Vì 0<z≤10<z\le10<z≤1 nên 1+z/2≤3/21+z/2\le3/21+z/2≤3/2. Do đó V≤z(1−z/2)(1+z/2)/6≤z/4−z2/8=1/8−(z−1)2/8V\le z(1-z/2)(1+z/2)/6\le z/4-z^2/8=1/8-(z-1)^2/8V≤z(1−z/2)(1+z/2)/6≤z/4−z2/8=1/8−(z−1)2/8.