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Today's Calculation Of Integral
2013 Today's Calculation Of Integral
2013 Today's Calculation Of Integral
Part of
Today's Calculation Of Integral
Subcontests
(39)
899
1
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Today's calculation of Integral 899
Find the limit as below.
lim
n
→
∞
(
1
2
+
2
2
+
⋯
+
n
2
)
(
1
3
+
2
3
+
⋯
+
n
3
)
(
1
4
+
2
4
+
⋯
+
n
4
)
(
1
5
+
2
5
+
⋯
+
n
5
)
2
\lim_{n\to\infty} \frac{(1^2+2^2+\cdots +n^2)(1^3+2^3+\cdots +n^3)(1^4+2^4+\cdots +n^4)}{(1^5+2^5+\cdots +n^5)^2}
n
→
∞
lim
(
1
5
+
2
5
+
⋯
+
n
5
)
2
(
1
2
+
2
2
+
⋯
+
n
2
)
(
1
3
+
2
3
+
⋯
+
n
3
)
(
1
4
+
2
4
+
⋯
+
n
4
)
898
1
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Today's calculation of Integral 898
Let
a
,
b
a,\ b
a
,
b
be positive constants.Evaluate
∫
0
1
ln
(
x
+
a
)
x
+
a
(
x
+
b
)
x
+
b
(
x
+
a
)
(
x
+
b
)
ln
(
x
+
a
)
ln
(
x
+
b
)
d
x
.
\int_0^1 \frac{\ln \frac{(x+a)^{x+a}}{(x+b)^{x+b}}}{(x+a)(x+b)\ln (x+a)\ln (x+b)}\ dx.
∫
0
1
(
x
+
a
)
(
x
+
b
)
ln
(
x
+
a
)
ln
(
x
+
b
)
ln
(
x
+
b
)
x
+
b
(
x
+
a
)
x
+
a
d
x
.
897
1
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Today's calculation of Integral 897
Find the volume
V
V
V
of the solid formed by a rotation of the region enclosed by the curve
y
=
2
x
−
1
y=2^{x}-1
y
=
2
x
−
1
and two lines
x
=
0
,
y
=
1
x=0,\ y=1
x
=
0
,
y
=
1
around the
y
y
y
axis.
896
1
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Today's calculation of Integral 896
Given sequences
a
n
=
1
n
2
n
P
n
n
,
b
n
=
1
n
2
4
n
P
2
n
n
a_n=\frac{1}{n}{\sqrt[n] {_{2n}P_n}},\ b_n=\frac{1}{n^2}{\sqrt[n] {_{4n}P_{2n}}}
a
n
=
n
1
n
2
n
P
n
,
b
n
=
n
2
1
n
4
n
P
2
n
and
c
n
=
8
n
P
4
n
6
n
P
4
n
n
c_n=\sqrt[n]{\frac{_{8n}P_{4n}}{_{6n}P_{4n}}}
c
n
=
n
6
n
P
4
n
8
n
P
4
n
, find
lim
n
→
∞
a
n
,
lim
n
→
∞
b
n
\lim_{n\to\infty} a_n,\ \lim_{n\to\infty} b_n
lim
n
→
∞
a
n
,
lim
n
→
∞
b
n
and
lim
n
→
∞
c
n
.
\lim_{n\to\infty} c_n.
lim
n
→
∞
c
n
.
895
1
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Today's calculation of Integral 895
In the coordinate plane, suppose that the parabola
C
:
y
=
−
p
2
x
2
+
q
(
p
>
0
,
q
>
0
)
C: y=-\frac{p}{2}x^2+q\ (p>0,\ q>0)
C
:
y
=
−
2
p
x
2
+
q
(
p
>
0
,
q
>
0
)
touches the circle with radius 1 centered on the origin at distinct two points. Find the minimum area of the figure enclosed by the part of
y
≥
0
y\geq 0
y
≥
0
of
C
C
C
and the
x
x
x
-axis.
894
1
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Today's calculation of Integral 894
Let
a
a
a
be non zero real number. Find the area of the figure enclosed by the line
y
=
a
x
y=ax
y
=
a
x
, the curve
y
=
x
ln
(
x
+
1
)
.
y=x\ln (x+1).
y
=
x
ln
(
x
+
1
)
.
893
1
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Today's calculation of Integral 893
Find the minimum value of
f
(
x
)
=
∫
0
π
4
∣
tan
t
−
x
∣
d
t
.
f(x)=\int_0^{\frac{\pi}{4}} |\tan t-x|dt.
f
(
x
)
=
∫
0
4
π
∣
tan
t
−
x
∣
d
t
.
891
1
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Today's calculation of Integral 891
Given a triangle
O
A
B
OAB
O
A
B
with the vetices
O
(
0
,
0
,
0
)
,
A
(
1
,
0
,
0
)
,
B
(
1
,
1
,
0
)
O(0,\ 0,\ 0),\ A(1,\ 0,\ 0),\ B(1,\ 1,\ 0)
O
(
0
,
0
,
0
)
,
A
(
1
,
0
,
0
)
,
B
(
1
,
1
,
0
)
in the
x
y
z
xyz
x
yz
space. Let
V
V
V
be the cone obtained by rotating the triangle around the
x
x
x
-axis. Find the volume of the solid obtained by rotating the cone
V
V
V
around the
y
y
y
-axis.
890
1
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Today's calculation of Integral 890
A function
f
n
(
x
)
(
n
=
1
,
2
,
⋯
)
f_n(x)\ (n=1,\ 2,\ \cdots)
f
n
(
x
)
(
n
=
1
,
2
,
⋯
)
is defined by
f
1
(
x
)
=
x
f_1(x)=x
f
1
(
x
)
=
x
and
f
n
(
x
)
=
x
+
e
2
∫
0
1
f
n
−
1
(
t
)
e
x
−
t
d
t
(
n
=
2
,
3
,
⋯
)
f_n(x)=x+\frac{e}{2}\int_0^1 f_{n-1}(t)e^{x-t}dt\ (n=2,\ 3,\ \cdots)
f
n
(
x
)
=
x
+
2
e
∫
0
1
f
n
−
1
(
t
)
e
x
−
t
d
t
(
n
=
2
,
3
,
⋯
)
. Find
f
n
(
x
)
f_n(x)
f
n
(
x
)
.
889
1
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Today's calculation of Integral 889
Find the area
S
S
S
of the region enclosed by the curve
y
=
∣
x
−
1
x
∣
(
x
>
0
)
y=\left|x-\frac{1}{x}\right|\ (x>0)
y
=
x
−
x
1
(
x
>
0
)
and the line
y
=
2
y=2
y
=
2
.
888
1
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Today's calculation of Integral 888
In the coordinate plane, given a circle
K
:
x
2
+
y
2
=
1
,
C
:
y
=
x
2
−
2
K: x^2+y^2=1,\ C: y=x^2-2
K
:
x
2
+
y
2
=
1
,
C
:
y
=
x
2
−
2
. Let
l
l
l
be the tangent line of
K
K
K
at
P
(
cos
θ
,
sin
θ
)
(
π
<
θ
<
2
π
)
.
P(\cos \theta,\ \sin \theta)\ (\pi<\theta <2\pi).
P
(
cos
θ
,
sin
θ
)
(
π
<
θ
<
2
π
)
.
Find the minimum area of the part enclosed by
l
l
l
and
C
C
C
.
887
1
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Today's calculation of Integral 887
For the function
f
(
x
)
=
∫
0
x
d
t
1
+
t
2
f(x)=\int_0^x \frac{dt}{1+t^2}
f
(
x
)
=
∫
0
x
1
+
t
2
d
t
, answer the questions as follows.Note : Please solve the problems without using directly the formula
∫
1
1
+
x
2
d
x
=
tan
−
1
x
+
C
\int \frac{1}{1+x^2}\ dx=\tan^{-1}x +C
∫
1
+
x
2
1
d
x
=
tan
−
1
x
+
C
for Japanese High School students those who don't study arc sin x, arc cos x, arc tanx.(1) Find
f
(
3
)
f(\sqrt{3})
f
(
3
)
(2) Find
∫
0
3
x
f
(
x
)
d
x
\int_0^{\sqrt{3}} xf(x)\ dx
∫
0
3
x
f
(
x
)
d
x
(3) Prove that for
x
>
0
x>0
x
>
0
.
f
(
x
)
+
f
(
1
x
)
f(x)+f\left(\frac{1}{x}\right)
f
(
x
)
+
f
(
x
1
)
is constant, then find the value.
886
1
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Today's calculation of Integral 886
Find the functions
f
(
x
)
,
g
(
x
)
f(x),\ g(x)
f
(
x
)
,
g
(
x
)
such that
f
(
x
)
=
e
x
sin
x
+
∫
0
π
u
g
(
u
)
d
u
f(x)=e^{x}\sin x+\int_0^{\pi} ug(u)\ du
f
(
x
)
=
e
x
sin
x
+
∫
0
π
ug
(
u
)
d
u
g
(
x
)
=
e
x
cos
x
+
∫
0
π
u
f
(
u
)
d
u
g(x)=e^{x}\cos x+\int_0^{\pi} uf(u)\ du
g
(
x
)
=
e
x
cos
x
+
∫
0
π
u
f
(
u
)
d
u
885
1
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Today's calculation of Integral 885
Find the infinite integrals as follows.(1) 2013 Hiroshima City University entrance exam/Informatic Science
∫
x
2
2
−
x
2
d
x
\int \frac{x^2}{2-x^2}dx
∫
2
−
x
2
x
2
d
x
(2) 2013 Kanseigakuin University entrance exam/Science and Technology
∫
x
4
ln
x
d
x
\int x^4\ln x\ dx
∫
x
4
ln
x
d
x
(3) 2013 Shinsyu University entrance exam/Textile Science and Technology, Second-exam
∫
cos
3
x
sin
2
x
d
x
\int \frac{\cos ^ 3 x}{\sin ^ 2 x}\ dx
∫
s
i
n
2
x
c
o
s
3
x
d
x
884
1
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Today's calculation of Integral 884
Prove that :
π
(
e
−
1
)
<
∫
0
π
e
∣
cos
4
x
∣
d
x
<
2
(
e
π
2
−
1
)
\pi (e-1)<\int_0^{\pi} e^{|\cos 4x|}dx<2(e^{\frac{\pi}{2}}-1)
π
(
e
−
1
)
<
∫
0
π
e
∣
c
o
s
4
x
∣
d
x
<
2
(
e
2
π
−
1
)
883
1
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Today's calculation of Integral 883
Prove that for each positive integer
n
n
n
4
n
2
+
1
4
n
2
−
1
∫
0
π
(
e
x
−
e
−
x
)
cos
2
n
x
d
x
>
e
π
−
e
−
π
−
2
4
ln
(
2
n
+
1
)
2
(
2
n
−
1
)
(
n
+
3
)
.
\frac{4n^2+1}{4n^2-1}\int_0^{\pi} (e^{x}-e^{-x})\cos 2nx\ dx>\frac{e^{\pi}-e^{-\pi}-2}{4}\ln \frac{(2n+1)^2}{(2n-1)(n+3)}.
4
n
2
−
1
4
n
2
+
1
∫
0
π
(
e
x
−
e
−
x
)
cos
2
n
x
d
x
>
4
e
π
−
e
−
π
−
2
ln
(
2
n
−
1
)
(
n
+
3
)
(
2
n
+
1
)
2
.
882
1
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Today's calculation of Integral 882
Find
lim
n
→
∞
∑
k
=
1
n
1
n
+
k
(
ln
(
n
+
k
)
−
ln
n
)
\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{n+k}(\ln (n+k)-\ln\ n)
lim
n
→
∞
∑
k
=
1
n
n
+
k
1
(
ln
(
n
+
k
)
−
ln
n
)
.
881
1
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Today's calculation of Integral 881
Evaluate
∫
−
π
π
(
∑
k
=
1
2013
sin
k
x
)
2
d
x
\int_{-\pi}^{\pi} \left(\sum_{k=1}^{2013} \sin kx\right)^2dx
∫
−
π
π
(
∑
k
=
1
2013
sin
k
x
)
2
d
x
.
880
1
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Today's calculation of Integral 880
For
a
>
2
a>2
a
>
2
, let
f
(
t
)
=
sin
2
a
t
+
t
2
a
t
sin
a
t
,
g
(
t
)
=
sin
2
a
t
−
t
2
a
t
sin
a
t
(
0
<
∣
t
∣
<
π
2
a
)
f(t)=\frac{\sin ^ 2 at+t^2}{at\sin at},\ g(t)=\frac{\sin ^ 2 at-t^2}{at\sin at}\ \left(0<|t|<\frac{\pi}{2a}\right)
f
(
t
)
=
a
t
s
i
n
a
t
s
i
n
2
a
t
+
t
2
,
g
(
t
)
=
a
t
s
i
n
a
t
s
i
n
2
a
t
−
t
2
(
0
<
∣
t
∣
<
2
a
π
)
andlet
C
:
x
2
−
y
2
=
4
a
2
(
x
≥
2
a
)
.
C: x^2-y^2=\frac{4}{a^2}\ \left(x\geq \frac{2}{a}\right).
C
:
x
2
−
y
2
=
a
2
4
(
x
≥
a
2
)
.
Answer the questions as follows.(1) Show that the point
(
f
(
t
)
,
g
(
t
)
)
(f(t),\ g(t))
(
f
(
t
)
,
g
(
t
))
lies on the curve
C
C
C
.(2) Find the normal line of the curve
C
C
C
at the point
(
lim
t
→
0
f
(
t
)
,
lim
t
→
0
g
(
t
)
)
.
\left(\lim_{t\rightarrow 0} f(t),\ \lim_{t\rightarrow 0} g(t)\right).
(
lim
t
→
0
f
(
t
)
,
lim
t
→
0
g
(
t
)
)
.
(3) Let
V
(
a
)
V(a)
V
(
a
)
be the volume of the solid generated by a rotation of the part enclosed by the curve
C
C
C
, the nornal line found in (2) and the
x
x
x
-axis. Express
V
(
a
)
V(a)
V
(
a
)
in terms of
a
a
a
, then find
lim
a
→
∞
V
(
a
)
\lim_{a\to\infty} V(a)
lim
a
→
∞
V
(
a
)
.
879
1
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Today's calculation of Integral 879
Evaluate the integrals as follows.(1)
∫
x
2
2
−
x
d
x
\int \frac{x^2}{2-x}\ dx
∫
2
−
x
x
2
d
x
(2)
∫
x
5
+
x
3
3
d
x
\int \sqrt[3]{x^5+x^3}\ dx
∫
3
x
5
+
x
3
d
x
(3)
∫
0
1
(
1
−
x
)
cos
π
x
d
x
\int_0^1 (1-x)\cos \pi x\ dx
∫
0
1
(
1
−
x
)
cos
π
x
d
x
878
1
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Today's calculation of Integral 878
A cubic function
f
(
x
)
f(x)
f
(
x
)
satisfies the equation
sin
3
t
=
f
(
sin
t
)
\sin 3t=f(\sin t)
sin
3
t
=
f
(
sin
t
)
for all real numbers
t
t
t
.Evaluate
∫
0
1
f
(
x
)
2
1
−
x
2
d
x
\int_0^1 f(x)^2\sqrt{1-x^2}\ dx
∫
0
1
f
(
x
)
2
1
−
x
2
d
x
.
877
1
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Today's calculation of Integral 877
Let
f
(
x
)
=
lim
n
→
∞
sin
n
+
2
x
+
cos
n
+
2
x
sin
n
x
+
cos
n
x
f(x)=\lim_{n\to\infty} \frac{\sin^{n+2}x+\cos^{n+2}x}{\sin^n x+\cos^n x}
f
(
x
)
=
lim
n
→
∞
s
i
n
n
x
+
c
o
s
n
x
s
i
n
n
+
2
x
+
c
o
s
n
+
2
x
for
0
≤
x
≤
π
2
.
0\leq x\leq \frac{\pi}2.
0
≤
x
≤
2
π
.
Evaluate
∫
0
π
2
f
(
x
)
d
x
.
\int_0^{\frac{\pi}2} f(x)\ dx.
∫
0
2
π
f
(
x
)
d
x
.
876
1
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Today's calculation of Integral 876
Suppose a function
f
(
x
)
f(x)
f
(
x
)
is continuous on
[
−
1
,
1
]
[-1,\ 1]
[
−
1
,
1
]
and satisfies the condition :1)
f
(
−
1
)
≥
f
(
1
)
.
f(-1)\geq f(1).
f
(
−
1
)
≥
f
(
1
)
.
2)
x
+
f
(
x
)
x+f(x)
x
+
f
(
x
)
is non decreasing function.3)
∫
−
1
1
f
(
x
)
d
x
=
0.
\int_{-1}^ 1 f(x)\ dx=0.
∫
−
1
1
f
(
x
)
d
x
=
0.
Show that
∫
−
1
1
f
(
x
)
2
d
x
≤
2
3
.
\int_{-1}^1 f(x)^2dx\leq \frac 23.
∫
−
1
1
f
(
x
)
2
d
x
≤
3
2
.
875
1
Hide problems
Today's calculation of Integral 875
Evaluate
∫
0
1
x
2
+
x
+
1
x
4
+
x
3
+
x
2
+
x
+
1
d
x
.
\int_0^1 \frac{x^2+x+1}{x^4+x^3+x^2+x+1}\ dx.
∫
0
1
x
4
+
x
3
+
x
2
+
x
+
1
x
2
+
x
+
1
d
x
.
874
1
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Today's calculation of Integral 874
Given a parabola
C
:
y
=
1
−
x
2
C : y=1-x^2
C
:
y
=
1
−
x
2
in
x
y
xy
x
y
-palne with the origin
O
O
O
. Take two points
P
(
p
,
1
−
p
2
)
,
Q
(
q
,
1
−
q
2
)
(
p
<
q
)
P(p,\ 1-p^2),\ Q(q,\ 1-q^2)\ (p<q)
P
(
p
,
1
−
p
2
)
,
Q
(
q
,
1
−
q
2
)
(
p
<
q
)
on
C
C
C
.(1) Express the area
S
S
S
of the part enclosed by two segments
O
P
,
O
Q
OP,\ OQ
OP
,
OQ
and the parabalola
C
C
C
in terms of
p
,
q
p,\ q
p
,
q
.(2) If
q
=
p
+
1
q=p+1
q
=
p
+
1
, then find the minimum value of
S
S
S
.(3) If
p
q
=
−
1
pq=-1
pq
=
−
1
, then find the minimum value of
S
S
S
.
873
1
Hide problems
Today's calculation of Integral 873
Let
a
,
b
a,\ b
a
,
b
be positive real numbers. Consider the circle
C
1
:
(
x
−
a
)
2
+
y
2
=
a
2
C_1: (x-a)^2+y^2=a^2
C
1
:
(
x
−
a
)
2
+
y
2
=
a
2
and the ellipse
C
2
:
x
2
+
y
2
b
2
=
1.
C_2: x^2+\frac{y^2}{b^2}=1.
C
2
:
x
2
+
b
2
y
2
=
1.
(1) Find the condition for which
C
1
C_1
C
1
is inscribed in
C
2
C_2
C
2
.(2) Suppose
b
=
1
3
b=\frac{1}{\sqrt{3}}
b
=
3
1
and
C
1
C_1
C
1
is inscribed in
C
2
C_2
C
2
. Find the coordinate
(
p
,
q
)
(p,\ q)
(
p
,
q
)
of the point of tangency in the first quadrant for
C
1
C_1
C
1
and
C
2
C_2
C
2
.(3) Under the condition in (1), find the area of the part enclosed by
C
1
,
C
2
C_1,\ C_2
C
1
,
C
2
for
x
≥
p
x\geq p
x
≥
p
.60 point
872
1
Hide problems
Today's calculation of Integral 872
Let
n
n
n
be a positive integer.(1) For a positive integer
k
k
k
such that
1
≤
k
≤
n
1\leq k\leq n
1
≤
k
≤
n
, Show that :
∫
k
−
1
2
n
π
k
2
n
π
sin
2
n
t
cos
t
d
t
=
(
−
1
)
k
+
1
2
n
4
n
2
−
1
(
cos
k
2
n
π
+
cos
k
−
1
2
n
π
)
.
\int_{\frac{k-1}{2n}\pi}^{\frac{k}{2n}\pi} \sin 2nt\cos t\ dt=(-1)^{k+1}\frac{2n}{4n^2-1}(\cos \frac{k}{2n}\pi +\cos \frac{k-1}{2n}\pi).
∫
2
n
k
−
1
π
2
n
k
π
sin
2
n
t
cos
t
d
t
=
(
−
1
)
k
+
1
4
n
2
−
1
2
n
(
cos
2
n
k
π
+
cos
2
n
k
−
1
π
)
.
(2) Find the area
S
n
S_n
S
n
of the part expressed by a parameterized curve
C
n
:
x
=
sin
t
,
y
=
sin
2
n
t
(
0
≤
t
≤
π
)
.
C_n: x=\sin t,\ y=\sin 2nt\ (0\leq t\leq \pi).
C
n
:
x
=
sin
t
,
y
=
sin
2
n
t
(
0
≤
t
≤
π
)
.
If necessary, you may use
∑
k
=
1
n
−
1
cos
k
2
n
π
=
1
2
(
1
tan
π
4
n
−
1
)
(
n
≥
2
)
.
{\sum_{k=1}^{n-1} \cos \frac{k}{2n}\pi =\frac 12(\frac{1}{\tan \frac{\pi}{4n}}-1})\ (n\geq 2).
∑
k
=
1
n
−
1
cos
2
n
k
π
=
2
1
(
t
a
n
4
n
π
1
−
1
)
(
n
≥
2
)
.
(3) Find
lim
n
→
∞
S
n
.
\lim_{n\to\infty} S_n.
lim
n
→
∞
S
n
.
871
1
Hide problems
Today's calculation of Integral 871
Define sequences
{
a
n
}
,
{
b
n
}
\{a_n\},\ \{b_n\}
{
a
n
}
,
{
b
n
}
by
a
n
=
∫
−
π
6
π
6
e
n
sin
θ
d
θ
,
b
n
=
∫
−
π
6
π
6
e
n
sin
θ
cos
θ
d
θ
(
n
=
1
,
2
,
3
,
⋯
)
.
a_n=\int_{-\frac {\pi}6}^{\frac{\pi}6} e^{n\sin \theta}d\theta,\ b_n=\int_{-\frac {\pi}6}^{\frac{\pi}6} e^{n\sin \theta}\cos \theta d\theta\ (n=1,\ 2,\ 3,\ \cdots).
a
n
=
∫
−
6
π
6
π
e
n
s
i
n
θ
d
θ
,
b
n
=
∫
−
6
π
6
π
e
n
s
i
n
θ
cos
θ
d
θ
(
n
=
1
,
2
,
3
,
⋯
)
.
(1) Find
b
n
b_n
b
n
.(2) Prove that for each
n
n
n
,
b
n
≤
a
n
≤
2
3
b
n
.
b_n\leq a_n\leq \frac 2{\sqrt{3}}b_n.
b
n
≤
a
n
≤
3
2
b
n
.
(3) Find
lim
n
→
∞
1
n
ln
(
n
a
n
)
.
\lim_{n\to\infty} \frac 1{n}\ln (na_n).
lim
n
→
∞
n
1
ln
(
n
a
n
)
.
870
1
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Today's calculation of Integral 870
Consider the ellipse
E
:
3
x
2
+
y
2
=
3
E: 3x^2+y^2=3
E
:
3
x
2
+
y
2
=
3
and the hyperbola
H
:
x
y
=
3
4
.
H: xy=\frac 34.
H
:
x
y
=
4
3
.
(1) Find all points of intersection of
E
E
E
and
H
H
H
.(2) Find the area of the region expressed by the system of inequality
{
3
x
2
+
y
2
≤
3
e
m
s
p
;
x
y
≥
3
4
,
e
m
s
p
;
\left\{ \begin{array}{ll} 3x^2+y^2\leq 3 &  \\ xy\geq \frac 34 , &  \end{array} \right.
{
3
x
2
+
y
2
≤
3
x
y
≥
4
3
,
e
m
s
p
;
e
m
s
p
;
869
1
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Today's calculation of Integral 869
Let
I
n
=
1
n
+
1
∫
0
π
x
(
sin
n
x
+
n
π
cos
n
x
)
d
x
(
n
=
1
,
2
,
⋯
)
.
I_n=\frac{1}{n+1}\int_0^{\pi} x(\sin nx+n\pi\cos nx)dx\ \ (n=1,\ 2,\ \cdots).
I
n
=
n
+
1
1
∫
0
π
x
(
sin
n
x
+
nπ
cos
n
x
)
d
x
(
n
=
1
,
2
,
⋯
)
.
Answer the questions below.(1) Find
I
n
.
I_n.
I
n
.
(2) Find
∑
n
=
1
∞
I
n
.
\sum_{n=1}^{\infty} I_n.
∑
n
=
1
∞
I
n
.
868
1
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Today's calculation of Integral 868
In the coordinate space, define a square
S
S
S
, defined by the inequality
∣
x
∣
≤
1
,
∣
y
∣
≤
1
|x|\leq 1,\ |y|\leq 1
∣
x
∣
≤
1
,
∣
y
∣
≤
1
on the
x
y
xy
x
y
-plane, with four vertices
A
(
−
1
,
1
,
0
)
,
B
(
1
,
1
,
0
)
,
C
(
1
,
−
1
,
0
)
,
D
(
−
1
,
−
1
,
0
)
A(-1,\ 1,\ 0),\ B(1,\ 1,\ 0),\ C(1,-1,\ 0), D(-1,-1,\ 0)
A
(
−
1
,
1
,
0
)
,
B
(
1
,
1
,
0
)
,
C
(
1
,
−
1
,
0
)
,
D
(
−
1
,
−
1
,
0
)
. Let
V
1
V_1
V
1
be the solid by a rotation of the square
S
S
S
about the line
B
D
BD
B
D
as the axis of rotation, and let
V
2
V_2
V
2
be the solid by a rotation of the square
S
S
S
about the line
A
C
AC
A
C
as the axis of rotation.(1) For a real number
t
t
t
such that
0
≤
t
<
1
0\leq t<1
0
≤
t
<
1
, find the area of cross section of
V
1
V_1
V
1
cut by the plane
x
=
t
x=t
x
=
t
.(2) Find the volume of the common part of
V
1
V_1
V
1
and
V
2
V_2
V
2
.
867
1
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Today's calculation of Integral 867
Express
∫
0
2
f
(
x
)
d
x
\int_0^2 f(x)dx
∫
0
2
f
(
x
)
d
x
for any quadratic functions
f
(
x
)
f(x)
f
(
x
)
in terms of
f
(
0
)
,
f
(
1
)
f(0),\ f(1)
f
(
0
)
,
f
(
1
)
and
f
(
2
)
.
f(2).
f
(
2
)
.
866
1
Hide problems
Today's calculation of Integral 866
Given a solid
R
R
R
contained in a semi cylinder with the hight
1
1
1
which has a semicircle with radius
1
1
1
as the base. The cross section at the hight
x
(
0
≤
x
≤
1
)
x\ (0\leq x\leq 1)
x
(
0
≤
x
≤
1
)
is the form combined with two right-angled triangles as attached figure as below. Answer the following questions.(1) Find the cross-sectional area
S
(
x
)
S(x)
S
(
x
)
at the hight
x
x
x
.(2) Find the volume of
R
R
R
. If necessary, when you integrate, set
x
=
sin
t
.
x=\sin t.
x
=
sin
t
.
865
1
Hide problems
Todaｙ's calculation of Integral 865
Find the volume of the solid generated by a rotation of the region enclosed by the curve
y
=
x
3
−
x
y=x^3-x
y
=
x
3
−
x
and the line
y
=
x
y=x
y
=
x
about the line
y
=
x
y=x
y
=
x
as the axis of rotation.
864
1
Hide problems
Today's calculation of Integral 864
Let
m
,
n
m,\ n
m
,
n
be positive integer such that
2
≤
m
<
n
2\leq m<n
2
≤
m
<
n
.(1) Prove the inequality as follows.
n
+
1
−
m
m
(
n
+
1
)
<
1
m
2
+
1
(
m
+
1
)
2
+
⋯
+
1
(
n
−
1
)
2
+
1
n
2
<
n
+
1
−
m
n
(
m
−
1
)
\frac{n+1-m}{m(n+1)}<\frac{1}{m^2}+\frac{1}{(m+1)^2}+\cdots +\frac{1}{(n-1)^2}+\frac{1}{n^2}<\frac{n+1-m}{n(m-1)}
m
(
n
+
1
)
n
+
1
−
m
<
m
2
1
+
(
m
+
1
)
2
1
+
⋯
+
(
n
−
1
)
2
1
+
n
2
1
<
n
(
m
−
1
)
n
+
1
−
m
(2) Prove the inequality as follows.
3
2
≤
lim
n
→
∞
(
1
+
1
2
2
+
⋯
+
1
n
2
)
≤
2
\frac 32\leq \lim_{n\to\infty} \left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)\leq 2
2
3
≤
n
→
∞
lim
(
1
+
2
2
1
+
⋯
+
n
2
1
)
≤
2
(3) Prove the inequality which is made precisely in comparison with the inequality in (2) as follows.
29
18
≤
lim
n
→
∞
(
1
+
1
2
2
+
⋯
+
1
n
2
)
≤
61
36
\frac {29}{18}\leq \lim_{n\to\infty} \left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)\leq \frac{61}{36}
18
29
≤
n
→
∞
lim
(
1
+
2
2
1
+
⋯
+
n
2
1
)
≤
36
61
863
1
Hide problems
Today's calculation of Integral 863
For
0
<
t
≤
1
0<t\leq 1
0
<
t
≤
1
, let
F
(
t
)
=
1
t
∫
0
π
2
t
∣
cos
2
x
∣
d
x
.
F(t)=\frac{1}{t}\int_0^{\frac{\pi}{2}t} |\cos 2x|\ dx.
F
(
t
)
=
t
1
∫
0
2
π
t
∣
cos
2
x
∣
d
x
.
(1) Find
lim
t
→
0
F
(
t
)
.
\lim_{t\rightarrow 0} F(t).
lim
t
→
0
F
(
t
)
.
(2) Find the range of
t
t
t
such that
F
(
t
)
≥
1.
F(t)\geq 1.
F
(
t
)
≥
1.
862
1
Hide problems
Today's calculation of Integral 862
Draw a tangent with positive slope to a parabola
y
=
x
2
+
1
y=x^2+1
y
=
x
2
+
1
. Find the
x
x
x
-coordinate such that the area of the figure bounded by the parabola, the tangent and the coordinate axisis is
11
3
.
\frac{11}{3}.
3
11
.
861
1
Hide problems
Today's calculation of Integral 861
Answer the questions as below.(1) Find the local minimum of
y
=
x
(
1
−
x
2
)
e
x
2
.
y=x(1-x^2)e^{x^2}.
y
=
x
(
1
−
x
2
)
e
x
2
.
(2) Find the total area of the part bounded the graph of the function in (1) and the
x
x
x
-axis.
860
1
Hide problems
Today's calculation of Integral 860
For a function
f
(
x
)
(
x
≥
1
)
f(x)\ (x\geq 1)
f
(
x
)
(
x
≥
1
)
satisfying
f
(
x
)
=
(
log
e
x
)
2
−
∫
1
e
f
(
t
)
t
d
t
f(x)=(\log_e x)^2-\int_1^e \frac{f(t)}{t}dt
f
(
x
)
=
(
lo
g
e
x
)
2
−
∫
1
e
t
f
(
t
)
d
t
, answer the questions as below.(a) Find
f
(
x
)
f(x)
f
(
x
)
and the
y
y
y
-coordinate of the inflection point of the curve
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
.(b) Find the area of the figure bounded by the tangent line of
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
at the point
(
e
,
f
(
e
)
)
(e,\ f(e))
(
e
,
f
(
e
))
, the curve
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
and the line
x
=
1
x=1
x
=
1
.