Applied calc question 2 and 3 

Question 2. In this problem the rate of change of the volume ($\frac{dV}{dt}$) of the water is given and we want to compute the volume ($V$). We can do so by integrating. Indeed 
(a)
\[\frac{dV}{dt}=2+\sin (\frac{\pi t}{180})\]
\[\Rightarrow V(t)=\int 2+\sin (\frac{\pi t}{180})dt=2t-\frac{180}{\pi}\cos (\frac{\pi t}{180})+C\]
\[(1)   \Rightarrow V(t)=2t-\frac{180}{\pi}\cos (\frac{\pi t}{180})+C.\]
Since in the begining of the year there were $200$ million litters, substituting in equation (1) we get 
\[V(0)=200=2 \times 0-\frac{180}{\pi}\cos (\frac{\pi \times 0}{180})+C=-\frac{180}{\pi}+C\]
\[\Rightarrow C=\frac{180}{\pi}+200.\]
Substituting in (1) again, we have
\[V(t)=2t-\frac{180}{\pi}\cos (\frac{\pi t}{180})+\frac{180}{\pi}+200.\]
(b) at the end of April (begining of May) we have $t=5$ so we have
\[V(5)=2\times 5-\frac{180}{\pi}\cos (\frac{\pi \times 5}{180})+\frac{180}{\pi}+200=210.2180    \text{million litters}.\]

Question 3. We simply need to integrate the weight function $w$ from $0$ to $6$:

\[\int_0^{6}900-\frac{100}{9}x^2 dx=(900x-\frac{100}{9}\frac{x^3}{3})\big|_0^{6}\]
\[=900\times6 -\frac{100}{9}\times\frac{6^3}{3}=4200   N. \]