Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 534: 10

Answer

$$Convergent,\frac{1}{ln2}$$

Work Step by Step

$$\int_{-\infty }^{0}2^{r}dr=\lim_{t\rightarrow -\infty}\int_{t}^{0}2^{r}dr$$ $$=\lim_{t\rightarrow -\infty}\left| \frac{2^{r}}{ln2} \right |_{t}^{0}$$ $$= \lim_{t\rightarrow -\infty} \frac{1-2^{t}}{ln2}$$ $$=\frac{1}{ln2}$$ Convergent.
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