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).

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 $g(b)=\int_{a}^{b}f'dx+f(a)$.