Intergal? work done by a non constant force

[Chuck 86]

1. The problem statement, all variables and given/known data
A force F = (4.38 x i + 2.84 y j) N acts on an object as it moves in the x direction from the origin to x = 5.52 m. Calculate the work done on the object by the force.


2. Relevant equations
intergal of F from 0 to 5.52m

3. The attempt at a solution

not sure if im suppose to intergrate, or what im supposed to do to start the process

[Donaldos]

W=\int\limits_{x=0}^{x=5.52}\vec{F}.{\rm d}\vec{x}

where \vec{x}=x\vec{i} is the position vector.

[Chuck 86]

ok so u intergrate the sum of X and Y right?

[Gear300]

Not exactly. By the looks of the problem, the object only moved in the x direction. So there is no net work in the y-direction. You only integrate the x-component.

[Chuck 86]

So then it would be the intergal of 4x?

[Donaldos]

Yes.

{\rm d}\vec{x}={\rm d}x \vec{i}

so

\vec{F}.{\rm d}\vec{x}=\left(4.38x \vec{i}+2.84y \vec{j}\right). \vec{i} {\rm d}x=4.38x{\rm d}x

W=\int\limits_0^{5.52} 4.38x{\rm d}x