(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 ] .

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Chapter 5.1, Problem 53E

(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 52E

Chapter 5.1, Problem 54E

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Ch. 5.1 - What is the displacement of an object that travels...Ch. 5.1 - Prob. 2QC Ch. 5.1 - If the interval [1, 9] is partitioned into 4...Ch. 5.1 - Prob. 4QC Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - The velocity in ft/s of an object moving along a...Ch. 5.1 - Prob. 6E

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

Solution Summary: The author explains the form of Riemann sum, sigma notation, and midpoint sum for n=75.

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

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 Solutions