Answer
a) No
b) Yes
Work Step by Step
a-
$$G(x)=\int_{4}^{x}\sqrt{t^{3}+1}dt$$
$$G(4)=\int_{4}^{4}\sqrt{t^{3}+1}dt$$
$$G(4)=0$$ because the limits of integration are equal so the answer is no.
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b-
$$G(x)=\int_{4}^{x}\sqrt{t^{3}+1}dt$$
Using the $FTC II$ it follows:
$$G'(x)=\sqrt{x^{3}+1}$$
$$G'(4)=\sqrt{4^{3}+1}=\sqrt{65}$$
So the answer is yes.
