Having a problem in steps while solving integrals

  • #1
prakhargupta3301
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Member advised (again!) to use the formatting template for all homework help requests
Okay. So I'm new to calculus. And this is the first time I'm solving a physics problem using integration.
I understood that ∫dt (or ∫1dt) will be equal to just 't+C.' (Just like f ' (t+C) = 1).
Though, that's not the problem. The problem is when I apply it this way:

Question: The expression for velocity of a body is given as 3t+2. Displacement (x) is in metres and time (t) in seconds. Find the displacement (x) in time 't' using integration. (Note: take starting time to be 0 seconds)

My attempt: x= t0∫vdt
x = t0∫(3t+2)dt
x = [(3t2)/2 + 2t] ⋅ (t+C) ('t+C' is the integral of dt)
x = [(3t3)/2 +2t2] ⋅ (t+C)
x = [(3t2C)/2 + (3t3)/2]+ [(2tC+2t2)] t0|
x= [(3t2C)/2 + (3t3)/2]- 0
x [(3t2C)/2 + (3t3)/2] metres

However, as is evident (even to me) that this approach + method is wrong. The correct solution is:
x= t0∫vdt
x = t0∫(3t+2)dt
x = (3t2)/2 + 2t t0|
x= [(3t2)/2 + 2t]- 0
x= (3t2)/2 + 2t metres

Please tell me, why in step 3 of correct solution, integral of dt was omitted? (Please refer to same step of my wrong calculation to know the error in my approach)

Also, please enlighten me on the correct way.
Thank you for reading.
 

Answers and Replies

  • #2
BvU
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why in step 3 of correct solution, integral of dt was omitted?
It was not omitted, but executed.

The important thing for you is now to understand that in your own working, the factor (t+C) is not correct (too much).

Perhaps it helps if you look at ##\displaystyle \int_{t_0}^{t_1} 2 \,dt = 2 t |_{t_0}^{t_1} = 2( t_1 - t_0 )##
In other words: the 2t in [(3t2)/2 + 2t] is a primitive of 2 dt
(and the 3t2)/2 is the primitive of 3t)
 
  • #3
haruspex
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't+C' is the integral of dt)
There is no such thing as an integral of "dt".
∫dt = ∫1.dt is the integral of 1 with respect to t.
Likewise, ∫(3t+2).dt is the integral of 3t+2 with respect to t.

Secondly, you only need a constant of integration in indefinite integrals. It represents the fact that you do not know both bounds. Where you do know the bounds, the constant would be the same at each end of the range, and so cancels out:
ab2x.dx=[x2+C]ab=(b2+C)-(a2+C).
 
  • #4
Dr Dr news
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∫ dt = t + C or ∫ dt = Δt
 
  • #5
prakhargupta3301
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Thank you BvU, Haruspex and Dr Dr news, because if you my concept is clear.
 
  • #6
BvU
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:smile:
 

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