Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.2 Fluid Pressure and Force - Exercises - Page 475: 18

$$17640\ \text{N}$$

We calculate the force as follows: \begin{aligned} F&=\rho g \int_{a}^{b} y f(y) d y \\ &=(900)(9.8) \int_{-\infty}^{0}(1-y) e^{y} d y \\ &=8820 \int_{-\infty}^{0} e^{y}(1-y) d y\ \ \text{Integrate by parts }\\ &=8820\lim_{t\to -\infty}\left[(1-y) e^{y}+e^{y}\right]_{t}^{0} \\ & =8820(2)\\ &=17640\ \text{N} \end{aligned}