Question:

<p><strong>Step 1:</strong> Differentiate both sides of $\int_0^{f(x)} 4t^3\, dt = (x - 2)g(x)$ with respect to $x$.</p><p>Using Leibniz rule on the left: $4[f(x)]^3 \cdot f'(x) = (x-2)g'(x) + g(x)$</p><p><strong>Step 2:</strong> At $x = 2$: $4[f(2)]^3 \cdot f'(2) = 0 \cdot g'(2) + g(2)$</p><p>Substitute $f(2) = 6$ and $f'(2) = \frac{1}{48}$:</p><p>$4(6)^3 \cdot \frac{1}{48} = g(2)$</p><p>$4 \cdot 216 \cdot \frac{1}{48} = g(2)$</p><p>$\frac{864}{48} = g(2)$</p><p>$g(2) = 18$</p><p><strong>Step 3:</strong> Since $g(x) = \frac{\int_0^{f(x)} 4t^3\, dt}{x-2}$ and the original integral equals $[t^4]_0^{f(x)} = [f(x)]^4$, we have the indeterminate form $\frac{0}{0}$ at $x=2$. By continuity of $g$ (which follows from the C¹ nature of $f$), $\lim_{x \to 2} g(x) = g(2) = 18$.</p><p>∴ Answer: <strong>18</strong></p>