## Calculus: Early Transcendentals 8th Edition

$$\int_{-\infty}^{\infty }(y^{3}-3y^{2})dy=\int_{-\infty}^{0}(y^{3}-3y^{2})dy+\int_{0}^{\infty }(y^{3}-3y^{2})dy$$ $$=\lim_{t\rightarrow \infty}\int_{-t}^{0}(y^{3}-3y^{2})dy+\lim_{s\rightarrow \infty}\int_{0}^{s}(y^{3}-3y^{2})dy$$ $$=\lim_{t\rightarrow \infty} \left| \frac{y^{4}}{4}-y^{3} \right |_{-t}^{0}+\lim_{s\rightarrow \infty} \left| \frac{y^{4}}{4}-y^{3} \right |_{0}^{s}$$ $$=(\infty-\infty)+(-\infty-\infty)$$ This limit does not exist, so the integral is divergent.