Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.2 Fluid Pressure and Force - Exercises - Page 474: 10

$$ 13733.23 \ \text{N}$$

The fluid force on one side of the plate is \begin{aligned} F=\rho g \int_{a}^{b} y f(y) d y &=(850)(9.8) \int_{2}^{5} 2 y\left(1+y^{2}\right)^{-1} d y \\ &=8330 \int_{2}^{5} \frac{2 y}{\left(1+y^{2}\right)} d y\\ &= 8330\ln |1+y^2|\bigg|_{2}^{5}\\ &=8330[\ln (26)-\ln (5)]\\ &=8330\ln \frac{26}{5}\\ &\approx 13733.23 \ \text{N} \end{aligned}