University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 24



Work Step by Step

Re-arrange the given integral as follows: $\int_{-\infty}^{0} 2x e^{-x^2} dx+\int_{0}^{\infty} 2x e^{-x^2} dx=\lim\limits_{p \to -\infty}\int_{p}^{0} 2x e^{-x^2} dx+\lim\limits_{q \to \infty}\int_{0}^{q} 2x e^{-x^2} dx$ ....(1) Plug $u(x)=x^2 \implies du=2x dx$ and $u(p)=p^2; u(q)=q^2$ Then, we have $\int_{p^2}^{1} e^{-u} du+\int_{1}^{q^2} e^{-u} du=-1+e^{-p^2}+(-e^{-q^2}+1)$ Thus, equation (1) becomes $\lim\limits_{p \to -\infty}(-1+e^{-p^2})+\lim\limits_{q \to \infty}(-e^{-q^2}+1)=-1+1=0$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.