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Putnam
1941 Putnam
A6
A6
Part of
1941 Putnam
Problems
(1)
Putnam 1941 A6
Source: Putnam 1941
2/23/2022
If the
x
x
x
-coordinate
x
‾
\overline{x}
x
of the center of mass of the area lying between the
x
x
x
-axis and the curve
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
with
f
(
x
)
>
0
f(x)>0
f
(
x
)
>
0
, and between the lines
x
=
0
x=0
x
=
0
and
x
=
a
x=a
x
=
a
is given by
x
‾
=
g
(
a
)
,
\overline{x}=g(a),
x
=
g
(
a
)
,
show that
f
(
x
)
=
A
⋅
g
′
(
x
)
(
x
−
g
(
x
)
)
2
⋅
e
∫
1
t
−
g
(
t
)
d
t
,
f(x)=A\cdot \frac{g'(x)}{(x-g(x))^{2}} \cdot e^{\int \frac{1}{t-g(t)} dt},
f
(
x
)
=
A
⋅
(
x
−
g
(
x
)
)
2
g
′
(
x
)
⋅
e
∫
t
−
g
(
t
)
1
d
t
,
where
A
A
A
is a positive constant.
Putnam
geometry