Question:

Step 1: By the Mean Value Theorem applied to $f$ on the interval $[0,4]$, there exists some $C_1 \in (0,4)$ such that $$f'(C_1) = \frac{f(4)-f(0)}{4-0} = \frac{2-0}{4} = \frac{2}{4} = \frac{1}{2}.$$ Since $(0,4) \subset (0,8)$, this implies there exists some $C_1 \in (0,8)$ such that $f'(C_1) = \frac{1}{2}$. Step 2: By the Mean Value Theorem applied to $f$ on the interval $[0,4]$, there exists $C_A \in (0,4)$ such that $f'(C_A) = \frac{f(4)-f(0)}{4-0} = \frac{2-0}{4} = \frac{1}{2}$. By the Mean Value Theorem applied to $f$ on the interval $[4,8]$, there exists $C_B \in (4,8)$ such that $f'(C_B) = \frac{f(8)-f(4)}{8-4} = \frac{2-2}{4} = 0$. Since $f$ is differentiable on $[0,8]$, its derivative $f'$ exists on $[0,8]$ and satisfies the Intermediate Value Property (Darboux's Theorem). As $f'$ takes values $0$ and $\frac{1}{2}$ on $(0,8)$, and $0 < \frac{1}{10} < \frac{1}{2}$, there must exist some $C_1 \in (0,8)$ such that $f'(C_1) = \frac{1}{10}$. Step 3: By the Mean Value Theorem applied to $f$ on the interval $[0,8]$, there exists some $C_1 \in (0,8)$ such that $$f'(C_1) = \frac{f(8)-f(0)}{8-0} = \frac{2-0}{8} = \frac{2}{8} = \frac{1}{4}.$$ Since $f$ is differentiable on $[0,8]$, it is continuous on $[0,8]$. Given $f(0)=0$ and $f(4)=2$, by the Intermediate Value Theorem, for any value $k$ between $0$ and $2$, there exists some $C_2 \in (0,4)$ such that $f(C_2)=k$. Specifically, for $k=\frac{1}{2}$, there exists some $C_2 \in (0,4) \subset (0,8)$ such that $f(C_2) = \frac{1}{2}$. Therefore, there exist $C_1, C_2 \in (0,8)$ such that $$8f'(C_1) \cdot f(C_2) = 8 \cdot \frac{1}{4} \cdot \frac{1}{2} = 1.$$ Step 4: Let $I = \int_0^8 f(t)\,dt$. Perform a substitution: let $t = x^3$. Then $dt = 3x^2\,dx$. When $t=0$, $x=0$. When $t=8$, $x=2$. The integral becomes: $$I = \int_0^2 f(x^3) \cdot 3x^2\,dx = 3\int_0^2 x^2 f(x^3)\,dx.$$ Split the integral into two parts: $$I = 3\left[\int_0^1 x^2 f(x^3)\,dx + \int_1^2 x^2 f(x^3)\,dx\right].$$ Since $f$ is differentiable, it is continuous. Thus, the function $g(x) = x^2 f(x^3)$ is continuous on $[0,2]$. By the Mean Value Theorem for Integrals, for the interval $[0,1]$, there exists some $C_1 \in (0,1)$ such that $$\int_0^1 x^2 f(x^3)\,dx = g(C_1) \cdot (1-0) = C_1^2 f(C_1^3).$$ Similarly, for the interval $[1,2]$, there exists some $C_2 \in (1,2)$ such that $$\int_1^2 x^2 f(x^3)\,dx = g(C_2) \cdot (2-1) = C_2^2 f(C_2^3).$$ Substituting these back into the expression for $I$: $$I = 3\left(C_1^2 f(C_1^3) + C_2^2 f(C_2^3)\right).$$ Thus, there exist some $C_1 \in (0,1)$ and $C_2 \in (1,2)$ such that $\displaystyle\int_0^8 f(t)\,dt = 3(C_1^2 f(C_1^3) + C_2^2 f(C_2^3))$.