A particle moves parallel to the x-axis from x_i to x_f under a force F(x) = bx. What is the work done by the force, and what is W if b = −2.36 N/m, x_i = 2.77 m, and x_f = 5.48 m?

What you are doing when the force depends on position

When a force changes with position, the work is found by integrating the force along the path. Since everything is along the $x$-axis here, work is just the area under the $F$ vs. $x$ graph from $x_i$ to $x_f$.

Setting up the work integral

Work along the $x$ direction is $$W=\int_{x_i}^{x_f} F(x)\, dx.$$ With $F(x)=bx$, this becomes $$W=\int_{x_i}^{x_f} bx\, dx.$$

Integrating $bx$

Since $\int x\, dx=\tfrac{x^2}{2}$, $$W=b\left[\frac{x^2}{2}\right]_{x_i}^{x_f}=\frac{b}{2}\left(x_f^2-x_i^2\right).$$

Plugging in the numbers

Compute the squared positions: $$x_f^2=5.48^2=30.0304,\quad x_i^2=2.77^2=7.6729.$$ Difference: $$x_f^2-x_i^2=30.0304-7.6729=22.3575.$$ Now multiply by $\frac{b}{2}$: $$W=\frac{-2.36}{2}(22.3575)=(-1.18)(22.3575)\approx -26.38\,\text{J}.$$ Rounded to three significant figures: $$W\approx -26.4\,\text{J}.$$

Sign check (does a negative answer make sense?)

For $x>0$ and $b<0$, the force $F(x)=bx$ is negative (points in the $-x$ direction) while the displacement from $2.77$ m to $5.48$ m is in the $+x$ direction, so the force does negative work. That matches the negative result.