Step-by-Step Solution
Key Concept: Comparing integrals using inequalities between the integrands to establish bounds on definite integrals.
Since $I = \int_0^1 \frac{\sin x}{x} dx < \int_0^1 \frac{\sin x}{\sin x} dx$ and $\sin x \geq \sin z$ for $x \in (0,1)$, we get $I < \int_0^1 \sqrt{x} dx = \frac{2}{3}[x^{3/2}]_0^1 = I < \frac{4}{3}$. For $x \in (0,1)$, $\frac{\sin x}{x} < \sin x$, hence $I < \int_0^1 x dx = 2$ and $J < 2$.
Correct Answer: 2