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March 1, 2015

Posted by **Rachel** on Sunday, February 27, 2011 at 12:46am.

z(t)=4tsin(t^2), [0,π^1/2]

Any input?

- Calculus -
**MathMate**, Sunday, February 27, 2011 at 8:52amLet

u=t^2

du = 2tdt

∫z(t)dt

=∫4tsin(t^2)dt from 0 to sqrt(π)

=∫2sin(u)du from 0 to π

=-2cos(u) from 0 to π

=-2[-1 - (1)]

=4

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