# The force applied against an object moving horizontally on a linear path is described by F(x)=xe^x+ 1/x . By how much does the object's kinetic energy change as the object moves from  x in [ 1, 3 ]?

Mar 29, 2016

Approximately: $41.270 J$

#### Explanation:

The work done will be the integral of the force over the distance.
$W = {\int}_{1}^{3} F \left(x\right) \mathrm{dx} = {\int}_{1}^{3} \left(x {e}^{x} + \frac{1}{x}\right) \mathrm{dx}$
We'll leave the process of solving this integral to the student. The answer is:
$W = 2 \cdot {e}^{3} + \log \left(3\right) = 41.270 J$
Since we know that the work done is equal to the change in kinetic energy, we can say that:
$\Delta K E = W = 41.270 J$