1) 1e ylnx 0 1 x 从而面积S?
?
101xdx?
e?
e11(x?
lnx)dxe11111e?
?
x2?
?
x2?
e20e21?
e112?
e?
xlnx?
(e?
1) 1e21e?
(e)?
(e?
1)21e?
12?
e1lndx ?
?
27.求过曲线y?
lnx上的点(e,1)的法线与x