1. **State the problem:** Find the average value of the function $f(x) = \sqrt[3]{x}$ on the interval $[1,8]$. 2. **Formula for average value of a function:** The average value $f_{avg}$ of a continuous function $f(x)$ on the interval $[a,b]$ is given by $$f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx$$ 3. **Apply the formula:** Here, $a=1$ and $b=8$, so $$f_{avg} = \frac{1}{8-1} \int_1^8 x^{1/3} \, dx = \frac{1}{7} \int_1^8 x^{1/3} \, dx$$ 4. **Integrate the function:** Recall that $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ for $n \neq -1$. For $n=\frac{1}{3}$, $$\int x^{1/3} \, dx = \frac{x^{(1/3)+1}}{(1/3)+1} + C = \frac{x^{4/3}}{4/3} + C = \frac{3}{4} x^{4/3} + C$$ 5. **Evaluate the definite integral:** $$\int_1^8 x^{1/3} \, dx = \left[ \frac{3}{4} x^{4/3} \right]_1^8 = \frac{3}{4} (8^{4/3} - 1^{4/3})$$ 6. **Simplify powers:** Since $8 = 2^3$, $$8^{4/3} = (2^3)^{4/3} = 2^{3 \times \frac{4}{3}} = 2^4 = 16$$ Also, $1^{4/3} = 1$. 7. **Calculate the integral value:** $$\int_1^8 x^{1/3} \, dx = \frac{3}{4} (16 - 1) = \frac{3}{4} \times 15 = \frac{45}{4}$$ 8. **Find the average value:** $$f_{avg} = \frac{1}{7} \times \frac{45}{4} = \frac{45}{28}$$ **Final answer:** The average value of $f(x) = \sqrt[3]{x}$ on $[1,8]$ is $$\boxed{\frac{45}{28}} \approx 1.607$$