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