$$f(x) = \begin{cases} x^2\cdot\sin(\dfrac{1}{x^2}) & \text{if } x \in ( 0,1 ),\\ 0 & \text{if } x = 0. \end{cases} $$ is integrable on [ 0,1 ]?
I have tried to show that it is continuous as:
$$\displaystyle\lim_{x \to 0}x^2\cdot\sin(\dfrac{1}{x^2})=0$$
Also,
$$f(0)=0$$
And I showed that f is differantiable:
$$f^{\prime}(0)=\displaystyle\lim_{x \to 0}\dfrac{x^2\cdot\sin(\dfrac{1}{x^2})}{x}=\displaystyle\lim_{x \to 0}x\cdot\sin(\dfrac{1}{x^2})$$
It can be proven that $f^{\prime}(0)=0$ by Squeeze Theorem. And I stuck here, I can't go on. Thanks for any help.
We have the Riemann integrable theorem:a function f(x) defined on a compact subset of $\mathbb{R}$ is integrable if and only if f is bounded and the set of discontinuous points of f have Lebesgue measure zero.