The force applied against a moving object travelling on a linear path is given by #F(x)=3x^2+e^x #. How much work would it take to move the object over #x in [0, 3 ] #?

2 Answers
Sep 10, 2017

Answer:

The work is #=46.1J#

Explanation:

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

#inte^xdx=e^x+C#

The work is

#DeltaW=FDeltax#

#DeltaW=(3x^2+e^x)Deltax#

#W=int_0^3(3x^2+e^x)dx#

#=[x^3+e^x]_0^3#

#=(3^3+e^3)-(0+e^0)#

#=27+e^3-1#

#=46.1J#

Sep 10, 2017

Answer:

I got #46J#.

Explanation:

Here you have a variable force so for an infinitesimal (=very small) Work you have:

#dW=F(x)dx#

integrating:

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

in your case:

#W=int_(0)^(3)(3x^2+e^x)dx=cancel(3)x^3/cancel(3)+e^x=x^3+e^x|_0^3=#

apply the Fundamental Theorem of Calculus:

#=(3^3+e^3)-(0^3+e^0)=46J#