Chapter 6: Applications of Definite Integrals - Section 6.5 - Work and Fluid Forces - Exercises 6.5 - Page 347: 1

$400$

Let $k$ be a constant which defines a natural length with limits $0$ to $3$. This means that $\int_0^3 kx dx=1800$ Also, $\int_0^3 kx dx=1800$ Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$ Then $k[\dfrac{x^2}{2}]_0^3 =1800$ or, $k (\dfrac{9}{2}-0) =1800$ Hence, $k=\dfrac{(1800)(2)}{9}=400$