91Ó°ÊÓ

A mass hanging at the end of a spring oscillates up and down from its equilibrium position with velocity

$v\left(t\right)=3\cos \left(\frac{3t}{\sqrt{2}}\right)−3\sqrt{2}\sin \left(\frac{3t}{\sqrt{2}}\right)$

centimetres per second, as shown in the figure. The mass is at its equilibrium at $t=0$. Use definite integrals to determine whether the mass will be above or below its equilibrium position at times$t=4\mathrm{and}t=5$.

$$\begin{gathered} {∫}_4^{5}v\left(t\right)dt={∫}_4^5\left[3\cos \left(\frac{3t}{\sqrt{2}}\right)−3\sqrt{2}\sin \left(\frac{3t}{\sqrt{2}}\right)\right]dt \\ {∫}_4^{5}v\left(t\right)dt=3{∫}_4^5\cos \left(\frac{3t}{\sqrt{2}}\right)dt-3\sqrt{2}{∫}_4^5\sin \left(\frac{3t}{\sqrt{2}}\right)dt \end{gathered}$$

Let

$$\begin{gathered} u=\frac{3t}{\sqrt{2}} \\ \frac{du}{dt}=\frac{3}{\sqrt{2}} \\ du==\frac{3}{\sqrt{2}}dt \\ \frac{\sqrt{2}}{3}du=dt \end{gathered}$$

$$\begin{gathered} {∫}_4^{5}v\left(t\right)dt=3·\frac{\sqrt{2}}{3}{∫}_4^5\cos udu-3\sqrt{2}·\frac{\sqrt{2}}{3}{∫}_4^5\sin udu \\ {∫}_4^{5}v\left(t\right)dt=\sqrt{2}{∫}_4^5\cos udu-2{∫}_4^5\sin udu \\ {∫}_4^{5}v\left(t\right)dt=\sqrt{2}{\left[\sin u\right]}_4^5-2{[-\cos u]}_4^5 \\ {∫}_4^{5}v\left(t\right)dt=\sqrt{2}{\left[\sin \left(\frac{3t}{\sqrt{2}}\right)\right]}_4^5+2{\left[\cos \left(\frac{3t}{\sqrt{2}}\right)\right]}_4^5 \end{gathered}$$

$$\begin{gathered} {∫}_4^{5}v\left(t\right)dt=\sqrt{2}{\left[\sin \left(\frac{3t}{\sqrt{2}}\right)\right]}_4^5+2{\left[\cos \left(\frac{3t}{\sqrt{2}}\right)\right]}_4^5 \\ {∫}_4^{5}v\left(t\right)dt=\sqrt{2}\left[\sin \left(\frac{3×5}{\sqrt{2}}\right)-\sin \left(\frac{3×4}{\sqrt{2}}\right)\right]+2\left[\cos \left(\frac{3×5}{\sqrt{2}}\right)-\cos \left(\frac{3×4}{\sqrt{2}}\right)\right] \\ {∫}_4^{5}v\left(t\right)dt=\sqrt{2}\left[\sin \left(\frac{15}{\sqrt{2}}\right)-\sin \left(\frac{12}{\sqrt{2}}\right)\right]+2\left[\cos \left(\frac{15}{\sqrt{2}}\right)-\cos \left(\frac{12}{\sqrt{2}}\right)\right] \\ {∫}_4^{5}v\left(t\right)dt=\sqrt{2}\left[\sin \left(10.64\right)-\sin \left(8.51\right)\right]+2\left[\cos \left(10.64\right)-\cos \left(8.51\right)\right] \\ {∫}_4^{5}v\left(t\right)dt=\sqrt{2}\left[-0.94-0.79\right]+2\left[-0.35-(-0.61)\right] \\ {∫}_4^{5}v\left(t\right)dt=\sqrt{2}\left[-0.94-0.79\right]+2\left[-0.35+0.61\right] \\ {∫}_4^{5}v\left(t\right)dt=1.41×(-1.73)+2×0.26 \\ {∫}_4^{5}v\left(t\right)dt=-2.44+0.52 \\ {∫}_4^{5}v\left(t\right)dt=-1.92 \end{gathered}$$