Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 725: 8

Convergent

$\int_1^\infty f(x)dx=\int_1^\infty x^{2}e^{-x^{3}}dx$ $=\lim\limits_{t \to \infty}\int_1^t x^{2}e^{-x^{3}}dx$ Let $u=-x^{3}$, so $dx=-\frac{du}{3x^{2}}$ $=\lim\limits_{t \to \infty}-\frac{1}{3}\int_{-1}^{-t^{3}} e^{u}du$ $=-\frac{1}{3}\lim\limits_{t \to \infty}[e^{u}]_{-1}^{-t^{3}} $ $=\frac{1}{3e} $ Hence, the given series is convergent.