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equations and then with the help of the relevant calculus formula, it solves the final equation. This tool is so

efficient that the conversion and solving process goes almost simultaneously, means the time consume is

very less in getting the result.

This can be better clarified by taking individual examples of differentiation and integration. The examples are

as shown below:-

10 s.

Let us assume that the small displacement is denoted by dx

Therefore from above, dx = 5 cm

Also, it is given that this displacement is covered in time duration of 10 s.

This time duration is also very small just 10 s. So we can consider this time as a very small time period, so we

can denote it by dt.

Therefore from above, dt = 10 s.

Now we know that the velocity is the ratio of displacement travelled to the total time taken. And let ‘v’ is the

velocity of an object

V = dx / dt = 5/10 = ½ = 0.5 cm.

5 s.

Let us assume that the small displacement is denoted by dx

Therefore from above,dx = 20 cm

Also, it is given that this displacement is covered in time duration of5 s.

This time duration is also very small just 10 s. So we can consider this time as a very small time period, so we

can denote it by dt.

Therefore from above, dt = 5 s.

Now we know that the velocity is the ratio of displacement travelled to the total time taken. And let ‘v’ is the

velocity of an object’

V = dx/dt = 20/5 = 4 = 4 cm/s