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Integration/Polynomial/x^m (1-x)^n/Formula/Exercise
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Show by induction over
n
{\displaystyle {}n}
, using
integration by parts
, that
∫
0
1
x
m
(
1
−
x
)
n
d
x
=
m
!
n
!
(
m
+
n
+
1
)
!
{\displaystyle {}\int _{0}^{1}x^{m}(1-x)^{n}dx={\frac {m!n!}{(m+n+1)!}}\,}
holds.
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