Answer
$g'(x) = \sqrt {x+x^3}$
Work Step by Step
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $g(x) = \int ^{x}_{0}\sqrt {t+t^3}dt$
Since $g(x) = \int^{b}_{a}f(t)dt$
$g'(x) = f(x)$
Thus,
$g'(x) = \sqrt {x+x^3}$