b. False. The interval of integration [-1, 1] contains 0 at which function and the above theorem cannot be applied. is discontinuous

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
ChapterA: Appendix
SectionA.2: Geometric Constructions
Problem 10P: A soda can has a volume of 25 cubic inches. Let x denote its radius and h its height, both in...
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Using the fundamental of calculus f(x) dx = F(b) – F(a), we can evaluate the
following integral as follows A dx = -2
a. True. The interval of integration [-1, 1] contains 0 at which function is discontinuous
and the above theorem can be applied.
b. False. The interval of integration [-1, 1| contains 0 at which function is discontinuous
and the above theorem cannot be applied.
c. False. The interval of integration [1, -1] contains 0 at which function is discontinuous
and the above theorem cannot be applied.
d. False. The interval of integration [-1, 1] contains 0 at which function x² is discontinuous
and the above theorem cannot be applied.
Transcribed Image Text:Using the fundamental of calculus f(x) dx = F(b) – F(a), we can evaluate the following integral as follows A dx = -2 a. True. The interval of integration [-1, 1] contains 0 at which function is discontinuous and the above theorem can be applied. b. False. The interval of integration [-1, 1| contains 0 at which function is discontinuous and the above theorem cannot be applied. c. False. The interval of integration [1, -1] contains 0 at which function is discontinuous and the above theorem cannot be applied. d. False. The interval of integration [-1, 1] contains 0 at which function x² is discontinuous and the above theorem cannot be applied.
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