I am having trouble in confirming that the volume of a sphere $V=(4\pi/3) R^3$ is equal to the sum of the volumes of a set of contiguous, nested spherical shells.

I start by obtaining the following expression for the volume $W\_x$ of a single spherical shell with outer radius $x$ and radial thickness $t$ based upon the difference in volumes of the outer sphere and inner sphere:-

$$W\_{x(t)} = (4\pi/3) x^3 - (4/3)\pi (x-t)^3 $$ 
$$W\_{x(t)} = (4\pi/3) ( x^3 - (x-t)^3) $$ 
$$W\_{x(t)} = (4\pi/3) ( x^3 - (x^3-3tx^2+3t^2x-t^3)) $$ 
$$W\_{x(t)} = (4\pi/3) ( 3tx^2 -3t^2x +t^3) $$.

This formula can be checked by setting $t=x=R$ giving:-  

$$V = W\_{R(R)} = (4\pi/3) ( 3R^3 -3R^3 +R^3) = (4\pi/3) R^3 $$ which is the volume $V$ of a sphere with radius $R$, as expected.

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Next I wish to obtain the volume $V$ of the sphere (radius $R$) by summing the volumes of a series of contiguous, nested spherical shells of common thickness $t$.
$$V\_R = \Sigma\_t^R W\_{x(t)} = (4\pi/3) \Sigma\_t^R ( 3tx^2 -3t^2x +t^3) 
$$
$$V\_R = 
4\pi 
\Sigma\_t^R \left[ tx^2 
-t^2x 
+(1/3)t^3 
\right].
$$
This formula has been checked by numerical testing.
Now I wish to go from the above **"discrete shell formula"** to an integral expression of the form:-
$$
V = \int \_0^R G(x) dx = (4 \pi /3)R^3
$$
I realize that setting the function $G(x) = 4 \pi x^2$ will give an anti-derivative function $F(x) = (4 \pi/3)x^3$. 

But, here is my question: **how do I get from the "discrete shell formula" to a suitable form of the integral formula involving $G(x)$?**

In particular I am unsure how to handle the "trailing terms" containing $t^2$ and $t^3$.

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**Update**

I have learned that it is not essential to obtain an integral formula. The exact formula $V=(4/3)\pi R^3$ can be obtained using power sums (see Dr.MV's answer (Method 1) and my own answer).

A.G. referred to the multiple thin shell approximation method for deriving sphere volume using the formula for sphere surface area. I have applied the power sum approach to that method in [this answer](https://math.stackexchange.com/questions/164/why-is-the-volume-of-a-sphere-frac43-pi-r3/1367398#1367398).