Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.7 Improper Integrals - Exercises - Page 441: 72

Answer

The integral $\int_{1}^{\infty} \dfrac{dx}{\sqrt {x^{1/3}+x^3}}$ converges.

Work Step by Step

We are given the function $f(x)=\int_{1}^{\infty} \dfrac{dx}{\sqrt {x^{1/3}+x^3}}$ Since, $\sqrt {x^{1/3}+x^3} \geq \sqrt {x^3}$ $\dfrac{dx}{\sqrt {x^{1/3}+x^3}} \leq \dfrac{1}{x^{3/2}} $ But the integral $\int_{1}^{\infty} \dfrac{dx}{x^{3/2}} $ shows a p-type integral with $\dfrac{3}{2} \gt 1$. Thus, the integral $\int_{1}^{\infty} \dfrac{dx}{x^{3/2}}$ converges. Therefore, by the comparison test, the integral $\int_{1}^{\infty} \dfrac{dx}{\sqrt {x^{1/3}+x^3}}$ converges as well.
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