Second moment of area

I can easily explain the stress-moment relationship in beams. I am sure you are familiar with it, and the area moment $I$ comes out naturally.

$$ sigma_{max} = frac{M y_{max}}{I} $$

Take a section of the beam with internal moment $M$. This moment has to be supported by the stress distribution internally. For straight beams, there is a neutral axis in the middle of the beam, and the stress values vary linearly from the distance $y$ to the neutral axis $$ sigma(y) = k y $$ At the limit $sigma_{max} = k y_{max}$ or $$k = frac{sigma_{max}}{y_{max}}$$

Take a cross section and integrate slices along the height $y =y_{min} ldots y_{max}$ each with a thickness $b(y)$.

The area if the section is thus

$$ A = int {rm d} A = int limits_{y_{min}}^{y_{max}} b(y),{rm d}y $$

The moment is found by the product if the infinitesimal tension ${rm d}F = sigma(y) ,{rm d}A$ and the height $y$.

$$ M =int y {rm d}F = int y sigma {rm d} A = int limits_{y_{min}}^{y_{max}}y sigma b(y),{rm d}y = int limits_{y_{min}}^{y_{max}}k y^2 b(y),{rm d}y $$

or

$$ k = frac{M}{int limits_{y_{min}}^{y_{max}} y^2 b(y),{rm d}y} $$

The denominator we call the area moment because it contains any information from the geometry of the problem

$$ boxed{I = int limits_{y_{min}}^{y_{max}} y^2 b(y),{rm d}y} $$

To answer your question on why the $y^2$ term in the integral, one of the $y$ comes from the linearity of the stress distribution $sigma = k y$ and the other $y$ from the moment arm calculation ${rm d}M = y {rm d}F$.