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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=

x =

x = [(3t

x = [(3t

x = [(3t

x= [(3t

x [(3t

However, as is evident (even to me) that this approach + method is wrong. The correct solution is:

x=

x =

x = (3t

x= [(3t

x= (3t

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.

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=

^{t}_{0}∫vdtx =

^{t}_{0}∫(3t+2)dtx = [(3t

^{2})/2 + 2t] ⋅ (t+C) ('t+C' is the integral of dt)x = [(3t

^{3})/2 +2t^{2}] ⋅ (t+C)x = [(3t

^{2}C)/2 + (3t^{3})/2]+ [(2tC+2t^{2})]^{t}_{0}|x= [(3t

^{2}C)/2 + (3t^{3})/2]- 0x [(3t

^{2}C)/2 + (3t^{3})/2] metresHowever, as is evident (even to me) that this approach + method is wrong. The correct solution is:

x=

^{t}_{0}∫vdtx =

^{t}_{0}∫(3t+2)dtx = (3t

^{2})/2 + 2t^{t}_{0}|x= [(3t

^{2})/2 + 2t]- 0x= (3t

^{2})/2 + 2t metresPlease 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.