Integrating v^-2 and Comparing Solutions for vi to v and t to 0

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The integral from vi to v of v^-2dv results in -1/v + 1/vi, not just a constant C, due to the evaluation of the definite integral. The comparison with the equation -3t indicates that the integration limits must be applied correctly. The confusion arises from the misunderstanding of the constant of integration in definite integrals. The correct interpretation shows that the book's solution includes the term 1/vi, which accounts for the initial condition at vi. Understanding the limits of integration clarifies the discrepancy in results.
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Homework Statement


The integral from v=vi to v of v^-2dv = -3 integral from t=0 to t of dt


Homework Equations





The Attempt at a Solution


I am getting -1/v + C=-3t and the book is getting -1/v+1/vi=-3t. Not quite sure where they are getting 1/vi as I am getting C.
 
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atypical said:

The Attempt at a Solution


I am getting -1/v + C=-3t and the book is getting -1/v+1/vi=-3t. Not quite sure where they are getting 1/vi as I am getting C.

Your integral is from vi to v, so you can't really get C.

your integral on the left works out as

\left[ \frac{-1}{v} \right] _{v_i} ^{v} = \frac{-1}{v}- \left( -\frac{-1}{v_i}\right)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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