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

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SUMMARY

The discussion revolves around the integration of the function v^-2 from an initial velocity vi to a final velocity v, equating it to -3 times the integral of dt from 0 to t. The user attempts to solve the integral, arriving at -1/v + C = -3t, while the reference book presents -1/v + 1/vi = -3t. The discrepancy arises from the constant of integration, C, which should not be included when evaluating definite integrals. The correct evaluation leads to the conclusion that the integration from vi to v yields the term 1/vi.

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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

[tex]\left[ \frac{-1}{v} \right] _{v_i} ^{v} = \frac{-1}{v}- \left( -\frac{-1}{v_i}\right)[/tex]
 

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