## integrating over piecewise functions

given a function $f(x)$ that is piecewise smooth on interval $-L<x<L$ except at $N-1$ points, is $\int_{-L}^L f'(x)dx$ legal or would i have to $$\sum_{i=1}^N \int_{x_i}^{x_{i+1}} f'(x)dx‎‎$$
where $x_{N+1}=L$ and $x_{1}=-L$

also, am i correct that if $f(x)$ is piecewise smooth, then $f'(x)$ is piecewise continuous but not necessarily piecewise smooth?

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire

Blog Entries: 1
Recognitions:
Gold Member
Homework Help
 Quote by joshmccraney given a function $f(x)$ that is piecewise smooth on interval $-L The latter. Suppose for example that ##f(x) = |x|^{1/2}##. This is smooth (infinitely differentiable) everywhere except at ##x = 0##, and f'(x) = \begin{cases} \frac{1}{2|x|^{1/2}} & \textrm{ if }x > 0 \\ \frac{-1}{2|x|^{1/2}} & \textrm{ if }x < 0 \\ \end{cases} As ##f'## is unbounded on ##[-L,L]##, it's necessary to use two (improper) integrals to integrate it: \lim_{a \rightarrow 0^-} \int_{-L}^{a} f'(x) dx + \lim_{b \rightarrow 0^+}\int_{b}^{L} f'(x) dx Both limits exist and the answers have opposite signs, so the result is 0.  also, am i correct that if [itex]f(x)$ is piecewise smooth, then $f'(x)$ is piecewise continuous but not necessarily piecewise smooth?