Why Does a Heavy Spring Extend Differently Under Its Own Weight?

[randommanonea]

A spring of mass $M$ is suspended from the ceiling of a room. Find the extension in the spring due to its own weight if it has a spring constant of value $k$.

I am getting answer as $\frac{Mg}{2k}$, but the answer given in back of the book is $\frac{Mg}{3k}$. What I did was :

Let the natural (in un-stretched position) length of the spring be $L$. Let us consider an element of length $dx$ at a distance $x$ from the bottom of the spring. Then the spring constant of this small spring is $k_x = \frac{L}{dx} k$. Tension in this spring is $T_x= \frac{x}{L} Mg$ (due to the weight of the spring below it, neglecting the weight of this small spring). Therefore, the extension in this small spring, $dl = \frac{T_x}{k_x} = \frac{Mg}{k L^{2}} x dx$. Therefore total extension, $l = \frac{Mg}{k L^{2}} \int_{0}^{L} x dx = \frac{Mg}{2k}$.

So, where I am wrong; or is the answer in the book wrong ?

 

[Doc Al]

I'd say that you are correct. What book are you using?