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November 28, 2014

November 28, 2014

Posted by **Anonymous** on Friday, December 9, 2011 at 4:31am.

- calculus -
**Damon**, Friday, December 9, 2011 at 5:32amdisplacement = integral v dt

= -t^3/3 + 2 t^2 - 3 t

from -2 to +6

do the arithmetic

distance traveled

that means always use absolute value of any part of the integral

when is velocity negative and when is it positive?

0 = t^2 - 4 t + 3

0 = (t-3)(t-1)

velocity 0 at t = 1 and t = 3

so:

take absolute value of integral from

-2 to + 1

add to absolute value of integral from +1 to +3

add to absolute value of integral from +3 to + 6

- calculus -
**Anonymous**, Friday, December 9, 2011 at 1:15pmthank you

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