Answer
$$\left( {\text{a}} \right)F'\left( x \right) = {x^3} + 1{\text{ and }}\left( {\text{b}} \right)F'\left( x \right) = {x^3} + 1$$
Work Step by Step
$$\eqalign{
& {\text{Let }}F\left( x \right) = \int_1^x {\left( {{t^3} + 1} \right)dt} \cr
& \cr
& \left( {\text{a}} \right){\text{Use Part 2 of the Fundamental Theorem of Calculus to calculate }}F'\left( x \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right) \cr
& {\text{Then}} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \frac{d}{{dx}}\left[ {\int_1^x {\left( {{t^3} + 1} \right)dt} } \right],\,\,\,\,\left\{ {{\text{With }}f\left( t \right) = {t^3} + 1} \right\} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = f\left( x \right) = {\left( x \right)^3} + 1 \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = {x^3} + 1 \cr
& \cr
& \left( {\text{b}} \right){\text{Integrating}} \cr
& \,\,\,\,\,\,\,F\left( x \right) = \left[ {\frac{{{t^4}}}{4} + t} \right]_1^x \cr
& \,\,\,\,\,\,\,F\left( x \right) = \left[ {\frac{{{x^4}}}{4} + x} \right] - \left[ {\frac{{{1^4}}}{4} + 1} \right] \cr
& \,\,\,\,\,\,\,F\left( x \right) = \frac{{{x^4}}}{4} + x - \frac{5}{4} \cr
& {\text{Differentiating}} \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{4} + x - \frac{5}{4}} \right] \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = \frac{{4{x^3}}}{4} + 1 - 0 \cr
& \,\,\,\,\,\,\,\,F'\left( x \right) = {x^3} + 1 \cr} $$
