To solve this problem, we need to calculate the work done by the variable force and then explain how integration handles changes in the force's sign. Part 1: Calculate the work done by the force. Step 1: Identify the given force function and parameters. The force acting on the object is given by $F(x) = ax + bx^2$. The given values are $a = 5 \, \text{N/m}$ and $b = -2 \, \text{N/m}^2$. So, the force function is $F(x) = 5x - 2x^2$. The object moves from $x_1 = 1 \, \text{m}$ to $x_2 = 3 \, \text{m}$. Step 2: Recall the formula for work done by a variable force. The work done ($W$) by a variable force $F(x)$ as an object moves from $x_1$ to $x_2$ is given by the definite integral: $$W = \int_{x_1}^{x_2} F(x) \, dx$$ Step 3: Substitute the force function and limits of integration into the formula. $$W = \int_{1}^{3} (5x - 2x^2) \, dx$$ Step 4: Perform the integration. First, find the antiderivative of $F(x)$: $$\int (5x - 2x^2) \, dx = \frac{5x^2}{2} - \frac{2x^3}{3}$$ Now, evaluate the definite integral: $$W = \left[ \frac{5x^2}{2} - \frac{2x^3}{3} \right]_{1}^{3}$$ $$W = \left( \frac{5(3)^2}{2} - \frac{2(3)^3}{3} \right) - \left( \frac{5(1)^2}{2} - \frac{2(1)^3}{3} \right)$$ $$W = \left( \frac{5 \times 9}{2} - \frac{2 \times 27}{3} \right) - \left( \frac{5}{2} - \frac{2}{3} \right)$$ $$W = \left( \frac{45}{2} - \frac{54}{3} \right) - \left( \frac{5}{2} - \frac{2}{3} \right)$$ $$W = \left( 22.5 - 18 \right) - \left( 2.5 - \frac{2}{3} \right)$$ $$W = 4.5 - \left( \frac{15}{6} - \frac{4}{6} \right)$$ $$W = 4.5 - \frac{11}{6}$$ To subtract, find a common denominator: $$W = \frac{9}{2} - \frac{11}{6}$$ $$W = \frac{27}{6} - \frac{11}{6}$$ $$W = \frac{16}{6}$$ $$W = \frac{8}{3} \, \text{J}$$ The work done by the force is $\boxed{\frac{8}{3} \text{ J}}$. Part 2: Does the process of integration take into account that the force $F(x)$ changes sign in the interval? Step 1: Determine if the force changes sign in the interval $[1, 3]$. The force function is $F(x) = 5x - 2x^2 = x(5 - 2x)$. The force is zero when $x=0$ or $5-2x=0 \Rightarrow x=2.5$. Since $x=2.5$ is within the interval $[1, 3]$: • For $x < 2.5$ (e.g., $x=1$), $F(1) = 5(1) - 2(1)^2 = 3 \, \text{N}$ (positive). • For $x > 2.5$ (e.g., $x=3$), $F(3) = 5(3) - 2(3)^2 = 15 - 18 = -3 \, \text{N}$ (negative). Thus, the force $F(x)$ does change sign within the interval $[1, 3]$. Step 2: Explain how integration accounts for the sign change. Yes, the process of definite integration inherently accounts for the fact that the force $F(x)$ changes sign in the interval. When $F(x)$ is positive, it contributes positive work (force acts in the direction of displacement). When $F(x)$ is negative, it contributes negative work (force acts opposite to the direction of displacement). The definite integral sums these positive and negative contributions algebraically to yield the net work done over the entire interval.