The work done by a variable force needs some consideration. It will also help you understand the physical aspects of integration.

By a 'variable force' I mean a force that is changing with some proper equation. For example

F = (10t^2 + 3t)n 1) (n is any vector)

F = 5x^2 i + 7y j 2) (i and j are unit vectors in x and y directions).

Equation 1 is a time dependent force (or function of time) and equation 2 is a position dependent force (or a function of position).

There are some forces that are variable but cannot be written down in the form of an equation. We won't deal with such forces here for the simple reason that such forces cannot be dealt with the techniques of calculus explained in this post.

Since the force is variable therefore the formula of work viz.

W =

fails here as this formula is valid only when

(Note Equation A can also be written as follows

By a 'variable force' I mean a force that is changing with some proper equation. For example

F = (10t^2 + 3t)n 1) (n is any vector)

F = 5x^2 i + 7y j 2) (i and j are unit vectors in x and y directions).

Equation 1 is a time dependent force (or function of time) and equation 2 is a position dependent force (or a function of position).

There are some forces that are variable but cannot be written down in the form of an equation. We won't deal with such forces here for the simple reason that such forces cannot be dealt with the techniques of calculus explained in this post.

Since the force is variable therefore the formula of work viz.

W =

**F.S**, A)fails here as this formula is valid only when

**F**is constant in the displacement**S**.(Note Equation A can also be written as follows

where theta is the angle between F and S.)

You can see in a somewhat different fashion why equation A fails for a variable force. Suppose the magnitude of total displacement (S) = 10 m.

Suppose the force at S = 1 m is 2 N. At S = 4 m is 3 N. At S = 6m is 5 N.

Even if you substitute 10 for S in equation A, what will you substitute in place of F? 2? 3? 5? Or the average of all the forces?

You see the helplessness of equation A for variable forces?

Now let's use calculus.

The entire path can be assumed to be made up of infinitely many small displacements and the small work dW in each small displacement can calculated by the formula

dw =

**F**.d

**S .**B)

Since I've made dS to be infinitely small therefore I no longer need to worry about the variation of F in this small interval. dS Is so so small that F is almost a constant in this small displacement and hence one can safely use equation B.

Now if we add all such elements (infinite in number)* together then we should get the total work. This is achieved by integrating equation B.

*Smaller the value of dS greater will be the number of terms. Since dS is infinitesimally small therefore there will be infinite terms in the series.

It can be proved mathematically that the integral of F.dS is equivalent to the sum of the required infinite series.

This summing up of infinite series was the physical basis of invention of integration. You may be surprised to know that integration was discovered before differentiation!

It can be proved mathematically that the integral of F.dS is equivalent to the sum of the required infinite series.

This summing up of infinite series was the physical basis of invention of integration. You may be surprised to know that integration was discovered before differentiation!