Answer
$R'(y) = -y^3sin(y)$
Work Step by Step
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $R(y) = \int ^{2}_{y}t^3sin(t)dt$
Swap upper and lower bounds: (this makes the expression negative)
$R(y) = -\int ^{y}_{2}t^3sin(t)dt $
Since $R(y) = \int^{x}_{0}f(t)dt$
$R'(y) = f(y)$
Thus,
$R'(y) = -y^3sin(y)$