Given a piecewise function
$$ f(x)=\left\{ \begin{array}{ll} x & 0 \leq x \leq 1 \\ 3-x & 1 < x \leq 3\\ \end{array} \right. $$
and it is integrable in $[0,3]$. Then Can I apply the average value theorem ($\frac{1}{b-a} \int_{a}^{b}f(x)dx$)? I think so because it is integrable, so it is continuous and thus I can use it. But the next question is: is there multiple $c$, for which $f(c)= \frac{1}{3-0} \int_{3}^{0}f(x)dx$? Notice that $a=0$ and $b = 3$ since this is the interval I said is integrable. For this question I think I should divide the integral into 2 parts, $\int_{0}^{1}f(x)dx$ and $\int_{1}^{3}f(x)dx$ because of the piecewise function. I did that and still can't see if there is more than one $c$ value.
By definition, a function is continuous at some point if
$\lim_{x \rightarrow a}f(x) = f(a)$
You can see that it is not continuous if you apply the $\epsilon, \delta$ definition for example.
And then by definition of the Mean value theorem $f$ should be continuous, but it is not, therefore you need to partition the integral into two parts.