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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
716
Today's calculation of Integral 716
Today's calculation of Integral 716
Source: 2010 Miyazaki University entrance exam/Medicine
June 25, 2011
calculus
integration
logarithms
calculus computations
Problem Statement
Prove that :
∫
1
e
(
ln
x
)
n
d
x
=
(
−
1
)
n
−
1
n
!
+
e
∑
m
=
0
n
(
−
1
)
n
−
m
n
!
m
!
(
1
2
)
m
\int_1^{\sqrt{e}} (\ln x)^n\ dx=(-1)^{n-1}n!+\sqrt{e}\sum_{m=0}^{n} (-1)^{n-m}\frac{n!}{m!}\left(\frac 12\right)^{m}
∫
1
e
(
ln
x
)
n
d
x
=
(
−
1
)
n
−
1
n
!
+
e
m
=
0
∑
n
(
−
1
)
n
−
m
m
!
n
!
(
2
1
)
m
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