Let say $$ h(x) = \begin{cases}x^2,& x \in \mathbb {Q}\\-x^2,& x \notin \mathbb{Q}\end{cases} $$
Is there a difference between riemann integrable and integrable?
And can I just say contiuous function is integrable and hence the above function is not
integrable in [-1,1]?
A function is Riemann Intrgrable if and only if the set of its points of discontinuity has measure zero.
In your case, the function $h$ is everywhere discontinuous in $[-1,1]$, except from zero. The measure of $[-1,1]\setminus \{0\}$ which is the set of the points of discontinuity of $h$, is $1\neq 0$. Therefore the function is not Riemann integrable.
When we refer to an integrable function in the notions of measure theory, we usually mean a measurable function $f$, defined on a measure space $X$ such that $\int_X|f(x)|dx<\infty$ (where the last integral is Lebesgue integral). In your case the function $h$ is measurable, and bounded by $1$. Therefore: $$\int_{[-1,1]}|h|\leq \int_{[-1,1]}1=2<\infty$$ so the function is Lebesgue integrable
The function is equal almost everywhere to $k(x) = -x^2$, and (as you note) continuous functions are integrable. So your function is Lebesgue integrable.
Together with Dimitrios's answer we see: there is a difference between Riemann integrable and Lebesgue integrable.
