Answer
$$ V\approx 20~\text{acre ft} =871,200 \text{ ft}^3 $$
Work Step by Step
Since
$$V=\int_{0}^{18} A(h) d h$$
From the given table, $\Delta x=2$. Then, by using $T_9$, we get
\begin{align*}
T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+..+2f(x_{n-1})+f(x_n)\right]\Delta x\\
T_{9}&= \left[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{8})+f(x_9)\right]\Delta x\\
&= \left[f(0)+2f(2)+2f(4)+2f(6)+2f(8)+2f(10)+2f(12)+2f(14)+2f(16)+f(18)\right] \\
&\approx 20 \text{arce ft}
\end{align*}
Hence
$$ V\approx 20~ \text{acre ft} =871,200 \text{ ft}^3 $$
