Discussion
Let \(f : [0, \infty) \to \mathbb{R}\) be defined by
\[ f(x) = \int_{0}^{x} \sin^{2}(t^{2})\, dt. \]
Then the function \(f\) is
Answer: C
\(f\) is uniformly continuous on both \([0, 1)\) and \((0, \infty)\).
By the fundamental theorem of calculus, \(f'(x) = \sin^{2}(x^{2})\), so \(0 \le f'(x) \le 1\) for all \(x \ge 0\). By the mean value theorem,
\[ |f(x) - f(y)| \le |x - y| \]
so \(f\) is Lipschitz with constant 1 on the whole of \([0, \infty)\), and therefore uniformly continuous on every subset of it, including both \([0,1)\) and \((0, \infty)\).
The options denying uniform continuity on either set are ruled out by this bound; the growth of \(x^{2}\) inside the sine does not matter because the derivative stays bounded.
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