Putnam 2025, Problem B-2

Let \( f: [0, 1] \to [0, \infty) \) be strictly increasing and continuous. Let \( R \) be the region bounded by \( x = 0 \), \( x = 1 \), \( y = 0 \), and \( y = f(x) \). Let \( x_1 \) be the \( x \)-coordinate of the centroid of \( R \). Let \( x_2 \) be the \( x \)-coordinate of the centroid of the solid generated by rotating \( R \) around the \( x \)-axis. Prove that \( x_1 < x_2 \).

Solution:
**First solution.** By writing
\[
x_1 = \frac{\int_0^1 x f(x) dx}{\int_0^1 f(x) dx}, \quad x_2 = \frac{\int_0^1 x f(x)^2 dx}{\int_0^1 f(x)^2 dx},
\]
we see that it suffices to show that \( I_1 < I_2 \) for
\[
I_1 := \left( \int_0^1 x f(x) dx \right) \left( \int_0^1 f(x)^2 dx \right), \quad I_2 := \left( \int_0^1 f(x) dx \right) \left( \int_0^1 x f(x)^2 dx \right).
\]
By Fubini’s theorem, we can write each of \( I_1 \) and \( I_2 \) in terms of double integrals:
\[
I_1 = \iint_{[0,1]^2} x f(x) f(y)^2 dy dx = \iint_{[0,1]^2} y f(y) f(x)^2 dy dx, \quad I_2 = \iint_{[0,1]^2} f(x) y f(y)^2 dy dx = \iint_{[0,1]^2} f(y) x f(x)^2 dy dx,
\]
where in each case the second double integral is obtained from the first by switching the roles of \( x \) and \( y \). By adding the two expressions for \( I_2 \) and subtracting the two expressions for \( I_1 \), we find that
\[
2(I_2 - I_1) = \iint_{[0,1]^2} f(x) f(y) (y - x)(f(y) - f(x)) dy dx.
\]
But since \( f \) is strictly increasing, \( (y - x)(f(y) - f(x)) \geq 0 \) for all \( x, y \), with equality only if \( x = y \). We conclude that this final double integral is positive and thus that \( I_1 < I_2 \), as desired.