The force applied against an object moving horizontally on a linear path is described by F(x)= 4x^2+5x . By how much does the object's kinetic energy change as the object moves from  x in [ 1, 2 ]?

$\frac{35}{3}$unit
Kinetic energy gained by the object is equlal to the work done by the aplied foce .So KE= ${\int}_{1}^{2} F \left(x\right) \mathrm{dx} = {\int}_{1}^{2} \left(4 {x}^{2} + 5 x\right) \mathrm{dx}$
$= {\left[4 {x}^{3} / 3 + 5 {x}^{2} / 2\right]}_{1}^{2} = 4 \cdot {2}^{3} / 3 + 5 \cdot {2}^{2} / 2 - 4 \cdot {1}^{3} / 1 - 5 \cdot {1}^{2} / 1 = \frac{35}{3}$