A simple Note to Understand Integration in Motion

Suppose we have a function and we need to find the area of it. For that, integration comes into place. We can explore integration in mathematics deeply, but here, we will use it for the derivation of equations and for solving numericals.  

In general, we have three equations of motion as we all know. And to derive these three, we can easily use integration methods. Integration in motion helps the topic motion to simplify the toughness of the topic. By using it, we can solve problems easily and more importantly, there are many problems that can be solved by only using integration. 

Integration is basically of two types. The first is indefinite integration which we can define as any area that does not need bounds and its output is always a variable. It is not used in physics usually. On the other hand, the other type of integration is definite integration, which is widely used in the physics world. It needs bounds for calculation and gives a fixed number. 

Suppose we have three equations of motion such as

v = u+at                  ……….(i)

s = ut + ½ (at2)           …….(ii)

v2-u2= 2            as…………(iii)

We are going to derive these equations one by one by using integration techniques.

Equation 1: v=u +at

This equation gives us the relation between velocity, acceleration, and time.

We know, 

a= dv / dt,

or,  dv= a.dt

By integrating both sides (left side from initial velocity u to final velocity v and right side time from 0 to t);

i.e, uv dv  = 0t a.dt ,

or, uv dv =  a.0t dt,   (since uv x.dv = xv + c)

This gives, v-u = at +c,

or,v= at + u+ c,

Since u and c are both constant, let us combine both and take constant as u overall,

Ao,  v=u+at..

Equation 2: s= ut + ½ (at2)

This equation tells us about the derivative of displacement s with respect to time t as the velocity v.

We know, 

ds/dt = v,

or,ds= v.dt,

From the previous equation, we will substitute v=u+at

Then, ds= (u+at) . dt,

Integrating both sides (left side from 0 to s and right side from 0 to t),

0s ds = 0t (u+at)dt,

or, 0s ds =  0t u.dt + 0t at.dt   ( we know 0s s.ds= ½ s2),

s= ut + ½ at2

Equation 3: v2-u2= 2as

This equation tells that the acceleration is equal to velocity times the rate at which it changes with respect to displacement.

As we know,

a= dv/dt,

On multiplying and dividing by ds in RHS,

I.e, a= (dv/ ds). ( ds/dt)

We know ds/dt= v.. 

So,

a= v. dv/ds

Or, v. dv = a. ds

By integrating both sides where velocity varies from initial velocity u to final velocity v and displacement varies from 0 to s

i.e,uv v. dv = 0t a . ds

(v2/2)vu = a . (s)s0 ,

or, (v2 – u2)/2 = a. s ,

So,   (v2 – u2) = 2 a. s.

Some Standard Formulas of Integration

1. dx = x + c;

2. x dx = x2/2 + c;

3. xn dx =( xn+1/ n+1) + c;

4. sinx dx = -cosx + c;

5. cosx dx = sinx + c;

6. tanx dx = ln |cosx| + c;

7. cotx dx = ln |sinx| + c;

8. secx dx = ln | secx + tanx + c;

9.   cosec x dx = ln | cosec x – cotx + c;

10. f(x) . g(x) dx =  f(x). g(x)dx – { d/dx(f(x)) g(x)} dx+ c;

11. ln x dx = x ln x-x + c;

12. ex dx = ex+ c;

13. ax dx =  ax / log a + c;

14. 1/x dx = ln x  + c;

Conclusion

Integration is the most important tool for solving numericals as well as proving theorems. For the most part, they are just the reverse process of differentiation. When we are unable to use equations for variable acceleration, with the help of integration, we can simply solve problems. We can use integration calculators for larger calculations.