Answer
$$321,250,000 \mathrm{lb}$$
Work Step by Step
Let $f (y)$ denote the width of the dam wall at depth $y$ feet. Then the force on the dam wall is
$$ F=w\int_{0}^{100}yf(y)dy$$
Using the Trapezoidal Rule and the width and depth measurements in the figure, we get:
\begin{aligned}
F & \approx w \frac{20}{2}[0 \cdot f(0)+2 \cdot 20 \cdot f(20)+2 \cdot 40 \cdot f(40)+2 \cdot 60 \cdot f(60)+2 \cdot 80 \cdot f(80)+100 \cdot f(100)] \\
&=10 w(0+66,000+112,000+132,000+144,000+60,000)\\
&=321,250,000 \mathrm{lb}
\end{aligned}
