Let R be the rectangle 0 ≤ x ≤ a, 0 ≤ y, ≤ b. Evaluate ∫R ef(x, y) dx dy, where f(x, y) = max(b2x2, a2y2).
Divide the rectangle into two by the diagonal from (0, 0) to (a, b). The required value is evidently twice the integral over the lower half of eg(x), where g(x) = b2x2. But integrating over y just gives a factor bx/a, and the integration over x is then trivial to give (ek - 1)/(2ab), where k = a2b2. So the required value is twice this or (ek - 1)/(ab).
50th Putnam 1989
© John Scholes
14 Aug 2002