## Calculus: Early Transcendentals 8th Edition

$$R'(y) = -y^{3}sin(y)$$
$$R(y) = \int\limits_y^2{t^3}sin(t)dt$$ 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. $$R'(y) = -y^3sin(y)$$