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Putnam
1941 Putnam
B6
B6
Part of
1941 Putnam
Problems
(1)
Putnam 1941 B6
Source: Putnam 1941
2/23/2022
Assuming that
f
(
x
)
f(x)
f
(
x
)
is continuous in the interval
(
0
,
1
)
(0,1)
(
0
,
1
)
, prove that
∫
x
=
0
x
=
1
∫
y
=
x
y
=
1
∫
z
=
x
z
=
y
f
(
x
)
f
(
y
)
f
(
z
)
d
z
d
y
d
x
=
1
6
(
∫
0
1
f
(
t
)
d
t
)
3
.
\int_{x=0}^{x=1} \int_{y=x}^{y=1} \int_{z=x}^{z=y} f(x)f(y)f(z)\;dz dy dx= \frac{1}{6}\left(\int_{0}^{1} f(t)\; dt\right)^{3}.
∫
x
=
0
x
=
1
∫
y
=
x
y
=
1
∫
z
=
x
z
=
y
f
(
x
)
f
(
y
)
f
(
z
)
d
z
d
y
d
x
=
6
1
(
∫
0
1
f
(
t
)
d
t
)
3
.
Putnam
integration
Triple integral