Explanation Since sin x increases in the interval [ 0.2 , 0.3 ] , we can use the equation, 0.2 ≤ x ≤ 0.3 to prove that 0.198 ≤ sin x ≤ 0.296 and then apply integrals to the equation to finally conclude that 0.0198 ≤ ∫ 0.2 0.3 sin x d x ≤ 0.0296 Given: 0.0198 ≤ ∫ 0.2 0.3 sin x d x ≤ 0.0296 and a hint that we need to prove that 0.198 ≤ sin x ≤ 0.296 for x in [ 0.2 , 0.3 ] . Calculation: For 0 ≤ x ≤ π 6 ≈ 0.52 , we have d d x ( sin x ) = cos x > 0 . Hence, it is observed that sin x increases in the interval [ 0.2 , 0.3 ] . Therefore, applying sin to 0.2 ≤ x ≤ 0.3 , we have sin 0.2 ≤ sin x ≤ sin 0.3 Since sin 0...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Confidence Intervals

When we calculate intervals in probability and statistics, it can be simply termed as finding out a confidence interval. The confidence interval is usually calculated from the data set which is observed. The value of the parameter that needs to be calcul…

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Chapter 5.2, Problem 77E

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To prove that $0.0198 \leq \int_{0.2}^{0.3} \sin x d x \leq 0.0296$

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Chapter 5.2, Problem 76E

Chapter 5.2, Problem 78E

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Chapter 5 Solutions

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To prove that 0.0198 ≤ ∫ 0.2 0.3 sin x d x ≤ 0.0296

Solution Summary: The author explains that since sin x increases in the interval, they can use the equation to prove that 0.198 0.2 0.3 sin

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To prove that $0.0198 \leq \int_{0.2}^{0.3} \sin x d x \leq 0.0296$

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To prove that 0.0198 ≤ ∫ 0.2 0.3 sin x d x ≤ 0.0296

Solution Summary: The author explains that since sin x increases in the interval, they can use the equation to prove that 0.198 0.2 0.3 sin

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To prove that 0.0198≤∫0.20.3sinxdx≤0.0296

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To prove that $0.0198 \leq \int_{0.2}^{0.3} \sin x d x \leq 0.0296$