Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 357: 72

The mass of the rod is $20~grams$

Let $M(x)$ be the mass of the rod of length $x$ $M'(x) = \rho(x) = \frac{1}{\sqrt{x}} = x^{-1/2}$ We can find an expression for $M(x)$: $M(x) = 2~\sqrt{x}+C$ Since $M(0) = 0,$ then $C = 0$ $M(x) = 2~\sqrt{x}$ We can find $M(100)$: $M(x) = 2~\sqrt{100} = 20~grams$ The mass of the rod is $20~grams$