**Determination of work by variable force using integral calculus:**

Suppose, a variable force is acting along x-axis on a body. Magnitude of the force depends on distance travelled by the body i.e., F is the function of distance x. In the figure a graph has been plotted showing the variation of F(x) for different values of x.

Total displacement consists of equally divided n number of exceedingly small displacement ∆x, where ∆x is the displacement between x_{1} and x_{1} + ∆x. For this small change of displacement the magnitude of force is considered constant and the value of force is F_{1}. So work done by this force tin this small segment is,

**∆W _{1} = F_{1} ∆x**

Similarly, second segment is extended between x_{1} + ∆x to x_{1} + 2∆x and the small displacement is ∆x. For this small segment constant force is F_{2}. So work done by the force in the second segment is ∆W_{2} = F_{2} ∆x. So, total work done by the force F(x) to move the body from position x_{i} to x_{f} is given by,

W = ∆W_{1} + ∆W_{2} + ∆W_{3} + … … …+ ∆W_{N}

= F_{1}∆x + F_{2}∆x + F_{3}∆x + … … … + F_{N}∆x

= **^{N}∑_{K=1} F_{K} ∆x** … … … (1)

When ∆_{x} becomes smaller and smaller, that means the value of N becomes very large, calculated value of work done will be more correct. We can get correct value of work done by the force F(x) it ∆x approaches zero and N becomes infinity. Then correct value will be,

**W = lim _{∆x→0} ^{N}∑_{K=1} F_{K} ∆x** … … … (2)

But in the language of calculus, the quantity, **lim _{∆x→0} ^{N}∑_{K=1} F_{K} ∆x**; is written as:

^{xf}**∫ _{xi} F(x) dx;** which indicates the integration between x

_{i}and x

_{f}.

So, equation (2) becomes,

**W = ^{xf}∫_{xi} F(x) dx**

Numerically this quantity is the area subtended by the curved line of force and the displacement between x_{1} and x_{f}. So, by integration work and area can be determined.