Work is fundamentally defined as force times distance. When we push a box with constant force F through distance d, the work done is simply W equals F times d. This straightforward formula works perfectly for constant forces. However, in many real situations, the force varies with position. For example, when stretching a spring, the force increases as we pull further. In such cases, our simple formula breaks down, and we need calculus to find the correct answer.
Let's examine specific examples of variable forces. Spring force follows Hooke's law, F equals k x, where force increases linearly with displacement. Gravitational force varies as one over r squared, decreasing rapidly with distance. When force varies, we cannot simply multiply by total distance. Instead, we must divide the path into small segments where force is approximately constant. Each small work element is F times delta x. The total work becomes the sum of all these small contributions. But as we make the segments infinitesimally small, this sum approaches an integral.
The key insight is transitioning from discrete approximation to continuous integration. We start with a Riemann sum: the limit as n approaches infinity of the sum of F of x i times delta x. This limit becomes the definite integral from a to b of F of x dx. The integral sign represents the continuous sum, F of x is our force function, dx represents an infinitesimal displacement, and the limits a and b define our integration bounds. Geometrically, this integral equals the area under the force-displacement curve, which represents the total work done.
Let's solve the classic spring problem step by step. We want to find the work required to stretch a spring from its natural length to extension x. Using Hooke's law, the force is F equals k x. We set up the integral: W equals the integral from 0 to x of k t dt. Solving this integral, we get k times t squared over 2, evaluated from 0 to x. This gives us one half k x squared. Geometrically, this represents the area of a triangle under the linear force curve, with base x and height k x, confirming our result of one half k x squared.