What will be the area under the graph of variable force versus position?

$A$ particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to

Force acting on a body of mass $1 \ kg$ is related to its position $x$ as $F = x^3 - 3x \ N$. It is at rest at $x = 1$. Its velocity at $x = 3$ can be ......... $m/s$.

$A$ particle is moved along a path $A-B-C-D-E-F-A$,as shown in the figure,in the presence of a force $\vec{F} = (\alpha y \hat{i} + 2 \alpha x \hat{j}) \ N$,where $x$ and $y$ are in meters and $\alpha = -1 \ N/m$. The work done on the particle by this force $\vec{F}$ will be . . . . . . Joule.

$A$ block of mass $1 \ kg$,moving along the $x$-axis with an initial speed $v_{i} = 10 \ m/s$,enters a rough region ranging from $x = 0.1 \ m$ to $x = 1.9 \ m$. The retarding force acting on the block in this range is $F_{r} = -kx \ N$,where $k = 10 \ N/m$. Find the final speed of the block as it crosses the rough region.

In a one-dimensional case,the relationship between force and position is shown in the figure. The work done by the force in displacing a body from $x = 1\, cm$ to $x = 5\, cm$ is ............ $ergs$.