More complex distributed loads can be reduced to two distributed loads of uniform shape, or the integral of the distributed load function can be taken to obtain the area under \( w(x) \), which is \( F_R \).

$$ F_R = \int_{0}^{L} w(x) \,dx \ $$

The location of \( F_R \) on the beam can be found by solving for the moment \( M_R \) about a point. In this case we will take the moment about the point \( x \)=0 and setting it equivalent to the moment produced by the resultant force \( F_R \) at \( x \)=0.

$$ M_R = \int_{0}^{L} x*w(x) \,dx = \bar{x}*F_R\ $$

Now we can solve for the center of mass of the distributed load, \( \bar{x} \), using

$$ \bar{x} = \dfrac{M_R}{F_R}\ $$