The force applied against a moving object travelling on a linear path is given by #F(x)= x^2+xe^x #. How much work would it take to move the object over #x in [4, 5] #?

1 Answer
Dec 24, 2016

Answer:

The answer is #=450.16J#

Explanation:

The work done by a force #F(x)# from #x_1# to #x_2# is

#W=int_(x_1)^(x_2) F(x)dx#

Here #F(x)=x^2+xe^x#

#x_1=4# and #x_2=5#

So,

#W=int_(4)^(5) (x^2+xe^x)dx#

#=int_4^5x^2dx+int_4^5xe^xdx#

#=[x^3/3] _4^5+int_4^5xe^xdx#

We perform the second inegration by parts

#intuv'dx=uv-intu'vdx#

#u=x#, #=>#, #u'=1#

#v'=e^x#, #=>#, #v=e^x#

#intxe^xdx=xe^x-inte^xdx#

#=xe^x-e^x#

So,
#W=[x^3/3] _4^5+[xe^x-e^x] _4^5#

#=(5^3/3-4^3/3)+(5e^5-e^5-4e^4+e^4)#

#=61/3+e^4(4e-3)#

#=450.16J#