Answer:
.
Step-by-step explanation:
We have been given that a variable force of
pounds moves an object along a straight line when it is x feet from the origin. We are asked to find the work done in moving the object from x = 1 ft to x = 10 ft.
We know that
, where F represents force.
![dW=3x^{-2}dx](https://tex.z-dn.net/?f=dW%3D3x%5E%7B-2%7Ddx)
Now, we will integrate both sides of equation as:
![\int\limits{dw} =\int\limits^{10}_1 {3x^{-2} \, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%7Bdw%7D%20%3D%5Cint%5Climits%5E%7B10%7D_1%20%7B3x%5E%7B-2%7D%20%5C%2C%20dx)
Take the constant out:
![\int\limits{1dw} =3\int\limits^{10}_1 {x^{-2} \, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%7B1dw%7D%20%3D3%5Cint%5Climits%5E%7B10%7D_1%20%7Bx%5E%7B-2%7D%20%5C%2C%20dx)
![w =3\int\limits^{10}_1 {x^{-2} \, dx](https://tex.z-dn.net/?f=w%20%3D3%5Cint%5Climits%5E%7B10%7D_1%20%7Bx%5E%7B-2%7D%20%5C%2C%20dx)
Upon applying power rule of integrals, we will get:
![w =3\/[ \frac{x^{-2+1}}{-2+1} ]^{10}_1](https://tex.z-dn.net/?f=w%20%3D3%5C%2F%5B%20%5Cfrac%7Bx%5E%7B-2%2B1%7D%7D%7B-2%2B1%7D%20%5D%5E%7B10%7D_1)
![w =3[ \frac{x^{-1}}{-1} ]^{10}_1](https://tex.z-dn.net/?f=w%20%3D3%5B%20%5Cfrac%7Bx%5E%7B-1%7D%7D%7B-1%7D%20%5D%5E%7B10%7D_1)
![w =3[ -\frac{1}{x} ]^{10}_1](https://tex.z-dn.net/?f=w%20%3D3%5B%20-%5Cfrac%7B1%7D%7Bx%7D%20%5D%5E%7B10%7D_1)
![w =3[ -\frac{1}{10} -(-\frac{1}{1})]](https://tex.z-dn.net/?f=w%20%3D3%5B%20-%5Cfrac%7B1%7D%7B10%7D%20-%28-%5Cfrac%7B1%7D%7B1%7D%29%5D)
![w =3[ -\frac{1}{10}+\frac{10}{10})]](https://tex.z-dn.net/?f=w%20%3D3%5B%20-%5Cfrac%7B1%7D%7B10%7D%2B%5Cfrac%7B10%7D%7B10%7D%29%5D)
![w =3[\frac{9}{10}]](https://tex.z-dn.net/?f=w%20%3D3%5B%5Cfrac%7B9%7D%7B10%7D%5D)
![w=2.70](https://tex.z-dn.net/?f=w%3D2.70)
Therefore, the work done by the object is
.