Need derivation of a formula related to determination of moments of a object about $x$-axis and $y$-axis I'm given a plate,its density $\rho$ and it is the region  bounded by the two curves  and  on the interval [a,b].I want to find the center of mass of the region below.

I’ll first need the mass of this plate.  The mass is,
$M=\rho (Area of plate)$=$\rho \int_{a}^bf(x)-g(x)dx$.
Next I’ll need the moments of the region.  There are two moments, denoted by Mx and My.  The moments measure the tendency of the region to rotate about the x and y-axis respectively.  The moments are given by,
$M_x=\rho\int_{a}^b\frac{1}{2}([f(x)]^2-[g(x)]^2)dx$ and $M_y=\rho\int_{a}^bx(f(x)-g(x))dx$
My question is that how can I derive the above two equations.Can anyone help me?
 A: Before giving you some guidance, a comment about terminology, to avoid misundertandings.
You are trying to evaluate the center of mass, which is defined by
$$x_{CM} = \frac{1}{M}\int x dm $$
and
$$y_{CM} = \frac{1}{M}\int y dm $$
These are weighted averages of the coordinates, the weight being the mass. Let us call the integrals $\mu_x$ and $\mu_y$, so $x_{cm}=\mu_x/M$ and $y_{cm}=\mu_y/M$.
Now,  $\mu_x$ and $\mu_y$ are the first order momenta of the mass distribution.
You are using the term "momenta" in a different way, which if I understood is motivated as follows. Let's suppose that your plate is subject to an uniform gravitational field in the $z$ direction. The torque of this force with respect to the origin will be given by
$$\vec{M} = \int (x \hat{x}+y\hat{y}) \times \left( g \hat{z} dm \right)$$
which gives, after evaluating the cross products,
$$M_x = M g \left( \frac{1}{M} \int y dm \right)$$
$$M_y = -M g \left( \frac{1}{M} \int x dm \right)$$
$$M_z = 0$$
Neglecting the minus sign the expression between braces are the "momenta" you define, which are connected to the usual ones in an obvious way. 
Whatever the name, we can write the integrals you need in an explicit way by noting that $dm = \rho dx dy$. This gives
$$\mu_x = \int_a^b dx \int_{g(x)}^{f(x)} x dx dy$$
$$\mu_y = \int_a^b dx \int_{g(x)}^{f(x)} y dx dy$$
and you can check that these gives the equations you quote. 
I can give you a more intuitive derivation. Divide your plate in infinitesimally thin vertical stripes. The area of the stripe at $x$ will be $dA=\left[f(x)-g(x) \right]dx$,
and its mass $dm = \rho dA$. Now you can evaluate the weighted sum of $x$ obtaining your expression for $M_y$. For $M_x$ you need a different strategy: find the $y$ of the center of mass of each stripe, which will be at its middle $\left[f(x)+g(x) \right]/2$. Next evaluate the weighted sum of this, weighting with the mass of the stripe defined previously, and you're done. 
