In the main problem, we were trying to find a function, \(f(x)\), such that \[\int_{-1}^2 f(x)\,dx=1\]

Abi says, \[\int_{-1}^2 \frac{1}{x^2}\,dx = \left[\frac{x^{-1}}{-1}\right]_{-1}^2 = \frac{1/2}{-1} - \frac{-1}{-1} = -\frac{3}{2}\] so \(f(x)=-\dfrac{2}{3x^2}\) would satisfy the condition.

Ben says, \[\int_{-1}^2 \frac{1}{x^3}\,dx = \left[\frac{x^{-2}}{-2}\right]_{-1}^2 = \frac{1/4}{-2} - \frac{1}{-2} = \frac{3}{8}\] so \(f(x)=\dfrac{8}{3x^3}\) would satisfy the condition.

Charlie says, \[\int_{-1}^2 \frac{1}{x^{1/3}}\,dx = \left[\frac{3}{2}x^{2/3}\right]_{-1}^2 = \frac{3}{2} (\sqrt[3]4 -1)\] so \(f(x)=\dfrac{2}{3(\sqrt[3]4 -1)}\dfrac{1}{x^{1/3}}\) would satisfy the condition.

Are any of them right?