For every function $f$ with $f'$ integrable there is a function $g$ equal to $f$ everywhere but a point such that $\int_{a}^{b}g'dx=g(b)-g(a)$. Take g(x)=f(x) for x different from b and g(b)=\int_{a}^{b}f'dx+f(a).
O.R.
- 807
- 1
- 17
- 28