Chapter 6: Applications of Definite Integrals - Section 6.3 - Arc Length - Exercises 6.3 - Page 335: 1

$12$

We know that the formula to calculate the arc length is defined as: $L=\int_m^n \sqrt {1+[f'(x)]^2} dx$ This implies that $L=\int_0^3 \sqrt {1+[x(x^2+2)]^2}dx=\int_0^3 (x^4+2x^2+1) dx$ Use formula such as: $\int x^n dx=\dfrac{x^{(n+1)}}{(n+1)}+C$ Thus, $\int_0^3 (x^2+1) dx= [\dfrac{x^3}{3}+x]_0^3$ Hence, $L=\int_0^3 (x^2+1) dx=(\dfrac{27}{3})+3=12$