Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 6

$$2\sqrt{10}.$$

To find the arc length $s$ of the curve $y=9-3x$ between $x=1$ and $x=3$, we fist calculate $y'=-3$. Then we have the integral, $$s=\int_1^{3}\sqrt{1+(y')^2}dx=\int_1^{3}\sqrt{1+9}dx=\sqrt{10}x|_1^3=2\sqrt{10}.$$