# Chapter 6 - Exercises and Problems - Page 111: 83

a) The proof is below. b) $2b(\sqrt{x_2})$

#### Work Step by Step

a) We know that work is equal to the integral of the force with respect to x. Thus: $W = \int_{x_1}^{x_2} bx^{-.5}dx \\ W = 2b\sqrt{x}|_{x_1}^{x_2}= 2b(\sqrt{x_2}-\sqrt{x_1})$ When $x_1$ equals 0, this becomes: $2b(\sqrt{x_2})$, which is a finite value. b) As we found above, the work done as the limit of x approaches 0 is: $2b(\sqrt{x_2})$

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