Answer
$$F'\left( x \right) = \left| {x - 1} \right|$$
Work Step by Step
$$\eqalign{
& {\text{Calculate }}\frac{d}{{dx}}\left[ {\int_0^x {\left| {t - 1} \right|dt} } \right] \cr
& {\text{Use Part 2 of the Fundamental Theorem of Calculus to calculate }}F'\left( x \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& {\text{Then}} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \frac{d}{{dx}}\left[ {\int_0^x {\left| {t - 1} \right|dt} } \right],\,\,\,\,\left\{ {{\text{With }}f\left( t \right) = \left| {t - 1} \right|} \right\} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = f\left( x \right) = \left| {x - 1} \right| \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \left| {x - 1} \right| \cr} $$
