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Today's Calculation Of Integral
2012 Today's Calculation Of Integral
2012 Today's Calculation Of Integral
Part of
Today's Calculation Of Integral
Subcontests
(90)
859
1
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Today's calculation of Integral 859
In the
x
x
x
-
y
y
y
plane, for
t
>
0
t>0
t
>
0
, denote by
S
(
t
)
S(t)
S
(
t
)
the area of the part enclosed by the curve
y
=
e
t
2
x
y=e^{t^2x}
y
=
e
t
2
x
, the
x
x
x
-axis,
y
y
y
-axis and the line
x
=
1
t
.
x=\frac{1}{t}.
x
=
t
1
.
Show that
S
(
t
)
>
4
3
.
S(t)>\frac 43.
S
(
t
)
>
3
4
.
If necessary, you may use
e
3
>
20.
e^3>20.
e
3
>
20.
858
1
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Today's calculation of Integral 858
On the plane
S
S
S
in a space, given are unit circle
C
C
C
with radius 1 and the line
L
L
L
. Find the volume of the solid bounded by the curved surface formed by the point
P
P
P
satifying the following condition
(
a
)
,
(
b
)
(a),\ (b)
(
a
)
,
(
b
)
.
(
a
)
(a)
(
a
)
The point of intersection
Q
Q
Q
of the line passing through
P
P
P
and perpendicular to
S
S
S
are on the perimeter or the inside of
C
C
C
.
(
b
)
(b)
(
b
)
If
A
,
B
A,\ B
A
,
B
are the points of intersection of the line passing through
Q
Q
Q
and pararell to
L
L
L
, then
P
Q
‾
=
A
Q
‾
⋅
B
Q
‾
\overline{PQ}=\overline{AQ}\cdot \overline{BQ}
PQ
=
A
Q
⋅
BQ
.
857
1
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Today's calculation of Integral 857
Let
f
(
x
)
=
lim
n
→
∞
(
cos
n
x
+
sin
n
x
)
1
n
f(x)=\lim_{n\to\infty} (\cos ^ n x+\sin ^ n x)^{\frac{1}{n}}
f
(
x
)
=
lim
n
→
∞
(
cos
n
x
+
sin
n
x
)
n
1
for
0
≤
x
≤
π
2
.
0\leq x\leq \frac{\pi}{2}.
0
≤
x
≤
2
π
.
(1) Find
f
(
x
)
.
f(x).
f
(
x
)
.
(2) Find the volume of the solid generated by a rotation of the figure bounded by the curve
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
and the line
y
=
1
y=1
y
=
1
around the
y
y
y
-axis.
856
1
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Today's calculation of Integral 856
On the coordinate plane, find the area of the part enclosed by the curve
C
:
(
a
+
x
)
y
2
=
(
a
−
x
)
x
2
(
x
≥
0
)
C: (a+x)y^2=(a-x)x^2\ (x\geq 0)
C
:
(
a
+
x
)
y
2
=
(
a
−
x
)
x
2
(
x
≥
0
)
for
a
>
0
a>0
a
>
0
.
855
1
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Today's calculation of Integral 855
Let
f
(
x
)
f(x)
f
(
x
)
be a function which is differentiable twice and
f
′
′
(
x
)
>
0
f''(x)>0
f
′′
(
x
)
>
0
on
[
0
,
1
]
[0,\ 1]
[
0
,
1
]
.For a positive integer
n
n
n
, find
lim
n
→
∞
n
{
∫
0
1
f
(
x
)
d
x
−
1
n
∑
k
=
0
n
−
1
f
(
k
n
)
}
.
\lim_{n\to\infty} n\left\{\int_0^1 f(x)\ dx-\frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right)\right\}.
lim
n
→
∞
n
{
∫
0
1
f
(
x
)
d
x
−
n
1
∑
k
=
0
n
−
1
f
(
n
k
)
}
.
854
1
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Today's calculation of Integral 854
Given a figure
F
:
x
2
+
y
2
3
=
1
F: x^2+\frac{y^2}{3}=1
F
:
x
2
+
3
y
2
=
1
on the coordinate plane. Denote by
S
n
S_n
S
n
the area of the common part of the
n
+
1
′
s
n+1' s
n
+
1
′
s
figures formed by rotating
F
F
F
of
k
2
n
π
(
k
=
0
,
1
,
2
,
⋯
,
n
)
\frac{k}{2n}\pi\ (k=0,\ 1,\ 2,\ \cdots,\ n)
2
n
k
π
(
k
=
0
,
1
,
2
,
⋯
,
n
)
radians counterclockwise about the origin. Find
lim
n
→
∞
S
n
\lim_{n\to\infty} S_n
lim
n
→
∞
S
n
.
853
1
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Today's calculation of Integral 853
Let
0
<
a
<
π
2
.
0<a<\frac {\pi}2.
0
<
a
<
2
π
.
Find
lim
a
→
+
0
1
a
3
∫
0
a
ln
(
1
+
tan
a
tan
x
)
d
x
.
\lim_{a\rightarrow +0} \frac{1}{a^3}\int_0^a \ln\ (1+\tan a\tan x)\ dx.
lim
a
→
+
0
a
3
1
∫
0
a
ln
(
1
+
tan
a
tan
x
)
d
x
.
852
1
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Today's calculation of Integral 852
Let
f
(
x
)
f(x)
f
(
x
)
be a polynomial. Prove that if
∫
0
1
f
(
x
)
g
n
(
x
)
d
x
=
0
(
n
=
0
,
1
,
2
,
⋯
)
\int_0^1 f(x)g_n(x)\ dx=0\ (n=0,\ 1,\ 2,\ \cdots)
∫
0
1
f
(
x
)
g
n
(
x
)
d
x
=
0
(
n
=
0
,
1
,
2
,
⋯
)
, then all coefficients of
f
(
x
)
f(x)
f
(
x
)
are 0 for each case as follows.(1)
g
n
(
x
)
=
(
1
+
x
)
n
g_n(x)=(1+x)^n
g
n
(
x
)
=
(
1
+
x
)
n
(2)
g
n
(
x
)
=
sin
n
π
x
g_n(x)=\sin n\pi x
g
n
(
x
)
=
sin
nπ
x
(3)
g
n
(
x
)
=
e
n
x
g_n(x)=e^{nx}
g
n
(
x
)
=
e
n
x
851
1
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Today's calculation of Integral 851
Let
T
T
T
be a period of a function
f
(
x
)
=
∣
cos
x
∣
sin
x
(
−
∞
,
∞
)
.
f(x)=|\cos x|\sin x\ (-\infty,\ \infty).
f
(
x
)
=
∣
cos
x
∣
sin
x
(
−
∞
,
∞
)
.
Find
lim
n
→
∞
∫
0
n
T
e
−
x
f
(
x
)
d
x
.
\lim_{n\to\infty} \int_0^{nT} e^{-x}f(x)\ dx.
lim
n
→
∞
∫
0
n
T
e
−
x
f
(
x
)
d
x
.
850
1
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Today's calculation of Integral 850
Evaluate
∫
0
π
{
(
1
−
x
sin
2
x
)
e
cos
2
x
+
(
1
+
x
sin
2
x
)
e
sin
2
x
}
d
x
.
\int_0^{\pi} \{(1-x\sin 2x)e^{\cos ^2 x}+(1+x\sin 2x)e^{\sin ^ 2 x}\}\ dx.
∫
0
π
{(
1
−
x
sin
2
x
)
e
c
o
s
2
x
+
(
1
+
x
sin
2
x
)
e
s
i
n
2
x
}
d
x
.
849
1
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Today's calculation of Integral 849
Evaluate
∫
1
e
2
(
2
x
2
+
2
x
+
1
)
e
x
x
d
x
.
\int_1^{e^2} \frac{(2x^2+2x+1)e^{x}}{\sqrt{x}}\ dx.
∫
1
e
2
x
(
2
x
2
+
2
x
+
1
)
e
x
d
x
.
848
1
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Today's calculation of Integral 848
Evaluate
∫
0
π
4
sin
θ
−
2
ln
1
−
sin
θ
cos
θ
(
1
+
cos
2
θ
)
ln
1
+
sin
θ
cos
θ
d
θ
.
\int_0^{\frac {\pi}{4}} \frac {\sin \theta -2\ln \frac{1-\sin \theta}{\cos \theta}}{(1+\cos 2\theta)\sqrt{\ln \frac{1+\sin \theta}{\cos \theta}}}d\theta .
∫
0
4
π
(
1
+
c
o
s
2
θ
)
l
n
c
o
s
θ
1
+
s
i
n
θ
s
i
n
θ
−
2
l
n
c
o
s
θ
1
−
s
i
n
θ
d
θ
.
847
1
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Today's calculation of Integral 847
Consider a right-angled triangle with
A
B
=
1
,
A
C
=
3
,
∠
B
A
C
=
π
2
.
AB=1,\ AC=\sqrt{3},\ \angle{BAC}=\frac{\pi}{2}.
A
B
=
1
,
A
C
=
3
,
∠
B
A
C
=
2
π
.
Let
P
1
,
P
2
,
⋯
⋯
,
P
n
−
1
(
n
≥
2
)
P_1,\ P_2,\ \cdots\cdots,\ P_{n-1}\ (n\geq 2)
P
1
,
P
2
,
⋯⋯
,
P
n
−
1
(
n
≥
2
)
be the points which are closest from
A
A
A
, in this order and obtained by dividing
n
n
n
equally parts of the line segment
A
B
AB
A
B
. Denote by
A
=
P
0
,
B
=
P
n
A=P_0,\ B=P_n
A
=
P
0
,
B
=
P
n
, answer the questions as below.(1) Find the inradius of
△
P
k
C
P
k
+
1
(
0
≤
k
≤
n
−
1
)
\triangle{P_kCP_{k+1}}\ (0\leq k\leq n-1)
△
P
k
C
P
k
+
1
(
0
≤
k
≤
n
−
1
)
.(2) Denote by
S
n
S_n
S
n
the total sum of the area of the incircle for
△
P
k
C
P
k
+
1
(
0
≤
k
≤
n
−
1
)
\triangle{P_kCP_{k+1}}\ (0\leq k\leq n-1)
△
P
k
C
P
k
+
1
(
0
≤
k
≤
n
−
1
)
.Let
I
n
=
1
n
∑
k
=
0
n
−
1
1
3
+
(
k
n
)
2
I_n=\frac{1}{n}\sum_{k=0}^{n-1} \frac{1}{3+\left(\frac{k}{n}\right)^2}
I
n
=
n
1
∑
k
=
0
n
−
1
3
+
(
n
k
)
2
1
, show that
n
S
n
≤
3
π
4
I
n
nS_n\leq \frac {3\pi}4I_n
n
S
n
≤
4
3
π
I
n
, then find the limit
lim
n
→
∞
I
n
\lim_{n\to\infty} I_n
lim
n
→
∞
I
n
.(3) Find the limit
lim
n
→
∞
n
S
n
\lim_{n\to\infty} nS_n
lim
n
→
∞
n
S
n
.
846
1
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Today's calculation of Integral 846
For
a
>
0
a>0
a
>
0
, let f(a)=\lim_{t\to\+0} \int_{t}^1 |ax+x\ln x|\ dx. Let
a
a
a
vary in the range
0
<
a
<
+
∞
0 <a< +\infty
0
<
a
<
+
∞
, find the minimum value of
f
(
a
)
f(a)
f
(
a
)
.
845
1
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Today's calculation of Integral 845
Consider for a real number
t
>
1
t>1
t
>
1
,
I
(
t
)
=
∫
−
4
4
t
−
4
(
x
−
4
)
x
+
4
d
x
.
I(t)=\int_{-4}^{4t-4} (x-4)\sqrt{x+4}\ dx.
I
(
t
)
=
∫
−
4
4
t
−
4
(
x
−
4
)
x
+
4
d
x
.
Find the minimum value of
I
(
t
)
(
t
>
1
)
.
I(t)\ (t>1).
I
(
t
)
(
t
>
1
)
.
844
1
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Today's calculation of Integral 844
Let
α
\alpha
α
be a solution satisfying the equation
∣
x
∣
=
e
−
x
.
|x|=e^{-x}.
∣
x
∣
=
e
−
x
.
Let
I
n
=
∫
0
α
(
x
e
−
n
x
+
α
x
n
−
1
)
d
x
(
n
=
1
,
2
,
⋯
)
.
I_n=\int_0^{\alpha} (xe^{-nx}+\alpha x^{n-1})dx\ (n=1,\ 2,\ \cdots).
I
n
=
∫
0
α
(
x
e
−
n
x
+
α
x
n
−
1
)
d
x
(
n
=
1
,
2
,
⋯
)
.
Find
lim
n
→
∞
n
2
I
n
.
\lim_{n\to\infty} n^2I_n.
lim
n
→
∞
n
2
I
n
.
843
1
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Today's calculation of Integral 843
Let
f
(
x
)
f(x)
f
(
x
)
be a continuous function such that
∫
0
1
f
(
x
)
d
x
=
1.
\int_0^1 f(x)\ dx=1.
∫
0
1
f
(
x
)
d
x
=
1.
Find
f
(
x
)
f(x)
f
(
x
)
for which
∫
0
1
(
x
2
+
x
+
1
)
f
(
x
)
2
d
x
\int_0^1 (x^2+x+1)f(x)^2dx
∫
0
1
(
x
2
+
x
+
1
)
f
(
x
)
2
d
x
is minimized.
842
1
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Today's calculation of Integral 842
Let
S
n
=
∫
0
π
sin
n
x
d
x
(
n
=
1
,
2
,
,
⋯
)
.
S_n=\int_0^{\pi} \sin ^ n x\ dx\ (n=1,\ 2,\ ,\ \cdots).
S
n
=
∫
0
π
sin
n
x
d
x
(
n
=
1
,
2
,
,
⋯
)
.
Find
lim
n
→
∞
n
S
n
S
n
+
1
.
\lim_{n\to\infty} nS_nS_{n+1}.
lim
n
→
∞
n
S
n
S
n
+
1
.
841
1
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Today's calculation of Integral 841
Find
∫
0
x
d
t
1
+
t
2
+
∫
0
1
x
d
t
1
+
t
2
(
x
>
0
)
.
\int_0^x \frac{dt}{1+t^2}+\int_0^{\frac{1}{x}} \frac{dt}{1+t^2}\ (x>0).
∫
0
x
1
+
t
2
d
t
+
∫
0
x
1
1
+
t
2
d
t
(
x
>
0
)
.
840
1
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Today's calculation of Integral 840
Let
x
,
y
x,\ y
x
,
y
be real numbers. For a function
f
(
t
)
=
x
sin
t
+
y
cos
t
f(t)=x\sin t+y\cos t
f
(
t
)
=
x
sin
t
+
y
cos
t
, draw the domain of the points
(
x
,
y
)
(x,\ y)
(
x
,
y
)
for which the following inequality holds.
∣
∫
−
π
π
f
(
t
)
cos
t
d
t
∣
≤
∫
−
π
π
{
f
(
t
)
}
2
d
t
.
\left|\int_{-\pi}^{\pi} f(t)\cos t\ dt\right|\leq \int_{-\pi}^{\pi} \{f(t)\}^2dt.
∫
−
π
π
f
(
t
)
cos
t
d
t
≤
∫
−
π
π
{
f
(
t
)
}
2
d
t
.
839
1
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Today's calculation of Integral 839
Evaluate
∫
1
2
1
1
−
x
2
d
x
.
\int_{\frac 12}^1 \sqrt{1-x^2}\ dx.
∫
2
1
1
1
−
x
2
d
x
.
838
1
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Today's calculation of Integral 838
Prove that :
e
−
1
e
<
∫
0
1
e
−
x
2
d
x
<
π
4
.
\frac{e-1}{e}<\int_0^1 e^{-x^2}dx<\frac{\pi}{4}.
e
e
−
1
<
∫
0
1
e
−
x
2
d
x
<
4
π
.
837
1
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Today's calculation of Integral 837
Let
f
n
(
x
)
=
∑
k
=
1
n
(
−
1
)
k
+
1
(
x
2
k
−
1
2
k
−
1
+
x
2
k
2
k
)
.
f_n(x)=\sum_{k=1}^n (-1)^{k+1} \left(\frac{x^{2k-1}}{2k-1}+\frac{x^{2k}}{2k}\right).
f
n
(
x
)
=
∑
k
=
1
n
(
−
1
)
k
+
1
(
2
k
−
1
x
2
k
−
1
+
2
k
x
2
k
)
.
Find
lim
n
→
∞
f
n
(
1
)
.
\lim_{n\to\infty} f_n(1).
lim
n
→
∞
f
n
(
1
)
.
836
1
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Today's calculation of Integral 836
Evaluate
∫
0
π
e
sin
x
cos
2
(
sin
x
)
cos
x
d
x
\int_0^{\pi} e^{\sin x}\cos ^ 2(\sin x )\cos x\ dx
∫
0
π
e
s
i
n
x
cos
2
(
sin
x
)
cos
x
d
x
.
835
1
Hide problems
Today's calculation of Integral 835
Evaluate the following definite integrals.(a)
∫
1
2
x
−
1
x
2
−
2
x
+
2
d
x
\int_1^2 \frac{x-1}{x^2-2x+2}\ dx
∫
1
2
x
2
−
2
x
+
2
x
−
1
d
x
(b)
∫
0
1
e
4
x
e
2
x
+
2
d
x
\int_0^1 \frac{e^{4x}}{e^{2x}+2}\ dx
∫
0
1
e
2
x
+
2
e
4
x
d
x
(c)
∫
1
e
x
ln
x
d
x
\int_1^e x\ln \sqrt{x}\ dx
∫
1
e
x
ln
x
d
x
(d)
∫
0
π
3
(
cos
2
x
sin
3
x
−
1
4
sin
5
x
)
d
x
\int_0^{\frac{\pi}{3}} \left(\cos ^ 2 x\sin 3x-\frac 14\sin 5x\right)\ dx
∫
0
3
π
(
cos
2
x
sin
3
x
−
4
1
sin
5
x
)
d
x
834
1
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Today's calculation of Integral 834
Find the maximum and minimum areas of the region enclosed by the curve
y
=
∣
x
∣
e
∣
x
∣
y=|x|e^{|x|}
y
=
∣
x
∣
e
∣
x
∣
and the line
y
=
a
(
0
≤
a
≤
e
)
y=a\ (0\leq a\leq e)
y
=
a
(
0
≤
a
≤
e
)
at
[
−
1
,
1
]
[-1,\ 1]
[
−
1
,
1
]
.
833
1
Hide problems
Today's calculation of Integral 833
Let
f
(
x
)
=
∫
0
x
e
t
(
cos
t
+
sin
t
)
d
t
,
g
(
x
)
=
∫
0
x
e
t
(
cos
t
−
sin
t
)
d
t
.
f(x)=\int_0^{x} e^{t} (\cos t+\sin t)\ dt,\ g(x)=\int_0^{x} e^{t} (\cos t-\sin t)\ dt.
f
(
x
)
=
∫
0
x
e
t
(
cos
t
+
sin
t
)
d
t
,
g
(
x
)
=
∫
0
x
e
t
(
cos
t
−
sin
t
)
d
t
.
For a real number
a
a
a
, find
∑
n
=
1
∞
e
2
a
{
f
(
n
)
(
a
)
}
2
+
{
g
(
n
)
(
a
)
}
2
.
\sum_{n=1}^{\infty} \frac{e^{2a}}{\{f^{(n)}(a)\}^2+\{g^{(n)}(a)\}^2}.
∑
n
=
1
∞
{
f
(
n
)
(
a
)
}
2
+
{
g
(
n
)
(
a
)
}
2
e
2
a
.
832
1
Hide problems
Today's calculation of Integral 832
Find the limit
lim
n
→
∞
1
n
ln
n
∫
π
(
n
+
1
)
π
(
sin
2
t
)
(
ln
t
)
d
t
.
\lim_{n\to\infty} \frac{1}{n\ln n}\int_{\pi}^{(n+1)\pi} (\sin ^ 2 t)(\ln t)\ dt.
n
→
∞
lim
n
ln
n
1
∫
π
(
n
+
1
)
π
(
sin
2
t
)
(
ln
t
)
d
t
.
831
1
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Today's calculation of Integral 831
Let
n
n
n
be a positive integer. Answer the following questions.(1) Find the maximum value of
f
n
(
x
)
=
x
n
e
−
x
f_n(x)=x^{n}e^{-x}
f
n
(
x
)
=
x
n
e
−
x
for
x
≥
0
x\geq 0
x
≥
0
.(2) Show that
lim
x
→
∞
f
n
(
x
)
=
0
\lim_{x\to\infty} f_n(x)=0
lim
x
→
∞
f
n
(
x
)
=
0
.(3) Let
I
n
=
∫
0
x
f
n
(
t
)
d
t
I_n=\int_0^x f_n(t)\ dt
I
n
=
∫
0
x
f
n
(
t
)
d
t
. Find
lim
x
→
∞
I
n
(
x
)
\lim_{x\to\infty} I_n(x)
lim
x
→
∞
I
n
(
x
)
.
826
1
Hide problems
Hyperelementary abelian groups
Let
G
G
G
be a hyper elementary abelian
p
−
p-
p
−
group and let
f
:
G
→
G
f : G \rightarrow G
f
:
G
→
G
be a homomorphism. Then prove that
ker
f
\ker f
ker
f
is isomorphic to
c
o
k
e
r
f
\mathrm{coker} f
coker
f
.
830
1
Hide problems
Today's calculation of Integral 830
Find
lim
n
→
∞
1
(
ln
n
)
2
∑
k
=
3
n
ln
k
k
.
\lim_{n\to\infty} \frac{1}{(\ln n)^2}\sum_{k=3}^n \frac{\ln k}{k}.
lim
n
→
∞
(
l
n
n
)
2
1
∑
k
=
3
n
k
l
n
k
.
829
1
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Today's calculation of Integral 829
Let
a
a
a
be a positive constant. Find the value of
ln
a
\ln a
ln
a
such that
∫
1
e
ln
(
a
x
)
d
x
∫
1
e
x
d
x
=
∫
1
e
ln
(
a
x
)
x
d
x
.
\frac{\int_1^e \ln (ax)\ dx}{\int_1^e x\ dx}=\int_1^e \frac{\ln (ax)}{x}\ dx.
∫
1
e
x
d
x
∫
1
e
ln
(
a
x
)
d
x
=
∫
1
e
x
ln
(
a
x
)
d
x
.
828
1
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Today's calculation of Integral 828
Find a function
f
(
x
)
f(x)
f
(
x
)
, which is differentiable and
f
′
(
x
)
f'(x)
f
′
(
x
)
is continuous, such that
∫
0
x
f
(
t
)
cos
(
x
−
t
)
d
t
=
x
e
2
x
.
\int_0^x f(t)\cos (x-t)\ dt=xe^{2x}.
∫
0
x
f
(
t
)
cos
(
x
−
t
)
d
t
=
x
e
2
x
.
827
1
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Today's calculation of Integral 827
Find
lim
n
→
∞
∑
k
=
0
∞
∫
2
k
π
(
2
k
+
1
)
π
x
e
−
x
sin
x
d
x
.
\lim_{n\to\infty}\sum_{k=0}^{\infty} \int_{2k\pi}^{(2k+1)\pi} xe^{-x}\sin x\ dx.
lim
n
→
∞
∑
k
=
0
∞
∫
2
kπ
(
2
k
+
1
)
π
x
e
−
x
sin
x
d
x
.
825
1
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Today's calculation of Integral 825
Answer the following questions.(1) For
x
≥
0
x\geq 0
x
≥
0
, show that
x
−
x
3
6
≤
sin
x
≤
x
.
x-\frac{x^3}{6}\leq \sin x\leq x.
x
−
6
x
3
≤
sin
x
≤
x
.
(2) For
x
≥
0
x\geq 0
x
≥
0
, show that
x
3
3
−
x
5
30
≤
∫
0
x
t
sin
t
d
t
≤
x
3
3
.
\frac{x^3}{3}-\frac{x^5}{30}\leq \int_0^x t\sin t\ dt\leq \frac{x^3}{3}.
3
x
3
−
30
x
5
≤
∫
0
x
t
sin
t
d
t
≤
3
x
3
.
(3) Find the limit
lim
x
→
0
sin
x
−
x
cos
x
x
3
.
\lim_{x\rightarrow 0} \frac{\sin x-x\cos x}{x^3}.
x
→
0
lim
x
3
sin
x
−
x
cos
x
.
824
1
Hide problems
Today's calculation of Integral 824
In the
x
y
xy
x
y
-plane, for
a
>
1
a>1
a
>
1
denote by
S
(
a
)
S(a)
S
(
a
)
the area of the figure bounded by the curve
y
=
(
a
−
x
)
ln
x
y=(a-x)\ln x
y
=
(
a
−
x
)
ln
x
and the
x
x
x
-axis.Find the value of integer
n
n
n
for which
lim
a
→
∞
S
(
a
)
a
n
ln
a
\lim_{a\rightarrow \infty} \frac{S(a)}{a^n\ln a}
lim
a
→
∞
a
n
l
n
a
S
(
a
)
is non-zero real number.
823
1
Hide problems
Today's calculation of Integral 823
Let
C
C
C
be the curve expressed by
x
=
sin
t
,
y
=
sin
2
t
(
0
≤
t
≤
π
2
)
.
x=\sin t,\ y=\sin 2t\ \left(0\leq t\leq \frac{\pi}{2}\right).
x
=
sin
t
,
y
=
sin
2
t
(
0
≤
t
≤
2
π
)
.
(1) Express
y
y
y
in terms of
x
x
x
.(2) Find the area of the figure
D
D
D
enclosed by the
x
x
x
-axis and
C
C
C
.(3) Find the volume of the solid generated by a rotation of
D
D
D
about the
y
y
y
-axis.
822
1
Hide problems
Today's calculation of Integral 822
For
n
=
0
,
1
,
2
,
⋯
n=0,\ 1,\ 2,\ \cdots
n
=
0
,
1
,
2
,
⋯
, let
a
n
=
∫
n
n
+
1
{
x
e
−
x
−
(
n
+
1
)
e
−
n
−
1
(
x
−
n
)
}
d
x
,
a_n=\int_{n}^{n+1} \{xe^{-x}-(n+1)e^{-n-1}(x-n)\}\ dx,
a
n
=
∫
n
n
+
1
{
x
e
−
x
−
(
n
+
1
)
e
−
n
−
1
(
x
−
n
)}
d
x
,
b
n
=
∫
n
n
+
1
{
x
e
−
x
−
(
n
+
1
)
e
−
n
−
1
}
d
x
.
b_n=\int_{n}^{n+1} \{xe^{-x}-(n+1)e^{-n-1}\}\ dx.
b
n
=
∫
n
n
+
1
{
x
e
−
x
−
(
n
+
1
)
e
−
n
−
1
}
d
x
.
Find
lim
n
→
∞
∑
k
=
0
n
(
a
k
−
b
k
)
.
\lim_{n\to\infty} \sum_{k=0}^n (a_k-b_k).
lim
n
→
∞
∑
k
=
0
n
(
a
k
−
b
k
)
.
821
1
Hide problems
Today's calculation of Integral 821
Prove that :
ln
11
27
<
∫
1
4
3
4
1
ln
(
1
−
x
)
d
x
<
ln
7
15
.
\ln \frac{11}{27}<\int_{\frac 14}^{\frac 34} \frac{1}{\ln (1-x)}\ dx<\ln \frac{7}{15}.
ln
27
11
<
∫
4
1
4
3
l
n
(
1
−
x
)
1
d
x
<
ln
15
7
.
820
1
Hide problems
Today's calculation of Integral 820
Let
P
k
P_k
P
k
be a point whose
x
x
x
-coordinate is
1
+
k
n
(
k
=
1
,
2
,
⋯
,
n
)
1+\frac{k}{n}\ (k=1,\ 2,\ \cdots,\ n)
1
+
n
k
(
k
=
1
,
2
,
⋯
,
n
)
on the curve
y
=
ln
x
y=\ln x
y
=
ln
x
. For
A
(
1
,
0
)
A(1,\ 0)
A
(
1
,
0
)
, find the limit
lim
n
→
∞
1
n
∑
k
=
1
n
A
P
k
‾
2
.
\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} \overline{AP_k}^2.
lim
n
→
∞
n
1
∑
k
=
1
n
A
P
k
2
.
819
1
Hide problems
Today's calculation of Integral 819
For real numbers
a
,
b
a,\ b
a
,
b
with
0
≤
a
≤
π
,
a
<
b
0\leq a\leq \pi,\ a<b
0
≤
a
≤
π
,
a
<
b
, let
I
(
a
,
b
)
=
∫
a
b
e
−
x
sin
x
d
x
.
I(a,\ b)=\int_{a}^{b} e^{-x} \sin x\ dx.
I
(
a
,
b
)
=
∫
a
b
e
−
x
sin
x
d
x
.
Determine the value of
a
a
a
such that
lim
b
→
∞
I
(
a
,
b
)
=
0.
\lim_{b\rightarrow \infty} I(a,\ b)=0.
lim
b
→
∞
I
(
a
,
b
)
=
0.
818
1
Hide problems
Today's calculation of Integral 818
For a function
f
(
x
)
=
x
3
−
x
2
+
x
f(x)=x^3-x^2+x
f
(
x
)
=
x
3
−
x
2
+
x
, find the limit
lim
n
→
∞
∫
n
2
n
1
f
−
1
(
x
)
3
+
∣
f
−
1
(
x
)
∣
d
x
.
\lim_{n\to\infty} \int_{n}^{2n}\frac{1}{f^{-1}(x)^3+|f^{-1}(x)|}\ dx.
lim
n
→
∞
∫
n
2
n
f
−
1
(
x
)
3
+
∣
f
−
1
(
x
)
∣
1
d
x
.
817
1
Hide problems
Today's calculation of Integral 817
Define two functions
f
(
t
)
=
1
2
(
t
+
1
t
)
,
g
(
t
)
=
t
2
−
2
ln
t
f(t)=\frac 12\left(t+\frac{1}{t}\right),\ g(t)=t^2-2\ln t
f
(
t
)
=
2
1
(
t
+
t
1
)
,
g
(
t
)
=
t
2
−
2
ln
t
. When real number
t
t
t
moves in the range of
t
>
0
t>0
t
>
0
, denote by
C
C
C
the curve by which the point
(
f
(
t
)
,
g
(
t
)
)
(f(t),\ g(t))
(
f
(
t
)
,
g
(
t
))
draws on the
x
y
xy
x
y
-plane. Let
a
>
1
a>1
a
>
1
, find the area of the part bounded by the line
x
=
1
2
(
a
+
1
a
)
x=\frac 12\left(a+\frac{1}{a}\right)
x
=
2
1
(
a
+
a
1
)
and the curve
C
C
C
.
816
1
Hide problems
Today's calculation of Integral 816
Find the volume of the solid of a circle
x
2
+
(
y
−
1
)
2
=
4
x^2+(y-1)^2=4
x
2
+
(
y
−
1
)
2
=
4
generated by a rotation about the
x
x
x
-axis.
815
1
Hide problems
Today's calculation of Integral 815
Prove that :
∣
∑
i
=
0
n
(
1
−
π
sin
i
π
4
n
cos
i
π
4
n
)
∣
<
1.
\left|\sum_{i=0}^n \left(1-\pi \sin \frac{i\pi}{4n}\cos \frac{i\pi}{4n}\right)\right|<1.
∑
i
=
0
n
(
1
−
π
sin
4
n
iπ
cos
4
n
iπ
)
<
1.
814
1
Hide problems
Today's calculation of Integral 814
Find the area of the region bounded by
C
:
y
=
−
x
4
+
8
x
3
−
18
x
2
+
11
C: y=-x^4+8x^3-18x^2+11
C
:
y
=
−
x
4
+
8
x
3
−
18
x
2
+
11
and the tangent line which touches
C
C
C
at distinct two points.
813
1
Hide problems
Today's calculation of Integral 813
Let
a
a
a
be a real number. Find the minimum value of
∫
0
1
∣
a
x
−
x
3
∣
d
x
\int_0^1 |ax-x^3|dx
∫
0
1
∣
a
x
−
x
3
∣
d
x
.How many solutions (including University Mathematics )are there for the problem?Any advice would be appreciated. :)
812
1
Hide problems
Today's calculation of Integral 812
Let
f
(
x
)
=
cos
2
x
−
(
a
+
2
)
cos
x
+
a
+
1
sin
x
.
f(x)=\frac{\cos 2x-(a+2)\cos x+a+1}{\sin x}.
f
(
x
)
=
s
i
n
x
c
o
s
2
x
−
(
a
+
2
)
c
o
s
x
+
a
+
1
.
For constant
a
a
a
such that
lim
x
→
0
f
(
x
)
x
=
1
2
\lim_{x\rightarrow 0} \frac{f(x)}{x}=\frac 12
lim
x
→
0
x
f
(
x
)
=
2
1
, evaluate
∫
π
3
π
2
1
f
(
x
)
d
x
.
\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{f(x)}dx.
∫
3
π
2
π
f
(
x
)
1
d
x
.
811
1
Hide problems
Today's calculation of Integral 811
Let
a
a
a
be real number. Evaluate
∫
a
a
+
π
∣
x
∣
cos
x
d
x
.
\int_a^{a+\pi} |x|\cos x\ dx.
∫
a
a
+
π
∣
x
∣
cos
x
d
x
.
810
1
Hide problems
Today's calculation of Integral 810
Given the functions
f
(
x
)
=
x
e
x
+
2
x
∫
0
2
∣
g
(
t
)
∣
d
t
−
1
,
g
(
x
)
=
x
2
−
x
∫
0
1
f
(
t
)
d
t
f(x)=xe^{x}+2x\int_0^2 |g(t)|dt-1,\ g(x)=x^2-x\int_0^1 f(t)dt
f
(
x
)
=
x
e
x
+
2
x
∫
0
2
∣
g
(
t
)
∣
d
t
−
1
,
g
(
x
)
=
x
2
−
x
∫
0
1
f
(
t
)
d
t
, evaluate
∫
0
2
∣
g
(
t
)
∣
d
t
.
\int_0^2 |g(t)|dt.
∫
0
2
∣
g
(
t
)
∣
d
t
.
809
1
Hide problems
Today's calculation of Integral 809
For
a
>
0
a>0
a
>
0
, denote by
S
(
a
)
S(a)
S
(
a
)
the area of the part bounded by the parabolas
y
=
1
2
x
2
−
3
a
y=\frac 12x^2-3a
y
=
2
1
x
2
−
3
a
and
y
=
−
1
2
x
2
+
2
a
x
−
a
3
−
a
2
y=-\frac 12x^2+2ax-a^3-a^2
y
=
−
2
1
x
2
+
2
a
x
−
a
3
−
a
2
. Find the maximum area of
S
(
a
)
S(a)
S
(
a
)
.
808
1
Hide problems
Today's calculation of Integral 808
For a constant
c
c
c
, a sequence
a
n
a_n
a
n
is defined by
a
n
=
∫
c
1
n
x
n
−
1
(
ln
(
1
x
)
)
n
d
x
(
n
=
1
,
2
,
3
,
⋯
)
.
a_n=\int_c^1 nx^{n-1}\left(\ln \left(\frac{1}{x}\right)\right)^n dx\ (n=1,\ 2,\ 3,\ \cdots).
a
n
=
∫
c
1
n
x
n
−
1
(
ln
(
x
1
)
)
n
d
x
(
n
=
1
,
2
,
3
,
⋯
)
.
Find
lim
n
→
∞
a
n
\lim_{n\to\infty} a_n
lim
n
→
∞
a
n
.
807
1
Hide problems
Today's calculation of Integral 807
Define a sequence
a
n
a_n
a
n
satisfying :
a
1
=
1
,
a
n
+
1
=
n
a
n
2
+
n
(
a
n
+
1
)
(
n
=
1
,
2
,
3
,
⋯
)
.
a_1=1,\ \ a_{n+1}=\frac{na_n}{2+n(a_n+1)}\ (n=1,\ 2,\ 3,\ \cdots).
a
1
=
1
,
a
n
+
1
=
2
+
n
(
a
n
+
1
)
n
a
n
(
n
=
1
,
2
,
3
,
⋯
)
.
Find
lim
m
→
∞
m
∑
n
=
m
+
1
2
m
a
n
.
\lim_{m\to\infty} m\sum_{n=m+1}^{2m} a_n.
lim
m
→
∞
m
∑
n
=
m
+
1
2
m
a
n
.
806
1
Hide problems
Today's calculation of Integral 806
Let
n
n
n
be positive integers and
t
t
t
be a positive real number. Evaluate
∫
0
2
n
t
π
∣
x
sin
t
x
∣
d
x
.
\int_0^{\frac{2n}{t}\pi} |x\sin\ tx|\ dx.
∫
0
t
2
n
π
∣
x
sin
t
x
∣
d
x
.
805
1
Hide problems
Today's calculation of Integral 805
Prove the following inequalities:(1) For
0
≤
x
≤
1
0\leq x\leq 1
0
≤
x
≤
1
,
1
−
1
3
x
≤
1
1
+
x
2
≤
1.
1-\frac 13x\leq \frac{1}{\sqrt{1+x^2}}\leq 1.
1
−
3
1
x
≤
1
+
x
2
1
≤
1.
(2)
π
3
−
1
6
≤
∫
0
3
2
1
1
−
x
4
d
x
≤
π
3
.
\frac{\pi}{3}-\frac 16\leq \int_0^{\frac{\sqrt{3}}{2}} \frac{1}{\sqrt{1-x^4}}dx\leq \frac{\pi}{3}.
3
π
−
6
1
≤
∫
0
2
3
1
−
x
4
1
d
x
≤
3
π
.
804
1
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Today's calculation of Integral 804
For
a
>
0
a>0
a
>
0
, find the minimum value of
I
(
a
)
=
∫
1
e
∣
ln
a
x
∣
d
x
.
I(a)=\int_1^e |\ln ax|\ dx.
I
(
a
)
=
∫
1
e
∣
ln
a
x
∣
d
x
.
803
1
Hide problems
Today's calculation of Integral 803
Answer the following questions:(1) Evaluate
∫
−
1
1
(
1
−
x
2
)
e
−
2
x
d
x
.
\int_{-1}^1 (1-x^2)e^{-2x}dx.
∫
−
1
1
(
1
−
x
2
)
e
−
2
x
d
x
.
(2) Find
lim
n
→
∞
{
(
2
n
)
!
n
!
n
n
}
1
n
.
\lim_{n\to\infty} \left\{\frac{(2n)!}{n!n^n}\right\}^{\frac{1}{n}}.
lim
n
→
∞
{
n
!
n
n
(
2
n
)!
}
n
1
.
802
1
Hide problems
Today's calculation of Integral 802
Let
k
k
k
and
a
a
a
are positive constants. Denote by
V
1
V_1
V
1
the volume of the solid generated by a rotation of the figure enclosed by the curve
C
:
y
=
x
x
+
k
(
x
≥
0
)
C: y=\frac{x}{x+k}\ (x\geq 0)
C
:
y
=
x
+
k
x
(
x
≥
0
)
, the line
x
=
a
x=a
x
=
a
and the
x
x
x
-axis around the
x
x
x
-axis, and denote by
V
2
V_2
V
2
that of the solid by a rotation of the figure enclosed by the curve
C
C
C
, the line
y
=
a
a
+
k
y=\frac{a}{a+k}
y
=
a
+
k
a
and the
y
y
y
-axis around the
y
y
y
-axis. Find the ratio
V
2
V
1
.
\frac{V_2}{V_1}.
V
1
V
2
.
801
1
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Today's calculation of Integral 801
Answer the following questions:(1) Let
f
(
x
)
f(x)
f
(
x
)
be a function such that
f
′
′
(
x
)
f''(x)
f
′′
(
x
)
is continuous and
f
′
(
a
)
=
f
′
(
b
)
=
0
f'(a)=f'(b)=0
f
′
(
a
)
=
f
′
(
b
)
=
0
for some
a
<
b
a<b
a
<
b
.Prove that
f
(
b
)
−
f
(
a
)
=
∫
a
b
(
a
+
b
2
−
x
)
f
′
′
(
x
)
d
x
f(b)-f(a)=\int_a^b \left(\frac{a+b}{2}-x\right)f''(x)dx
f
(
b
)
−
f
(
a
)
=
∫
a
b
(
2
a
+
b
−
x
)
f
′′
(
x
)
d
x
.(2) Consider the running a car on straight road. After a car which is at standstill at a traffic light started at time 0, it stopped again at the next traffic light apart a distance
L
L
L
at time
T
T
T
. During the period, prove that there is an instant for which the absolute value of the acceleration of the car is more than or equal to
4
L
T
2
.
\frac{4L}{T^2}.
T
2
4
L
.
800
1
Hide problems
Today's calculation of Integral 800
For a positive constant
a
a
a
, find the minimum value of
f
(
x
)
=
∫
0
π
2
∣
sin
t
−
a
x
cos
t
∣
d
t
.
f(x)=\int_0^{\frac{\pi}{2}} |\sin t-ax\cos t|dt.
f
(
x
)
=
∫
0
2
π
∣
sin
t
−
a
x
cos
t
∣
d
t
.
799
1
Hide problems
Today's calculation of Integral 799
Let
n
n
n
be positive integer. Define a sequence
{
a
k
}
\{a_k\}
{
a
k
}
by
a
1
=
1
n
(
n
+
1
)
,
a
k
+
1
=
−
1
k
+
n
+
1
+
n
k
∑
i
=
1
k
a
i
(
k
=
1
,
2
,
3
,
⋯
)
.
a_1=\frac{1}{n(n+1)},\ a_{k+1}=-\frac{1}{k+n+1}+\frac{n}{k}\sum_{i=1}^k a_i\ \ (k=1,\ 2,\ 3,\ \cdots).
a
1
=
n
(
n
+
1
)
1
,
a
k
+
1
=
−
k
+
n
+
1
1
+
k
n
i
=
1
∑
k
a
i
(
k
=
1
,
2
,
3
,
⋯
)
.
(1) Find
a
2
a_2
a
2
and
a
3
a_3
a
3
.(2) Find the general term
a
k
a_k
a
k
.(3) Let
b
n
=
∑
k
=
1
n
a
k
b_n=\sum_{k=1}^n \sqrt{a_k}
b
n
=
∑
k
=
1
n
a
k
. Prove that
lim
n
→
∞
b
n
=
ln
2
\lim_{n\to\infty} b_n=\ln 2
lim
n
→
∞
b
n
=
ln
2
.50 points
798
1
Hide problems
Today's calculation of Integral 798
Denote by
C
,
l
C,\ l
C
,
l
the graphs of the cubic function
C
:
y
=
x
3
−
3
x
2
+
2
x
C: y=x^3-3x^2+2x
C
:
y
=
x
3
−
3
x
2
+
2
x
, the line
l
:
y
=
a
x
l: y=ax
l
:
y
=
a
x
.(1) Find the range of
a
a
a
such that
C
C
C
and
l
l
l
have intersection point other than the origin.(2) Denote
S
(
a
)
S(a)
S
(
a
)
by the area bounded by
C
C
C
and
l
l
l
. If
a
a
a
move in the range found in (1), then find the value of
a
a
a
for which
S
(
a
)
S(a)
S
(
a
)
is minimized.50 points
797
1
Hide problems
Today's calculation of Integral 797
In the
x
y
z
xyz
x
yz
-space take four points
P
(
0
,
0
,
2
)
,
A
(
0
,
2
,
0
)
,
B
(
3
,
−
1
,
0
)
,
C
(
−
3
,
−
1
,
0
)
P(0,\ 0,\ 2),\ A(0,\ 2,\ 0),\ B(\sqrt{3},-1,\ 0),\ C(-\sqrt{3},-1,\ 0)
P
(
0
,
0
,
2
)
,
A
(
0
,
2
,
0
)
,
B
(
3
,
−
1
,
0
)
,
C
(
−
3
,
−
1
,
0
)
. Find the volume of the part satifying
x
2
+
y
2
≥
1
x^2+y^2\geq 1
x
2
+
y
2
≥
1
in the tetrahedron
P
A
B
C
PABC
P
A
BC
.50 points
796
1
Hide problems
Today's calculation of Integral 796
Answer the following questions:(1) Let
a
a
a
be non-zero constant. Find
∫
x
2
cos
(
a
ln
x
)
d
x
.
\int x^2 \cos (a\ln x)dx.
∫
x
2
cos
(
a
ln
x
)
d
x
.
(2) Find the volume of the solid generated by a rotation of the figures enclosed by the curve
y
=
x
cos
(
ln
x
)
y=x\cos (\ln x)
y
=
x
cos
(
ln
x
)
, the
x
x
x
-axis and the lines
x
=
1
,
x
=
e
π
4
x=1,\ x=e^{\frac{\pi}{4}}
x
=
1
,
x
=
e
4
π
about the
x
x
x
-axis.
795
1
Hide problems
Today's calculation of Integral 795
Evaluate
∫
π
3
π
2
2
+
sin
x
1
+
cos
x
d
x
.
\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{2+\sin x}{1+\cos x}\ dx.
∫
3
π
2
π
1
+
c
o
s
x
2
+
s
i
n
x
d
x
.
794
1
Hide problems
Today's calculation of Integral 794
Define a function
f
(
x
)
=
∫
0
π
2
cos
∣
t
−
x
∣
1
+
sin
∣
t
−
x
∣
d
t
f(x)=\int_0^{\frac{\pi}{2}} \frac{\cos |t-x|}{1+\sin |t-x|}dt
f
(
x
)
=
∫
0
2
π
1
+
s
i
n
∣
t
−
x
∣
c
o
s
∣
t
−
x
∣
d
t
for
0
≤
x
≤
π
0\leq x\leq \pi
0
≤
x
≤
π
.Find the maximum and minimum value of
f
(
x
)
f(x)
f
(
x
)
in
0
≤
x
≤
π
0\leq x\leq \pi
0
≤
x
≤
π
.
793
1
Hide problems
Today's calculation of Integral 793
Find the area of the figure bounded by two curves
y
=
x
4
,
y
=
x
2
+
2
y=x^4,\ y=x^2+2
y
=
x
4
,
y
=
x
2
+
2
.
792
1
Hide problems
Today's calculation of Integral 792
Answer the following questions:(1) Let
a
a
a
be positive real number. Find
lim
n
→
∞
(
1
+
a
n
)
1
n
.
\lim_{n\to\infty} (1+a^{n})^{\frac{1}{n}}.
lim
n
→
∞
(
1
+
a
n
)
n
1
.
(2) Evaluate
∫
1
3
1
x
2
ln
1
+
x
2
d
x
.
\int_1^{\sqrt{3}} \frac{1}{x^2}\ln \sqrt{1+x^2}dx.
∫
1
3
x
2
1
ln
1
+
x
2
d
x
.
35 points
791
1
Hide problems
Today's calculation of Integral 791
Let
S
S
S
be the domain in the coordinate plane determined by two inequalities:
y
≥
1
2
x
2
,
x
2
4
+
4
y
2
≤
1
8
.
y\geq \frac 12x^2,\ \ \frac{x^2}{4}+4y^2\leq \frac 18.
y
≥
2
1
x
2
,
4
x
2
+
4
y
2
≤
8
1
.
Denote by
V
1
V_1
V
1
the volume of the solid by a rotation of
S
S
S
about the
x
x
x
-axis and by
V
2
V_2
V
2
, by a rotation of
S
S
S
about the
y
y
y
-axis.(1) Find the values of
V
1
,
V
2
V_1,\ V_2
V
1
,
V
2
.(2) Compare the size of the value of
V
2
V
1
\frac{V_2}{V_1}
V
1
V
2
and 1.
790
1
Hide problems
Today's calculation of Integral 790
Define a parabola
C
C
C
by
y
=
x
2
+
1
y=x^2+1
y
=
x
2
+
1
on the coordinate plane. Let
s
,
t
s,\ t
s
,
t
be real numbers with
t
<
0
t<0
t
<
0
. Denote by
l
1
,
l
2
l_1,\ l_2
l
1
,
l
2
the tangent lines drawn from the point
(
s
,
t
)
(s,\ t)
(
s
,
t
)
to the parabola
C
C
C
.(1) Find the equations of the tangents
l
1
,
l
2
l_1,\ l_2
l
1
,
l
2
.(2) Let
a
a
a
be positive real number. Find the pairs of
(
s
,
t
)
(s,\ t)
(
s
,
t
)
such that the area of the region enclosed by
C
,
l
1
,
l
2
C,\ l_1,\ l_2
C
,
l
1
,
l
2
is
a
a
a
.
789
1
Hide problems
Today's calculation of Integral 789
Find the non-constant function
f
(
x
)
f(x)
f
(
x
)
such that
f
(
x
)
=
x
2
−
∫
0
1
(
f
(
t
)
+
x
)
2
d
t
.
f(x)=x^2-\int_0^1 (f(t)+x)^2dt.
f
(
x
)
=
x
2
−
∫
0
1
(
f
(
t
)
+
x
)
2
d
t
.
788
1
Hide problems
Today's calculation of Integral 788
For a function
f
(
x
)
=
ln
(
1
+
1
−
x
2
)
−
1
−
x
2
−
ln
x
(
0
<
x
<
1
)
f(x)=\ln (1+\sqrt{1-x^2})-\sqrt{1-x^2}-\ln x\ (0<x<1)
f
(
x
)
=
ln
(
1
+
1
−
x
2
)
−
1
−
x
2
−
ln
x
(
0
<
x
<
1
)
, answer the following questions:(1) Find
f
′
(
x
)
f'(x)
f
′
(
x
)
.(2) Sketch the graph of
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
.(3) Let
P
P
P
be a mobile point on the curve
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
and
Q
Q
Q
be a point which is on the tangent at
P
P
P
on the curve
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
and such that
P
Q
=
1
PQ=1
PQ
=
1
. Note that the
x
x
x
-coordinate of
Q
Q
Q
is les than that of
P
P
P
. Find the locus of
Q
Q
Q
.
787
1
Hide problems
Today's calculation of Integral 787
Take two points
A
(
−
1
,
0
)
,
B
(
1
,
0
)
A\ (-1,\ 0),\ B\ (1,\ 0)
A
(
−
1
,
0
)
,
B
(
1
,
0
)
on the
x
y
xy
x
y
-plane. Let
F
F
F
be the figure by which the whole points
P
P
P
on the plane satisfies
π
4
≤
∠
A
P
B
≤
π
\frac{\pi}{4}\leq \angle{APB}\leq \pi
4
π
≤
∠
A
PB
≤
π
and the figure formed by
A
,
B
A,\ B
A
,
B
. Answer the following questions:(1) Illustrate
F
F
F
.(2) Find the volume of the solid generated by a rotation of
F
F
F
around the
x
x
x
-axis.
786
1
Hide problems
Today's calculation of Integral 786
For each positive integer
n
n
n
, define
H
n
(
x
)
=
(
−
1
)
n
e
x
2
d
n
d
x
n
e
−
x
2
.
H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}.
H
n
(
x
)
=
(
−
1
)
n
e
x
2
d
x
n
d
n
e
−
x
2
.
(1) Find
H
1
(
x
)
,
H
2
(
x
)
,
H
3
(
x
)
H_1(x),\ H_2(x),\ H_3(x)
H
1
(
x
)
,
H
2
(
x
)
,
H
3
(
x
)
.(2) Express
d
d
x
H
n
(
x
)
\frac{d}{dx}H_n(x)
d
x
d
H
n
(
x
)
interms of
H
n
(
x
)
,
H
n
+
1
(
x
)
.
H_n(x),\ H_{n+1}(x).
H
n
(
x
)
,
H
n
+
1
(
x
)
.
Then prove that
H
n
(
x
)
H_n(x)
H
n
(
x
)
is a polynpmial with degree
n
n
n
by induction.(3) Let
a
a
a
be real number. For
n
≥
3
n\geq 3
n
≥
3
, express
S
n
(
a
)
=
∫
0
a
x
H
n
(
x
)
e
−
x
2
d
x
S_n(a)=\int_0^a xH_n(x)e^{-x^2}dx
S
n
(
a
)
=
∫
0
a
x
H
n
(
x
)
e
−
x
2
d
x
in terms of
H
n
−
1
(
a
)
,
H
n
−
2
(
a
)
,
H
n
−
2
(
0
)
H_{n-1}(a),\ H_{n-2}(a),\ H_{n-2}(0)
H
n
−
1
(
a
)
,
H
n
−
2
(
a
)
,
H
n
−
2
(
0
)
.(4) Find
lim
a
→
∞
S
6
(
a
)
\lim_{a\to\infty} S_6(a)
lim
a
→
∞
S
6
(
a
)
. If necessary, you may use
lim
x
→
∞
x
k
e
−
x
2
=
0
\lim_{x\to\infty}x^ke^{-x^2}=0
lim
x
→
∞
x
k
e
−
x
2
=
0
for a positive integer
k
k
k
.
785
1
Hide problems
Today's calculation of Integral 785
For a positive real number
x
x
x
, find the minimum value of
f
(
x
)
=
∫
x
2
x
(
t
ln
t
−
t
)
d
t
.
f(x)=\int_x^{2x} (t\ln t-t)dt.
f
(
x
)
=
∫
x
2
x
(
t
ln
t
−
t
)
d
t
.
784
1
Hide problems
Today's calculation of Integral 784
Define for positive integer
n
n
n
, a function
f
n
(
x
)
=
ln
x
x
n
(
x
>
0
)
.
f_n(x)=\frac{\ln x}{x^n}\ (x>0).
f
n
(
x
)
=
x
n
l
n
x
(
x
>
0
)
.
In the coordinate plane, denote by
S
n
S_n
S
n
the area of the figure enclosed by
y
=
f
n
(
x
)
(
x
≤
t
)
y=f_n(x)\ (x\leq t)
y
=
f
n
(
x
)
(
x
≤
t
)
, the
x
x
x
-axis and the line
x
=
t
x=t
x
=
t
and denote by
T
n
T_n
T
n
the area of the rectagle with four vertices
(
1
,
0
)
,
(
t
,
0
)
,
(
t
,
f
n
(
t
)
)
(1,\ 0),\ (t,\ 0),\ (t,\ f_n(t))
(
1
,
0
)
,
(
t
,
0
)
,
(
t
,
f
n
(
t
))
and
(
1
,
f
n
(
t
)
)
(1,\ f_n(t))
(
1
,
f
n
(
t
))
.(1) Find the local maximum
f
n
(
x
)
f_n(x)
f
n
(
x
)
.(2) When
t
t
t
moves in the range of
t
>
1
t>1
t
>
1
, find the value of
t
t
t
for which
T
n
(
t
)
−
S
n
(
t
)
T_n(t)-S_n(t)
T
n
(
t
)
−
S
n
(
t
)
is maximized.(3) Find
S
1
(
t
)
S_1(t)
S
1
(
t
)
and
S
n
(
t
)
(
n
≥
2
)
S_n(t)\ (n\geq 2)
S
n
(
t
)
(
n
≥
2
)
.(4) For each
n
≥
2
n\geq 2
n
≥
2
, prove that there exists the only
t
>
1
t>1
t
>
1
such that
T
n
(
t
)
=
S
n
(
t
)
T_n(t)=S_n(t)
T
n
(
t
)
=
S
n
(
t
)
.Note that you may use
lim
x
→
∞
ln
x
x
=
0.
\lim_{x\to\infty} \frac{\ln x}{x}=0.
lim
x
→
∞
x
l
n
x
=
0.
783
1
Hide problems
Today's calculation of Integral 783
Define a sequence
a
1
=
0
,
1
1
−
a
n
+
1
−
1
1
−
a
n
=
2
n
+
1
(
n
=
1
,
2
,
3
,
⋯
)
a_1=0,\ \frac{1}{1-a_{n+1}}-\frac{1}{1-a_n}=2n+1\ (n=1,\ 2,\ 3,\ \cdots)
a
1
=
0
,
1
−
a
n
+
1
1
−
1
−
a
n
1
=
2
n
+
1
(
n
=
1
,
2
,
3
,
⋯
)
.(1) Find
a
n
a_n
a
n
.(2) Let
b
k
=
k
+
1
k
(
1
−
a
k
+
1
)
{b_k=\sqrt{\frac{k+1}{k}}\ (1-\sqrt{a_{k+1}}})
b
k
=
k
k
+
1
(
1
−
a
k
+
1
)
for
k
=
1
,
2
,
3
,
⋯
k=1,\ 2,\ 3,\ \cdots
k
=
1
,
2
,
3
,
⋯
.Prove that
∑
k
=
1
n
b
k
<
2
−
1
\sum_{k=1}^n b_k<\sqrt{2}-1
∑
k
=
1
n
b
k
<
2
−
1
for each
n
n
n
.Last Edited
782
1
Hide problems
Today's calculation of Integral 782
Let
C
C
C
be the part of the graph
y
=
1
x
(
x
>
0
)
y=\frac{1}{x}\ (x>0)
y
=
x
1
(
x
>
0
)
. Take a point
P
(
t
,
1
t
)
(
t
>
0
)
P\left(t,\ \frac{1}{t}\right)\ (t>0)
P
(
t
,
t
1
)
(
t
>
0
)
on
C
C
C
.(i) Find the equation of the tangent
l
l
l
at the point
A
(
1
,
1
)
A(1,\ 1)
A
(
1
,
1
)
on the curve
C
C
C
.(ii) Let
m
m
m
be the line passing through the point
P
P
P
and parallel to
l
l
l
. Denote
Q
Q
Q
be the intersection point of the line
m
m
m
and the curve
C
C
C
other than
P
P
P
. Find the coordinate of
Q
Q
Q
. (iii) Express the area
S
S
S
of the part bounded by two line segments
O
P
,
O
Q
OP,\ OQ
OP
,
OQ
and the curve
C
C
C
for the origin
O
O
O
in terms of
t
t
t
.(iv) Express the volume
V
V
V
of the solid generated by a rotation of the part enclosed by two lines passing through the point
P
P
P
and pararell to the
y
y
y
-axis and passing through the point
Q
Q
Q
and pararell to
y
y
y
-axis, the curve
C
C
C
and the
x
x
x
-axis in terms of
t
t
t
.(v)
lim
t
→
1
−
0
S
V
.
\lim_{t\rightarrow 1-0} \frac{S}{V}.
lim
t
→
1
−
0
V
S
.
781
1
Hide problems
Today's calculation of Integral 781
Let
l
,
m
l,\ m
l
,
m
be the tangent lines passing through the point
A
(
a
,
a
−
1
)
A(a,\ a-1)
A
(
a
,
a
−
1
)
on the line
y
=
x
−
1
y=x-1
y
=
x
−
1
and touch the parabola
y
=
x
2
y=x^2
y
=
x
2
. Note that the slope of
l
l
l
is greater than that of
m
m
m
.(1) Exress the slope of
l
l
l
in terms of
a
a
a
.(2) Denote
P
,
Q
P,\ Q
P
,
Q
be the points of tangency of the lines
l
,
m
l,\ m
l
,
m
and the parabola
y
=
x
2
y=x^2
y
=
x
2
. Find the minimum area of the part bounded by the line segment
P
Q
PQ
PQ
and the parabola
y
=
x
2
y=x^2
y
=
x
2
.(3) Find the minimum distance between the parabola
y
=
x
2
y=x^2
y
=
x
2
and the line
y
=
x
−
1
y=x-1
y
=
x
−
1
.
780
1
Hide problems
Today's calculation of Integral 780
Let
n
≥
3
n\geq 3
n
≥
3
be integer. Given a regular
n
n
n
-polygon
P
P
P
with side length 4 on the plane
z
=
0
z=0
z
=
0
in the
x
y
z
xyz
x
yz
-space.Llet
G
G
G
be a circumcenter of
P
P
P
. When the center of the sphere
B
B
B
with radius 1 travels round along the sides of
P
P
P
, denote by
K
n
K_n
K
n
the solid swept by
B
B
B
.Answer the following questions.(1) Take two adjacent vertices
P
1
,
P
2
P_1,\ P_2
P
1
,
P
2
of
P
P
P
. Let
Q
Q
Q
be the intersection point between the perpendicular dawn from
G
G
G
to
P
1
P
2
P_1P_2
P
1
P
2
, prove that
G
Q
>
1
GQ>1
GQ
>
1
.(2) (i) Express the area of cross section
S
(
t
)
S(t)
S
(
t
)
in terms of
t
,
n
t,\ n
t
,
n
when
K
n
K_n
K
n
is cut by the plane
z
=
t
(
−
1
≤
t
≤
1
)
z=t\ (-1\leq t\leq 1)
z
=
t
(
−
1
≤
t
≤
1
)
.(ii) Express the volume
V
(
n
)
V(n)
V
(
n
)
of
K
n
K_n
K
n
in terms of
n
n
n
.(3) Denote by
l
l
l
the line which passes through
G
G
G
and perpendicular to the plane
z
=
0
z=0
z
=
0
. Express the volume
W
(
n
)
W(n)
W
(
n
)
of the solid by generated by a rotation of
K
n
K_n
K
n
around
l
l
l
in terms of
n
n
n
.(4) Find
lim
n
→
∞
V
(
n
)
W
(
n
)
.
\lim_{n\to\infty} \frac{V(n)}{W(n)} .
lim
n
→
∞
W
(
n
)
V
(
n
)
.
779
1
Hide problems
Today's calculation of Integral 779
Consider parabolas
C
a
:
y
=
−
2
x
2
+
4
a
x
−
2
a
2
+
a
+
1
C_a: y=-2x^2+4ax-2a^2+a+1
C
a
:
y
=
−
2
x
2
+
4
a
x
−
2
a
2
+
a
+
1
and
C
:
y
=
x
2
−
2
x
C: y=x^2-2x
C
:
y
=
x
2
−
2
x
in the coordinate plane. When
C
a
C_a
C
a
and
C
C
C
have two intersection points, find the maximum area enclosed by these parabolas.
778
1
Hide problems
Today's calculation of Integral 778
In the
x
y
z
xyz
x
yz
space with the origin
O
O
O
, Let
K
1
K_1
K
1
be the surface and inner part of the sphere centered on the point
(
1
,
0
,
0
)
(1,\ 0,\ 0)
(
1
,
0
,
0
)
with radius 2 and let
K
2
K_2
K
2
be the surface and inner part of the sphere centered on the point
(
−
1
,
0
,
0
)
(-1,\ 0,\ 0)
(
−
1
,
0
,
0
)
with radius 2. For three points
P
,
Q
,
R
P,\ Q,\ R
P
,
Q
,
R
in the space, consider points
X
,
Y
X,\ Y
X
,
Y
defined by
O
X
→
=
O
P
→
+
O
Q
→
,
O
Y
→
=
1
3
(
O
P
→
+
O
Q
→
+
O
R
→
)
.
\overrightarrow{OX}=\overrightarrow{OP}+\overrightarrow{OQ},\ \overrightarrow{OY}=\frac 13(\overrightarrow{OP}+\overrightarrow{OQ}+\overrightarrow{OR}).
OX
=
OP
+
OQ
,
O
Y
=
3
1
(
OP
+
OQ
+
OR
)
.
(1) When
P
,
Q
P,\ Q
P
,
Q
move every cranny in
K
1
,
K
2
K_1,\ K_2
K
1
,
K
2
respectively, find the volume of the solid generated by the whole points of the point
X
X
X
.(2) Find the volume of the solid generated by the whole points of the point
R
R
R
for which for any
P
P
P
belonging to
K
1
K_1
K
1
and any
Q
Q
Q
belonging to
K
2
K_2
K
2
,
Y
Y
Y
belongs to
K
1
K_1
K
1
.(3) Find the volume of the solid generated by the whole points of the point
R
R
R
for which for any
P
P
P
belonging to
K
1
K_1
K
1
and any
Q
Q
Q
belonging to
K
2
K_2
K
2
,
Y
Y
Y
belongs to
K
1
∪
K
2
K_1\cup K_2
K
1
∪
K
2
.
777
1
Hide problems
Today's calculation of Integral 777
Given two points
P
,
Q
P,\ Q
P
,
Q
on the parabola
C
:
y
=
x
2
−
x
−
2
C: y=x^2-x-2
C
:
y
=
x
2
−
x
−
2
in the
x
y
xy
x
y
plane. Note that the
x
x
x
coodinate of
P
P
P
is less than that of
Q
Q
Q
.(a) If the origin
O
O
O
is the midpoint of the lines egment
P
Q
PQ
PQ
, then find the equation of the line
P
Q
PQ
PQ
.(b) If the origin
O
O
O
divides internally the line segment
P
Q
PQ
PQ
by 2:1, then find the equation of
P
Q
PQ
PQ
.(c) If the origin
O
O
O
divides internally the line segment
P
Q
PQ
PQ
by 2:1, find the area of the figure bounded by the parabola
C
C
C
and the line
P
Q
PQ
PQ
.
776
1
Hide problems
Today's calculation of Integral 776
Evaluate
∫
1
−
5
2
1
+
5
2
(
2
x
2
−
1
)
e
2
x
d
x
.
\int_{\frac{1-\sqrt{5}}{2}}^{\frac{1+\sqrt{5}}{2}} (2x^2-1)e^{2x}dx.
∫
2
1
−
5
2
1
+
5
(
2
x
2
−
1
)
e
2
x
d
x
.
775
1
Hide problems
Today's calculation of Integral 775
Let
a
a
a
be negative constant. Find the value of
a
a
a
and
f
(
x
)
f(x)
f
(
x
)
such that
∫
a
2
t
2
f
(
x
)
d
x
=
t
2
+
3
t
−
4
\int_{\frac{a}{2}}^{\frac{t}{2}} f(x)dx=t^2+3t-4
∫
2
a
2
t
f
(
x
)
d
x
=
t
2
+
3
t
−
4
holds for any real numbers
t
t
t
.
774
1
Hide problems
Today's calculation of Integral 774
Find the real number
a
a
a
such that
∫
0
a
e
x
+
e
−
x
2
d
x
=
12
5
.
\int_0^a \frac{e^x+e^{-x}}{2}dx=\frac{12}{5}.
∫
0
a
2
e
x
+
e
−
x
d
x
=
5
12
.
773
1
Hide problems
Today's calculation of Integral 773
For
x
≥
0
x\geq 0
x
≥
0
find the value of
x
x
x
by which
f
(
x
)
=
∫
0
x
3
t
(
3
t
−
4
)
(
x
−
t
)
d
t
f(x)=\int_0^x 3^t(3^t-4)(x-t)dt
f
(
x
)
=
∫
0
x
3
t
(
3
t
−
4
)
(
x
−
t
)
d
t
is minimized.
772
1
Hide problems
Today's calculation of Integral 772
Given are three points
A
(
2
,
0
,
2
)
,
B
(
1
,
1
,
0
)
,
C
(
0
,
0
,
3
)
A(2,\ 0,\ 2),\ B(1,\ 1,\ 0),\ C(0,\ 0,\ 3)
A
(
2
,
0
,
2
)
,
B
(
1
,
1
,
0
)
,
C
(
0
,
0
,
3
)
in the coordinate space. Find the volume of the solid of a triangle
A
B
C
ABC
A
BC
generated by a rotation about
z
z
z
-axis.
771
1
Hide problems
Today's calculation of Integral 771
(1) Find the range of
a
a
a
for which there exist two common tangent lines of the curve
y
=
8
27
x
3
y=\frac{8}{27}x^3
y
=
27
8
x
3
and the parabola
y
=
(
x
+
a
)
2
y=(x+a)^2
y
=
(
x
+
a
)
2
other than the
x
x
x
axis.(2) For the range of
a
a
a
found in the previous question, express the area bounded by the two tangent lines and the parabola
y
=
(
x
+
a
)
2
y=(x+a)^2
y
=
(
x
+
a
)
2
in terms of
a
a
a
.
770
1
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Today's calculation of Integral 770
Find the value of
a
a
a
such that :
101
a
=
6539
∫
−
1
1
x
12
+
31
1
+
201
1
x
d
x
.
101a=6539\int_{-1}^1 \frac{x^{12}+31}{1+2011^{x}}\ dx.
101
a
=
6539
∫
−
1
1
1
+
201
1
x
x
12
+
31
d
x
.