Derive the work-energy theorem for a variable force exerted on a body in one dimension.

(N/A) Consider a body of mass $m$ moving in one dimension along the $X$-axis under the influence of a variable force $F(x)$. The kinetic energy $K$ of the body is given by $K = \frac{1}{2}mv^2$.
Differentiating $K$ with respect to time $t$:
$\frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2}mv^2 \right) = \frac{1}{2}m \cdot 2v \cdot \frac{dv}{dt} = mv \cdot a$
Using Newton's second law,$F = ma$,we substitute $ma$ with $F$:
$\frac{dK}{dt} = Fv$
Since velocity $v = \frac{dx}{dt}$,we have:
$\frac{dK}{dt} = F \frac{dx}{dt}$
Multiplying both sides by $dt$:
$dK = F dx$
Integrating both sides from the initial position $x_i$ (with kinetic energy $K_i$) to the final position $x_f$ (with kinetic energy $K_f$):
$\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F(x) dx$
$K_f - K_i = \int_{x_i}^{x_f} F(x) dx$
Since the work done $W$ by a variable force is defined as $W = \int_{x_i}^{x_f} F(x) dx$,we get:
$W = K_f - K_i = \Delta K$
This is the work-energy theorem for a variable force. It states that the work done by the net force on a body equals the change in its kinetic energy.