Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 5

$3\sqrt{10}$

To find the arc length $s$ of the curve $y=3x+1$ between $x=0$ and $x=3$, we fist calculate $y'=3$. Then we have the integral, $$s=\int_0^{3}\sqrt{1+(y')^2}dx=\int_0^{3}\sqrt{1+9}dx=\sqrt{10}x|_0^3=3\sqrt{10}.$$