Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.8 Improper Integrals - 7.8 Exercises - Page 574: 1

Answer

$(a)$ infinite discontinuity $(b)$ Infinite interval $(c)$ Infinite interval $(d)$ Infinite discontinuity

Work Step by Step

$(a)\;\; I=\int_{1}^{2}\frac{x}{x-1}dx$ _____(1) Because integrand $\frac{x}{x-1}$ is discontinuous at $x=1$ That's why (1) is Type 2 improper integral. $(b)\;\; \int_{0}^{\infty}\frac{1}{1+x^3}dx$ ____(2) Because interval of integration is infinite That's why (2) is Type (1) improper integral. $(c)\;\; \int_{-\infty}^{\infty}x^2\:e^{-x^2}dx$ ____(3) Because interval of integration is infinte That's why (3) is Type (1) improper integral. $(d)\;\; \int_{0}^{\frac{π}{4}}\cot xdx=\int_{0}^{\frac{π}{4}}\frac{\cos x}{\sin x}dx$ __(4) Since $\sin (0)=0$ So integrand $\cot x$ is discontinuous at $x=0$ That's why (4) is Type (2) improper integral.
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