Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 8 - Techniques of Integration - 8.7 Improper Integrals - Exercises - Page 441: 54

Answer

The integral $\int_{1}^{\infty} \dfrac{dx}{x^2}$ converges to $1$. Hence, the integral $\int_1^\infty \dfrac{dx}{\sqrt {1+x^4}}$ converges as well by the comparison test.

Work Step by Step

Since, $\sqrt {x^4+1} \geq \sqrt {x^4}(=x^2)$, it follows that$\dfrac{1}{\sqrt {x^4+1}}\leq \dfrac{1}{x^2}$ We will compute the integral as follows: $\int_{1}^{\infty} \dfrac{dx}{x^2}= \lim\limits_{R \to \infty} \int_{1}^R \dfrac{dx}{x^2} \\=\lim\limits_{R \to \infty} [-\dfrac{1}{x}]_1^R\\=\lim\limits_{R \to \infty}(-\dfrac{1}{R}+1)\\=1$ So, the integral $\int_{1}^{\infty} \dfrac{dx}{x^2}$ converges to $1$. Hence, the integral $\int_1^\infty \dfrac{dx}{\sqrt {1+x^4}}$ converges as well by the comparison test.

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