A variable force of $5x^{-2}$ pounds moves an object along a straight line when it is $x$ feet from the origin

Question

A variable force of $\dfrac{5}{x^2}$ pounds moves an object along a straight line when it is $x$ feet from the origin. Calculate the work done in moving the object from $x=1$ ft to $x=10$ ft.

I tried doing $\displaystyle \int_1^{10} x\cdot \dfrac{5}{x^2}dx$

then got $\quad \displaystyle 5\int_1^{10} \dfrac{1}{x} dx $

then $5(\ln10-\ln1)$

then $5\ln10$ ft-lb

but the book says $4.5$ ft-lb....

Answer 1

work done = Force * distance

but if its variable force

then Work done = $\displaystyle \int_1^{10} Force \cdot dx$

$\displaystyle \int_1^{10} \dfrac{5}{x^2}dx$ = $\dfrac{-5}{x}$ between x=1 and x=10

so the answer is 4.5

Answer 2

use
$\displaystyle \int_1^{10} \dfrac{5}{x^2}dx$
instead of
$\displaystyle \int_1^{10} x\cdot \dfrac{5}{x^2}dx$

for more info: imagine that $F=5$ then with $\displaystyle \int_1^{10} x\cdot5dx$ formula, the result work will be different from 11 to 20 and from 1 to 10. with this paradox you will find that something is wrong.