Moment of inertia help Okay so lets say I have a lever with 2 objects on it, object one and object 2.
Object 1 has a mass of $m_1$ and is $r_1$ away from the fulcrum; object 2 has a mass of $m_2$ and is $r_2$ away from the fulcrum.
The $\tau_{net} = g m_1 * r_1 + g m_2 * r_2 = I \alpha$
$I = m_1 * (r_1)^2 + m_2 * (r_2)^2 + I_{Lever} $
Now what is confusing me is how I would calculate the moment of inertia for the lever; I will just take a guess and say that $I_{\rm Lever}$ would be:
calling $L$ the length of lever and $r_1$ the length from the beginning of the lever to the fulcrum, saying that object 1 is on the tip of the lever, $M$ is the mass of the lever, and $s$ being the distance from the fulcrum
$$
I_{\rm Lever} = \int_{r_1-L}^{L-r_1} s^2 dM.
$$
Knowing that $dM = M*ds/L $ in this case, we have
$$
I_{\rm Lever} = \frac{M}{L}\int_{r_1-L}^{L-r_!} s^2 ds,
$$
which is then
$$
I_{\rm Lever} = \left(\frac{2M}{3L} \right)(r_1-L)^3 
$$
Would this be right? Also if possible direct me to a place where I could learn more about general cases of these kind of problems.
 A: Your maths is almost correct. I think you should treat the lever as two separate rods which each have a pivot at one end.
If the mass per unit length is $\sigma = M/L$, then the following applies to the two rods of length $r_1$ and $r_2$ (with $r_2=L-r_1$), where I split the rod up into infinitesimal chunks of length $dr$, mass $dm=\sigma dr$ and moment of inertia $dI = \sigma r^2 dr$:
$$ I_{\rm lever} = \int^{r_1}_{0} \sigma r^2 dr  + \int^{r_2}_{0} \sigma r^2 dr $$
In each case, $\sigma$ is constant, so the integrals come out as
$$ I = \frac{\sigma}{3}(r_1^{3} + r_{2}^3) $$
If you wish to express this solely in terms of $r_1$ and $L$:
$$I = \frac{M}{3L}(r_1^{3} + (L-r_1)^3) $$
So, the general approach you were thinking of was ok, but I think you got confused over the limits on the integrals and ended up with a negative moment of inertia! The moments of inertia each side of the fulcrum must add. For working out general moments of inertia it is almost always the case that you split the object up into small elements of similar mass density and whose moments of inertia you can calculate, and then integrate over all these small elements. 
A shortcut is to know that the moment of inertia of a rod about its end is $ml^2/3$. In this case you have "two rods" of mass $r_1 M/L$ and $r_2 M/L$ and lengths $r_1$ and $r_2$. Hence
$$I_{\rm lever} = \frac{M}{3L}(r_1^3 + r_2^3),$$
which is the same result.
