The other answers here are great, but if they don't work for you, here is one other thing you might consider...

In one dimension, we find the center of mass with an average over the masses $$ \bar x=\frac {\sum m_i x_i}{\sum m_i} $$ We find the gyro-radius (one dimension still) with the root mean square over the masses $$ x_{gyro}=\sqrt{\frac {\sum m_i x_i^2}{\sum m_i}} $$ The root mean square tells us how spread out something is. And the more spread out the mass is, the harder it is to "rev up", to get spinning and once it is spinning, the harder it is to stop.

So we do need the square in the integral, since this gives us how spread out the mass is rather than the first power which gives us the center of mass.

The other answers here are great, but if they don't work for you, here is one other thing you might consider...

In one dimension, we find the center of mass with an average over the masses $$ \bar x=\frac {\sum m_i x_i}{\sum m_i} $$ We find the gyro-radius (one dimension still) with the root mean square over the masses $$ x_{gyro}=\sqrt{\frac {\sum m_i x_i^2}{\sum m_i}} $$ The root mean square tells us how spread out something is. And the more spread out the mass is, the harder it is to "rev up", to get spinning and once it is spinning, the harder it is to stop. The gyro-radius gives us the distance from the center all the mass could be relocated to and have the same rotational inertia.

So we do need the square in the integral, since this gives us how spread out the mass is rather than the first power which gives us the center of mass.