The force applied against an object moving horizontally on a linear path is described by #F(x)=4e^x+ 1/x #. By how much does the object's kinetic energy change as the object moves from # x in [ 1, 3 ]#?

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Answer 1 · 2017-07-16T09:33:20

The change in kinetic energy is #=70.57Jf#

We need

#inte^xdx=e^x+C#

#int(1/x)dx=ln(|x|)+C#

The change in Kinetic Energy is equal to the work done.

#DeltaE=W=F*d#

#W=int_1^3(4e^x+1/x)dx#

#=[4e^x+lnx]_1^3#

#=(4e^3+ln3)-(4e+ln1)#

#=70.57J#