Essential University Physics: Volume 1 (4th Edition) Clone

$I=(\frac{Ma^2}{12})$
We know that the moment of inertia is $\frac{M}{a}x^2$. The book asks us to use integration, so we take the integral to obtain: $I = \int_{-a/2}^{a/2}\frac{M}{a}x^2 dx \\= \frac{M}{a}((\frac{a/2}{3})^3-(\frac{-a/2}{3})^3\\ = \frac{M}{a}(\frac{2a^3}{24})\\ =(\frac{Ma^2}{12})$