Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 435: 48

$91.2$ $ m^2$

If $x=$distance from left end of pool and $w=w(x)=$width at $x$,then the Midpoint Rule with $n=4$ and $\triangle x=\frac{b-a}{n}=\frac{8.2-0}{4}=4$ gives the Area $=\int_{0}^{16}w$ $ dx \approx4(6.2+6.8+5.0+4.8)$ $=4(22.8)$ $=91.2$ $ m^2$