It started out as a snowy week - it snowed for an entire dayl Then it warmed up for a day, snowed for two days straight, and has been dry since. An equatión that equation: Scos(2t) 0sts S(t) = 57 23.92. A graph of the snowfall rate is shown below Note 4. S(1)

The given graph is shown below. 

From the graph, it can be seen that in the interval $\left[0,-\frac{3}{2}\right] , S\left(t\right)$ is decreasing , in$\left[-\frac{3}{2},3\right] , S\left(t\right)$ is increasing and $\left[3,4\right]$ it is decreasing again. For $t\ge 4,S\left(t\right)=0$. It implies that between ay 0 and day$\frac{3}{2}$the snowfall rate is decreasing and from day $\frac{3}{2}$ to day 3, this rate is increasing.  From day 3, it is again decreasing and from day 4 and onwards, this rate becomes zero. 

The given model representing the snowfall rate is given by 

$$\begin{gathered} S\left(t\right)=\left\{\begin{array}{l}\cos 2t , 0\le t\le \frac{5\pi }{4}\\ 0 , t\ge \frac{5\pi }{4}\end{array}\right. \\ \frac{5\pi }{4}\approx 3.92 \end{gathered}$$

Hence, 

$\begin{array}{rcl}{\int }_0^{4}S\left(t\right)dt& =& {\int }_0^{\frac{5\pi }{4}}S\left(t\right)dt+{\int }_{\frac{5\pi }{4}}^{4}S\left(t\right)dt \\ & =& {\int }_0^{\frac{5\pi }{4}}\cos 2tdt\\ & =& \frac{1}{2}\sin \frac{5\pi }{2}\\ & =& \frac{1}{2} \left[as \sin \frac{5\pi }{2}=1\right]\end{array}$