1. **State the problem:** We have the function $f(x) = 2\sqrt{x}$ defined on the interval $[0,4]$. We need to find: (a) The average value $f_{ave}$ of $f$ on $[0,4]$. (b) The value(s) $c$ such that $f_{ave} = f(c)$. (c) Sketch the graph of $f$ and a rectangle with the same area as under the graph of $f$. 2. **Formula for average value of a function:** The average value of $f$ on $[a,b]$ is given by $$f_{ave} = \frac{1}{b-a} \int_a^b f(x) \, dx$$ 3. **Calculate the average value $f_{ave}$:** Here, $a=0$, $b=4$, and $f(x) = 2\sqrt{x} = 2x^{1/2}$. Calculate the integral: $$\int_0^4 2x^{1/2} \, dx = 2 \int_0^4 x^{1/2} \, dx = 2 \left[ \frac{2}{3} x^{3/2} \right]_0^4 = 2 \times \frac{2}{3} (4)^{3/2} - 0$$ Calculate $4^{3/2}$: $$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$$ So the integral is: $$2 \times \frac{2}{3} \times 8 = \frac{32}{3}$$ Now compute the average value: $$f_{ave} = \frac{1}{4-0} \times \frac{32}{3} = \frac{32}{3} \times \frac{1}{4} = \frac{8}{3}$$ 4. **Find $c$ such that $f_{ave} = f(c)$:** We want $f(c) = 2\sqrt{c} = \frac{8}{3}$. Solve for $c$: $$2\sqrt{c} = \frac{8}{3} \implies \sqrt{c} = \frac{4}{3} \implies c = \left( \frac{4}{3} \right)^2 = \frac{16}{9}$$ 5. **Summary:** - Average value $f_{ave} = \frac{8}{3}$. - Value $c = \frac{16}{9}$. 6. **Explanation of the rectangle:** The rectangle has base length $4$ (the interval length) and height equal to the average value $\frac{8}{3}$, so its area is: $$4 \times \frac{8}{3} = \frac{32}{3}$$ which matches the area under the curve.