K is the force constant of a spring. The work done in increasing its extension from l₁ to l₂ will be

The correct option is K(l₂²-l₁²)/2.

When a spring of force constant K is extended by a distance x, the restoring force acting on it is given by Hooke's Law:
$F=Kx$

The small amount of work done, dW, in increasing the extension of the spring by an infinitesimal distance, dx, is given by:
$dW=Fdx=Kxdx$

To find the total work done, W, in increasing the extension of the spring from l₁ to l₂, we integrate the expression for dW within the limits of integration from l₁ to l₂:
$W={\int }_{l_1}^{l_2}Kxdx$

Since the force constant K is a constant, we can pull it outside the integral:
$W=K{\int }_{l_1}^{l_2}xdx$

Using the basic power rule of integration, we get:
$W=K{\left[}_{\frac{x^2}{2}}^{\right]}$

Applying the upper and lower limits of integration:
$W=K\left(\frac{{l_2}^2}{2}-\frac{{l_1}^2}{2}\right)$

Simplifying the expression gives us the final formula for the work done:
$W=\frac{K({l_2}^2-{l_1}^2)}{2}$