Question 1083176
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Position is the antiderivative of velocity, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p(t)\ =\ \int\,v(t)\,dt\ =\ 3\int\,t^2\,dt\ =\ 3\,\cdot\,\frac{1}{3}t^3\ +\ C\ =\ t^3\ +\ C]


where the constant represents the initial position.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(2)\ =\ 8\ +\ C]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(3)\ =\ 27\ +\ C]


Distance over the interval is *[tex \Large p(3)\ -\ p(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (27\ +\ C)\ -\ (8\ +\ C)\ =\ 19\ \text{units}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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