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What is integreation and differentiation and how to apply it inn follwing problem.If x=6t2 +5t then find velocity

which class are you in ? please inform. do you need explanation for integration? do you understand the concept explanation i gave for derivative and velocity.



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Let us take a small time interval Δt like 0.001 sec..
  At time t, distance x is    x1  =  6 t² + 5 t
at  t + Δt ie.,  t+0.001 sec,  x2 = 6 (t +Δt)² + 5 t = 6 t² + 12 t Δt + (Δt)² + 5 t + 5 Δt
Distance travelled in Δt duration is : x2 - x1 =  12 t Δt + (Δt)² + 5 Δt
   This is denoted by Δx.  So Δx = x2 - x1
Now find  x2 - x1 / Δt  = Δx / Δt  =  12 t + Δt + 5
Velocity is defined as    Limiting value of  Δx / Δt  as Δt approaches 0
so velocity =  limit  Δt -> 0,  (12 t + Δt + 5 )  = 12 t + 5
That is you take Δt = 0.00001 sec, find Δx.  take Δt = 0.000000001 sec, find Δx. Then Δx / Δt  value approaches a fixed value. That value is 12 t + 5.
This is also called derivative of x with respect to t.    Limit as Δt -->0  the value of Δx / Δt.  it is denoted by  d x / d t.

the reverse process of derivative is called integration.

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do you understand the derivative concept now clearly? which class are you? do you want a simple explanation of integration? is it in your syllabus?
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