(a) Explanation Given: The function f ( x ) = x 2 − 1 defined on interval [ 2 , 7 ] . Formula used: (1). The left Riemann sum for the function f ( x ) defined on the interval [ a , b ] for n subinterval is ∑ k = 1 n f ( x k * ) Δ x = ∑ k = 1 n f [ a + ( k − 1 ) Δ x ] Δ x . (2). The right Riemann sum for the function f ( x ) defined on the interval [ a , b ] for n subinterval is ∑ k = 1 n f ( x k * ) Δ x = ∑ k = 1 n f [ a + ( k ) Δ x ] Δ x . (3). The midpoint Riemann sum for the function f ( x ) defined on the interval [ a , b ] for n subinterval is ∑ k = 1 n f ( x k * ) Δ x = ∑ k = 1 n f [ a + ( k − 1 2 ) Δ x ] Δ x Calculation: First find the length of the each subinterval, Δ x = 7 − 2 75 = 5 75 = 1 15 Then, the subintervals are as follows. [ 2 , 31 15 ] , [ 31 15 , 32 15 ] , [ 32 15 , 33 15 ] , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ , [ 104 15 , 7 ] The left end points of the above subintervals are 2 , 31 15 , 32 15 , 33 15 , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ , 104 15 and the right end points are 31 15 , 32 15 , 33 15 , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ , 104 15 , 7 and the mid points are 61 30 , 63 30 , 65 30 , ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ , 209 30 . Computation of the left Riemann sum as follows. ∑ k = 1 75 f ( x k * ) Δ x = ∑ k = 1 75 f [ 2 + ( k − 1 ) Δ x ] Δ x [ ∵ x k * = a + ( k − 1 ) Δ x ] = ∑ k = 1 75 f [ 2 + ( k − 1 ) 1 15 ] × 1 15 = ∑ k = 1 75 [ ( 2 + ( k − 1 ) 1 15 ) 2 − 1 ] × 1 15 [ ∵ f ( x ) = x 2 − 1 ] = ∑ k = 0 74 [ ( 2 + ( k 15 ) ) 2 − 1 ] × 1 15 On further simplification, ∑ k = 1 75 f ( x k * ) Δ x = ∑ k = 0 74 [ ( 2 + ( k 15 ) ) 2 − 1 ] × 1 15 = [ ( 2 ) 2 + ( 2 + ( 1 15 ) ) 2 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + ( 2 + ( 74 15 ) ) 2 ] × 1 15 − 75 15 = [ 4 + ( 31 15 ) 2 + ( 32 15 ) 2 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + ( 104 15 ) 2 ] × 1 15 − 5 = 105.170 Computation of the right Riemann sum as follows. ∑ k = 1 75 f ( x k * ) Δ x = ∑ k = 1 75 f [ 2 + ( k ) Δ x ] Δ x [ ∵ x k * = a + ( k ) Δ x ] = ∑ k = 1 75 f [ 2 + ( k ) 1 15 ] × 1 15 = ∑ k = 1 75 [ ( 2 + ( k ) 1 15 ) 2 − 1 ] × 1 15 [ ∵ f ( x ) = x 2 − 1 ] = ∑ k = 1 75 [ ( 2 + ( k 15 ) ) 2 + 1 ] × 1 15 On further simplification, ∑ k = 1 75 f ( x k * ) Δ x = ∑ k = 1 75 [ ( 2 + ( k 15 ) ) 2 − 1 ] × 1 15 = [ ( 2 + ( 1 15 ) ) 2 + ( 2 + ( 2 15 ) ) 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + ( 2 + ( 74 15 ) ) 2 + ( 2 + ( 75 15 ) ) 2 ] × 1 15 − 75 15 = [ ( 31 15 ) 2 + ( 32 15 ) 2 + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + ( 104 15 ) 2 + 7 2 ] × 1 15 − 5 = 108 (b) To determine To estimate: The area of the region bounded by the graph of the function f ( x ) and the x -axis on the interval [ 2 , 7 ] .

Single Variable Calculus: Early Transcendentals (2nd Edition) - Standalone book

2nd Edition

ISBN: 9780321954237

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Question

Chapter 5.1, Problem 45E

(a)

To determine

To write: The left Riemann sum, right Riemann sum and the midpoint Riemann sum in the form of sigma notation for n=75 .

(b)

To determine

To estimate: The area of the region bounded by the graph of the function f(x) and the x -axis on the interval [2,7] .

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Chapter 5.1, Problem 44E

Chapter 5.1, Problem 46E

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Complete the following steps for the given functionf and interval. a. For the given value of n, use sigma notation to write the left, right, and midpoint Riemann sums. Then evaluate each sum using a calculator. b. Based on the approximations found in part (a), estimate the area of the region bounded by the graph of f on the interval. f(x) = /x for [0,20]; n= 40 a. Write the left Riemann sum. Choose the correct answer below. 40 1 k- O A. Σ k= 1 40 1 2k - 1 OB. 4. k= 1 40 1 k+1 Σ 40 k= 1 40 k= 1

Please find the left rieman, right rieman and the midpoint

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