Explanation Given: Let A be the area under y = sin x over [ 0 , π 2 ] Calculation: Let f ( x ) = sin x over the interval [ 0 , π 2 ] . Because f is an increasing function over this interval, it follows that L 4 ≤ A ≤ R 4 . Taking n = 4, we have Δ x = π 8 and { x i } i = 0 4 = { 0 + i Δ x } = { 0 , π 8 , π 4 , 3 π 8 , π 2 } ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 5.1, Problem 79E

To determine

That the area A under $y = \sin x$ over the interval $[0, \frac{π}{2}]$ satisfies $0.79 \leq A \leq 1.19.$

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Chapter 5.1, Problem 78E

Chapter 5.1, Problem 80E

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Chapter 5 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 5.1 - Prob. 1PQ Ch. 5.1 - Prob. 2PQ Ch. 5.1 - Prob. 3PQ Ch. 5.1 - Prob. 4PQ Ch. 5.1 - Prob. 5PQ Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E

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Knowledge Booster

That the area A under y = sin x over the interval [ 0 , π 2 ] satisfies 0.79 ≤ A ≤ 1.19.

Solution Summary: The author determines that the area A under y = sin x over the interval satisfies 0.79 A 1.19.

That the area A under $y = \sin x$ over the interval $[0, \frac{π}{2}]$ satisfies $0.79 \leq A \leq 1.19.$

Want to see the full answer?

Chapter 5 Solutions

That the area A under y = sin x over the interval [ 0 , π 2 ] satisfies 0.79 ≤ A ≤ 1.19.

Solution Summary: The author determines that the area A under y = sin x over the interval satisfies 0.79 A 1.19.

That the area A under y=sinx over the interval [0, π2] satisfies 0.79≤A≤1.19.

Want to see the full answer?

Chapter 5 Solutions

That the area A under $y = \sin x$ over the interval $[0, \frac{π}{2}]$ satisfies $0.79 \leq A \leq 1.19.$