How to show the function $f(x) = x^2 \sin(1/x)$ has integrable derivative?

Question

Consider the function

$$f(x) = \begin{cases} x^2\sin(1/x) & \text { if } x \neq 0 \\ 0 & \text{ otherwise} \end{cases} $$

has integrable derivative on $(-1, 1)$.

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I found

$$f'(x) = \begin{cases} 2x\sin(1/x) - \cos(1/x) & \text { if } x \neq 0 \\ 0 & \text{ otherwise,} \end{cases} $$

but I have no clue how to show that it is integrable on $(-1, 1)$. Can someone please help me?

Answer 1

Your computations are correct and they show that $f'$ is bounded and it has a single discontinuity point. Therefore, it Riemann-integrable (and this would still be true if it had a countable set of discontinuity points).