Let f(z) = ∫_{0}^{z} √(x^{4} + (z - z^{2})^{2}) dx. Find the maximum value of f(z) in the range 0 ≤ z ≤ 1.

**Solution**

Moderately hard.

Answer: 1/3.

The integrand is strictly positive for 0 < x ≤ 1 and 0 ≤ z ≤ 1, so f(z) > 0 for 0 < z ≤ 1. However, it is not at all clear how to get the integral in closed form except for z = 0 or 1. So we have a strong suspicion that the maximum must be the higher of the two values we can actually calculate: f(1) = 1/3! To prove this we need to show that f '(z) > 0 for z > 0.

f '(z) = √(z^{4} + z^{2}(1 - z)^{2}) + z(1 - z)(1 - 2z) ∫_{0}^{z} (x^{4} + z^{2}(1 - z)^{2})^{-1/2} dx. The integrand is positive, so clearly f '(z) > 0 for 0 < z ≤ 1/2. The integrand is less than 1/√(z^{2}(1 - z)^{2}) = 1/z 1/(1 - z), so the integral is less than 1/(1 - z), and for z > 1/2, f '(z) > √(z^{4} + z^{2}(1 - z)^{2}) - z(2z - 1). But, squaring, we see that z(2z - 1) < √(z^{4} + z^{2}(1 - z)^{2}) for 1/2 < z < 1, which completes the proof.

© John Scholes

jscholes@kalva.demon.co.uk

21 Sep 1999