91Ó°ÊÓ

Consider the given question,

$$\begin{gathered} f\left(t\right)=800+3.1p\left(t\right)\mathit{,} \\ p\left(t\right)=\left(t\mathrm{-}\mathrm{90}\right)\left(\mathrm{195}\mathrm{-}t\right) \end{gathered}$$

Substitute the function $p\left(t\right)$in the function $f\left(t\right)$,

$$\begin{gathered} f\left(t\right)=800+3.1\left(t-90\right)\left(195-t\right) \\ =800+3.1\left(-t^2+195t+90t-17550\right) \\ =-3.1t^2+883.5t-53605 \end{gathered}$$

Integrate the function $f\left(t\right)$from $t_0$ to t,

${∫}_{t_0}^{t}f\left(t\right)dt=-{∫}_{t_0}^{t}3.1t^{2}dt+{∫}_{t_0}^{t}883.5tdt-{∫}_{t_0}^{t}53605dt$

For a function $f\left(x\right)$which is continuous on an interval $\left[a,b\right]$, the fundamental theorem of calculus states that ${∫}_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$, where Fis the antiderivative of the function f.

Using the fundamental theorem of calculus in the right hand side,

$$\begin{gathered} {∫}_{t_0}^{t}f\left(t\right)dt=-{∫}_{t_0}^{t}3.1t^{2}dt+{∫}_{t_0}^{t}883.5tdt-{∫}_{t_0}^{t}53605dt \\ {F}_{t_0}\left(t\right)=-3.1{{\left[\frac{t^{2+1}}{2+1}\right]}^t}_{t_0}+883.5{{\left[\frac{t^{1+1}}{2+1}\right]}^t}_{t_0}-53605{{\left[t\right]}^t}_{t_0} \\ {F}_{t_0}\left(t\right)=-\frac{3.1}{3}\left[t^3-{t_0}^3\right]+\frac{883.5}{3}\left[t^2-{t_0}^2\right]-53605\left[t-t_0\right] \\ {F}_{t_0}\left(t\right)=-1.03\left(t^3-{t_0}^3\right)+441.759\left(t^2-{t_0}^2\right)-53605\left(t-t_0\right) \end{gathered}$$

Multiply the $F_{t_0}\left(t\right)$by $86400$,

localid="1648832733606" $F_{t_0}\left(t\right)=86400\left[-1.03\left(t^3-{t_0}^3\right)+441.759\left(t^2-{t_0}^2\right)-53605\left(t-t_0\right)\right]......\left(i\right)$

Substitute $$\begin{gathered} t_0=90, \\ t=195 \end{gathered}$$in equation (i),

$$\begin{gathered} F_{90}\left(195\right)=86400\left[-1.03\left({195}^3-{90}^3\right)+441.759\left({195}^2-{90}^2\right)-53605\left(195-90\right)\right] \\ ≈60.8billioncubicfeet \end{gathered}$$

For a full year, $$\begin{gathered} t_0=1, \\ t=365 \end{gathered}$$.

The function $p\left(t\right)$is $0$ when t does not lie between $\left(90,195\right)$.

Therefore, the function $f\left(t\right)$is $800$.

Integrate the function $f\left(t\right)=800$from $t_0$ to t,

$$\begin{gathered} {∫}_{t_0}^{t}f\left(t\right)dt={∫}_{t_0}^{t}800dt \\ {F}_{t_0}\left(t\right)=800{{\left[t\right]}^t}_{t_0} \\ {F}_{t_0}\left(t\right)=800\left[t-t_0\right] \end{gathered}$$

Multiply $86400$by $F_{t_0}\left(t\right)$,

$$\begin{gathered} F_{t_0}\left(t\right)=86400\left(800\left[t-t_0\right]\right) \\ {F}_{t_0}\left(t\right)=69120000\left[t-t_0\right] \end{gathered}$$

Substitute $$\begin{gathered} t_0=1, \\ t=365 \end{gathered}$$in $F_{t_0}\left(t\right)$,

$$\begin{gathered} F_{t_0}=69120000\left[365-1\right] \\ =25159680000 \end{gathered}$$

Therefore, $25.16$billion cubic feet of water flowed in a year.

Then, to calculate the percentage,

role="math" localid="1648836291342" $$\begin{gathered} =\frac{25.16}{60.8}×100 \\ =41.38\% \end{gathered}$$