Explanation Given: ∫ 2 3 d x ln x , N=5 We will apply midpoint sum formula. ∫ a b f ( x ) d x ≅ Δ x ( f ( x 0 + x 1 2 ) + f ( x 1 + x 2 2 ) + f ( x 2 + x 3 2 ) + … + f ( x n − 2 + x n − 1 2 ) + f ( x n − 1 + x n 2 ) ) Where, Δ x = b − a n . Calculations: We are given: f ( x ) = 1 ln x , n = 5 , a = 2 and b = 3 First, we will find Δ x , for that we will plug n = 5 , a = 2 and b = 3 into Δ x formula. Then, Δ x = 3 − 2 5 . ⇨ Δ x = 1 5 Hence, our interval [2, 3] is dividing into n=5 subintervals with length Δ x = 1 5 as: [ ( 2 , 11 5 ) , ( 11 5 , 12 5 ) , ( 12 5 , 13 5 ) , ( 13 5 , 14 5 ) , ( 14 5 , 3 ) ] . Hence, we got x 0 = 2 , x 1 = 11 5 , x 2 = 12 5 , x 3 = 13 5 , x 4 = 14 5 , x 5 = 3. . Now, we will calculate function’s value at midpoint of each interval. f ( x 0 + x 1 2 ) = f ( ( 2 ) + ( 11 5 ) 2 ) = f ( 21 10 ) = 1 − ln ( 10 ) + ln ( 21 ) = 1... To determine T N :

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 8.8, Problem 10E

To determine

$M_{N}$

To determine

$T_{N}$ :

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Chapter 8.8, Problem 9E

Chapter 8.8, Problem 11E

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Chapter 8 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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