A force of $\left(4 x^{2}+3 x\right)$ N acts on a particle which displaces it from $x=2 m$ to $x=3 \mathrm{m}$.

The work done by the force is

Given that,

Force, $F=\left(4 x^{2}+3 x\right)$

To calculate work done from

$x=2 m \text { to } x=3 \mathrm{~m} \text {. }$

we know,

$\begin{aligned} W & =F \cdot S \\ w & =\int F \cdot d s \\ & =\int_{2}^{3}\left(4 x^{2}+3 x\right) \cdot d x . \end{aligned}$

(To calculate work done from $x=2$ to $x=3$ ).

$\begin{aligned} & =\left[\frac{4 x^{3}}{3}+\frac{3 x^{2}}{2}\right]_{2}^{3} \\ \left(\because \int 4 x^{2} d x=\frac{4 x^{3}}{3} \text { and } \int 3 x^{2} x=\frac{3 x^{2}}{2}\right] . & \\ & \Rightarrow\left(\frac{4(3)^{3}}{3}+\frac{3(3)^{3}}{2}\right)-\left(\frac{4(2)^{3}}{3}+\frac{3 \times(2)^{2}}{2}\right) \\ & =\left(4 \times 9+\frac{3 \times 9}{2}\right)-\left(\frac{4 \times 8}{3}+\frac{3 \times 2 \times 2}{2}\right) \\ & =(36+(3.5)-(10.6+6) \\ & =49.5-16.6 \\ & =32.8 \mathrm{~J} . \end{aligned}$

$\therefore$ Work done $=32.8 \mathrm{~J}$.