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 ]#?

Kinetic energy gained by the object is equlal to the work done by the aplied foce .So KE= #int_1^2F(x)dx=int_1^2(4x^2+5x)dx# #=[4x^3/3+5x^2/2]_1^2=4*2^3/3+5*2^2/2-4*1^3/1-5*1^2/1=35/3#