When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude $$F = ax + bx^2$$ where a and b are constants. The work done in stretching the unstretched rubber-band by L is:

We begin by recalling the basic definition of mechanical work. The work $$W$$ done in stretching (or compressing) a body from an initial extension $$x = 0$$ to a final extension $$x = L$$ against a variable restoring force $$F(x)$$ is given by the line integral

$$ W \;=\; \int_{0}^{L} F(x)\,dx . $$

In the present problem the magnitude of the restoring force supplied by the rubber-band is described by the expression

$$ F(x) \;=\; a\,x + b\,x^{2}, $$

where $$a$$ and $$b$$ are constants. Substituting this force function into the integral for work, we get

$$ W = \int_{0}^{L} \left(a\,x + b\,x^{2}\right) dx. $$

Now we evaluate the integral term by term. First, for the linear term $$a\,x$$ we have

$$ \int a\,x \;dx = a \int x \;dx = a \left(\frac{x^{2}}{2}\right) = \frac{a x^{2}}{2}. $$

Second, for the quadratic term $$b\,x^{2}$$ we have

$$ \int b\,x^{2} \;dx = b \int x^{2} \;dx = b \left(\frac{x^{3}}{3}\right) = \frac{b x^{3}}{3}. $$

Combining these two antiderivatives, the total work integral becomes

$$ W = \left[\,\frac{a x^{2}}{2} + \frac{b x^{3}}{3}\,\right]_{0}^{L}. $$

We now apply the limits. At the upper limit $$x = L$$ the expression is $$\dfrac{a L^{2}}{2} + \dfrac{b L^{3}}{3}$$, and at the lower limit $$x = 0$$ the expression is clearly zero. Therefore,

$$ W = \left(\frac{a L^{2}}{2} + \frac{b L^{3}}{3}\right) - 0 = \frac{a L^{2}}{2} + \frac{b L^{3}}{3}. $$

This result matches Option C in the given list.

Hence, the correct answer is Option C.