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

1 Answer
Aug 3, 2017

Answer:

The work is #=180.02J#

Explanation:

#inte^xdx=e^x+C#

#int(x^n)dx=x^(n+1)/(n+1)+C(n!=-1)#

The work is

#DeltaW=F(x)*Deltax#

#F(x)=x^2+e^x#

#W=int_2^(5)(x^2+e^x)dx#

#=[1/3x^3+e^x]_2^(5)#

#=(1/3*5^3+e^5)-(1/3*2^3+e^2)#

#=125/3-8/3+e^5-e^2#

#=117/3+141.02#

#=180.02J#