To determine the work done by the force [tex]\( F(x) \)[/tex] as the object moves along the [tex]\( x \)[/tex]-axis from [tex]\( x = 1 \)[/tex] m to [tex]\( x = 3 \)[/tex] m, we need to integrate the force function over this interval. The force function is given as:
[tex]\[ F(x) = ax + bx^2 \][/tex]
where [tex]\( a = 5 \, \text{N/m} \)[/tex] and [tex]\( b = -2 \, \text{N/m} \)[/tex].
### Step-by-Step Solution:
1. Substitute the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the force function:
[tex]\[ F(x) = 5x - 2x^2 \][/tex]
2. Set up the integral to find the work done:
The work [tex]\( W \)[/tex] done by the force as the object moves from [tex]\( x = 1 \, \text{m} \)[/tex] to [tex]\( x = 3 \, \text{m} \)[/tex] is given by:
[tex]\[
W = \int_{1}^{3} F(x) \, dx
\][/tex]
3. Write the integral with the given force function:
[tex]\[
W = \int_{1}^{3} (5x - 2x^2) \, dx
\][/tex]
4. Integrate the function [tex]\( 5x - 2x^2 \)[/tex]:
[tex]\[
\int (5x - 2x^2) \, dx = \int 5x \, dx - \int 2x^2 \, dx
\][/tex]
The antiderivative of [tex]\( 5x \)[/tex] is [tex]\( \frac{5x^2}{2} \)[/tex] and the antiderivative of [tex]\( -2x^2 \)[/tex] is [tex]\( \frac{-2x^3}{3} \)[/tex].
Thus, the integral becomes:
[tex]\[
\int (5x - 2x^2) \, dx = \frac{5x^2}{2} - \frac{2x^3}{3}
\][/tex]
5. Evaluate the definite integral from [tex]\( x = 1 \)[/tex] to [tex]\( x = 3 \)[/tex]:
[tex]\[
W = \left[ \frac{5x^2}{2} - \frac{2x^3}{3} \right]_{1}^{3}
\][/tex]
This means we substitute [tex]\( x = 3 \)[/tex] and [tex]\( x = 1 \)[/tex] into the antiderivative and subtract:
[tex]\[
W = \left( \frac{5(3)^2}{2} - \frac{2(3)^3}{3} \right) - \left( \frac{5(1)^2}{2} - \frac{2(1)^3}{3} \right)
\][/tex]
Simplify each term:
[tex]\[
\frac{5(3)^2}{2} = \frac{5 \cdot 9}{2} = \frac{45}{2} = 22.5
\][/tex]
[tex]\[
\frac{2(3)^3}{3} = \frac{2 \cdot 27}{3} = 18
\][/tex]
[tex]\[
\frac{5(1)^2}{2} = \frac{5 \cdot 1}{2} = \frac{5}{2} = 2.5
\][/tex]
[tex]\[
\frac{2(1)^3}{3} = \frac{2 \cdot 1}{3} = \frac{2}{3} \approx 0.6667
\][/tex]
Substitute these values back into the expression for [tex]\( W \)[/tex]:
[tex]\[
W = (22.5 - 18) - (2.5 - 0.6667)
\][/tex]
Simplify the expression:
[tex]\[
W = 4.5 - 1.8333 = 2.6667
\][/tex]
6. Conclusion:
The work done by the force [tex]\( F(x) \)[/tex] as the object moves from [tex]\( x = 1 \, \text{m} \)[/tex] to [tex]\( x = 3 \, \text{m} \)[/tex] is approximately [tex]\( 2.67 \)[/tex] Joules.
### Consideration of the Force Change in Sign:
The process of integration inherently takes into account the fact that the force [tex]\( F(x) \)[/tex] changes sign over the interval. When the force function changes sign, the integral calculates the net work done, effectively summing the contributions where the force is positive and subtracting the contributions where the force is negative.
