Answer
$$R'(y) = -y^{3}sin(y)$$
Work Step by Step
$$R(y) = \int\limits_y^2{t^3}sin(t)dt$$
Switch the bounds of the integral making it negative.
$$R(y) = -\int\limits_w^y{t^3sin(t)}dt$$
Use FTC 1 by plugging the upper bound into t and multiply by the derivative of the upper bound (chain rule).
$$R'(y) = -y^3sin(y) \times(y)'$$
Simplify.
$$R'(y) = -y^3sin(y)$$
