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 [2, 5] #?

1 Answer
Jan 22, 2016

Answer:

#4e^5-e^2+39#

Explanation:

For constant force

#W = F*x#

For non-constant force

#dW = F *dx#

Integrate by-parts

#W = int_2^5 F *dx#

#= int_2^5 (x^2 + xe^x) *dx#

#= int_2^5 x^2 *dx + int_2^5 xfrac{d}{dx}(e^x) *dx#

#= [x^3/3]_2^5 + [xe^x]_2^5 - int_2^5 e^xfrac{d}{dx}(x) *dx#

#= [5^3/3-2^3/3] + [5e^5-2e^2] - int_2^5 e^x *dx#

#= 39 + (5e^3-2)e^2 - [e^x]_2^5#

#= 39 + (5e^3-2)e^2 - (e^3-1)e^2#

#= 39 + (4e^3-1)e^2#