Answer
$$\eqalign{
& \left( {\text{a}} \right)0 \cr
& \left( {\text{b}} \right)\sqrt {13} \cr
& \left( {\text{c}} \right)\frac{6}{{\sqrt {13} }} \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}F\left( x \right) = \int_2^x {\sqrt {3{t^2} + 1} dt} \cr
& \left( {\text{a}} \right)F\left( 2 \right) = \int_2^2 {\sqrt {3{t^2} + 1} dt} = 0 \cr
& {\text{Differentiate}} \cr
& F'\left( x \right) = \frac{d}{{dx}}\left[ {\int_2^x {\sqrt {3{t^2} + 1} dt} } \right] = \sqrt {3{x^2} + 1} \cr
& \left( {\text{b}} \right)F'\left( 2 \right) = \sqrt {3{{\left( 2 \right)}^2} + 1} = \sqrt {13} \cr
& F''\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt {3{x^2} + 1} } \right] \cr
& F''\left( x \right) = \frac{{6x}}{{2\sqrt {3{x^2} + 1} }} = \frac{{3x}}{{\sqrt {3{x^2} + 1} }} \cr
& \left( {\text{c}} \right)F''\left( 2 \right) = \frac{{3\left( 2 \right)}}{{\sqrt {3{{\left( 2 \right)}^2} + 1} }} = \frac{6}{{\sqrt {13} }} \cr} $$
