The force acting on a particle in the \(x\)-direction is
\[
(3+2x)\ \text{N},
\]
where \(x\) is the displacement of the particle in metre. The work done in displacing the particle from
\[
x=1.5\ \text{m} \text{ to } x=3.5\ \text{m}
\]
is
Show Hint
When force varies with displacement,
\[
\boxed{W=\int_{x_1}^{x_2}F(x)\,dx.}
\]
The work done equals the area under the force--displacement graph.