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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
602
602
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 602
Source:
4/28/2010
Prove the following inequality.
e
−
1
n
+
1
≦
∫
1
e
(
log
x
)
n
d
x
≦
(
n
+
1
)
e
+
1
(
n
+
1
)
(
n
+
2
)
(
n
=
1
,
2
,
⋅
⋅
⋅
)
\frac{e-1}{n+1}\leqq\int^e_1(\log x)^n dx\leqq\frac{(n+1)e+1}{(n+1)(n+2)}\ (n=1,2,\cdot\cdot\cdot)
n
+
1
e
−
1
≦
∫
1
e
(
lo
g
x
)
n
d
x
≦
(
n
+
1
)
(
n
+
2
)
(
n
+
1
)
e
+
1
(
n
=
1
,
2
,
⋅
⋅
⋅
)
1994 Kyoto University entrance exam/Science
calculus
integration
logarithms
floor function
counting
derangement
function