Explanation Given: a) ∫ π 4 π 2 sin θ d θ M 4 b) n = 4 c) Mid-point approximation. d) a = π 4 and b = π 2 Formula Used: The Midpoint approximation is: ∫ a b f ( θ ) d θ ≈ h × [ f ( θ 0 + θ 1 2 ) + f ( θ 1 + θ 2 2 ) + f ( θ 2 + θ 3 2 ) + f ( θ 3 + θ 4 2 ) + ..... + f ( θ n − 2 + θ n − 1 2 ) + f ( θ n − 1 + θ n 2 ) ] Where, n + 1 Data are equally spaced. h = Width of interval. n = No. of intervals. h = b − a n θ 0 = a θ n = a + n h Calculation: ∫ π 4 π 2 sin θ d θ Where, f ( θ ) = sin θ a = π 4 And b = π 2 h = Width of interval ∴ h = b − a n h = π 2 − π 4 4 = π 16 h = π 16 Dividing the [ π 4 , π 2 ] into five intervals with each interval width of π 16 . So end points are, ∵ θ n = a + n h θ 0 = π 4 + ( 0 × π 16 ) = π 4 θ 1 = π 4 + ( 1 × π 16 ) = 5 π 16 θ 2 = π 4 + ( 2 × π 16 ) = 3 π 8 θ 3 = π 4 + ( 3 × π

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 8, Problem 95CRE

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To determine: approximation of the integral $\int_{\frac{π}{4}}^{\frac{π}{2}} \sqrt{\sin θ} d θ$

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Chapter 8, Problem 94CRE

Chapter 8, Problem 96CRE

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Chapter 8 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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