Obviously, f(x) has only 1 jump discontinuity x=0 in the interval [- 1], so according to the properties of definite integral, f(x) is continuously integrable in [- 1, 1].

And it is also easy to calculate ∫-/kloc-0 /→ xf (t) dt = | x |-1.

|x|- 1 is not derivable at x=0, although |x|- 1 is continuous at x=0.

So if f(x) has a jump discontinuity at [a, b], then ∫a→xf(t)dt is not derivable at this jump discontinuity. But it is continuous in the jumping interval.

In fact, the left and right derivatives of ∫a→x f(t)dt exist at the jumping discontinuity, but they are not equal. So it is continuously non-conductive.

Continuous must be integrable,

A continuous function on a closed interval must be bounded.