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I am struggling to get the correct answer for the question:

What is the magnitude of the displacement of a particle moving in the $x$-$y$ plane with the velocity vector given by $v(t) = (e^{\sin t}, 5t^2)$ for time $t \ge 0$ between time $t=1$ and $t=2$.

I attempted to take the integral of the speed function $\int_1^{2} (\sqrt{e^{2\sin t}+25t^4}) \, \mathrm{d}t \\$ since the velocity of the particle is always positive on the interval and got the answer $11.992$. However, apparently the correct answer is $11.954$, but I can't find a way to come about that answer.

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  • $\begingroup$ What you need is $ \left\| \int_1^2 v \mathrm{d}t\right\| $, not $ \int_1^2\left\| v \right\|\mathrm{d}t$. $\endgroup$ Mar 16, 2019 at 23:57
  • $\begingroup$ @coreyman317 how would I go about evaluating that integral since $v(t)$ is a parametric function? $\endgroup$
    – Ludwig
    Mar 17, 2019 at 4:11
  • $\begingroup$ @ChargeShivers how would I go about evaluating that integral since v(t) is a parametric function? $\endgroup$
    – Ludwig
    Mar 17, 2019 at 4:27
  • $\begingroup$ Let $v = (v_x, v_y)$, then $\int v \mathrm{d}t = \left( \int v_x \mathrm{d}t, \int v_y\mathrm{d}t\right)$, and so $\left\| \int v \mathrm{d}t \right\| = \sqrt{ \left( \int v_x \mathrm{d}t \right)^2 + \left( \int v_y \mathrm{d}t \right)^2 }$. $\endgroup$ Mar 18, 2019 at 3:39

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