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The given table is

Consider an integral function fon [a,b]and a positive even integer n.

Define $鈭�x=\frac{b-a}{n}and{x}_k=a+k鈭�x$. Then the integral is approximated using $\frac{n}{2}$parabola topped rectangles.

Take number of days in an year is 360. Here, we have, $a=0,b=360,n=6$

$$\begin{gathered} 鈭�x=\frac{b-a}{n} \\ =\frac{360-0}{6} \\ =60 \end{gathered}$$

Take $x_0=0,x_1=60,....x_6=360\mathrm{and}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}f\left(x_k\right),0鈮�k鈮�6$as the rate of water flow for the corresponding $x_k$value.

Approximate the annual water flow using simpson's rule as follows,

$$\begin{gathered} SIMP\left(6\right)=\left(f\left(x_0\right)+4\left(f\left(x_1\right)+f\left(x_3\right)+f\left(x_5\right)\right)+2\left(f\right(x_2)+f(x_4\left)\right)+x_6\right)\left(\frac{鈭�x}{3}\right) \\ =(700+4(1000)+2(6300)+4(4000)+2(500)+4(650)+700)\left(\frac{60}{3}\right) \\ =\left(700+4000+12600+16000+1000+2600+700\right)20 \\ =752000 \end{gathered}$$

Thus, the water flow of the river is 752000 cubic feet.

On day 0 and on day 360, the rate of flow of water is 700 cubic feet.

Assume that rate of flow on each of the extra 5 days is 700 cubic feet.

The total flow during the 5 days are $5脳700=3500$

On adding the flow of extra 5 days, we get, $752000+3500=755500$

The annual flow in the river is the sum total of water flow per day.

In a day there are 86400 seconds.

The annual water flow is calculated as follows,

$755500脳86400=6.52752脳{10}^{10}$