Explanation Given information: The interval 0 ≤ x ≤ 1 is divided into 100 subintervals of width Δ x = 0.01 . The sum is [ 3 e − 0.01 ] Δ x + [ 3 e − 0.02 ] Δ x + [ 3 e − 0.03 ] + ..... + [ 3 e − 1 ] Δ x . Formula used: The fundamental theorem of calculus: Let f be a continuous function on the interval [ a , b ] , with an antiderivative F . Then the Riemann sum f ( x 1 ) Δ x + f ( x 2 ) Δ x + ... + f ( x n ) Δ x approaches the definite integral of f on [ a , b ] , as the number of subintervals increases indefinitely. That is, lim Δ x → 0 [ f ( x 1 ) + f ( x 2 ) + ... + f ( x n ) ] Δ x = ∫ a b f ( x ) d x = F ( b ) − F ( a ) Proof: Let the interval 0 ≤ x ≤ 1 is divided into 100 subintervals of width Δ x = 0.01 . The sum is [ 3 e − 0.01 ] Δ x + [ 3 e − 0

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ISBN: 9780137590810

Author: Larry J. Goldstein, David C. Lay, David I. Schneider, Nakhle H. Asmar, William Edward Tavernetti

Publisher: PEARSON

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Chapter 6, Problem 54RE

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To prove: [3e−0.01]Δx+[3e−0.02]Δx+ [3e−0.03] +.....+[3e−1]Δx= 3(1 − e−1), where the interval 0≤x≤1 is divided into 100 subintervals of width Δx=0.01.

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Chapter 6, Problem 53RE

Chapter 6, Problem 55RE

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To prove: [ 3 e − 0.01 ] Δ x + [ 3 e − 0.02 ] Δ x + [ 3 e − 0.03 ] + ..... + [ 3 e − 1 ] Δ x = 3 ( 1 − e − 1 ) , where the interval 0 ≤ x ≤ 1 is divided into 100 subintervals of width Δ x = 0.01 .

Solution Summary: The author explains the fundamental theorem of calculus: Let f be a continuous function on the interval

To prove: [3e−0.01]Δx+[3e−0.02]Δx+ [3e−0.03] +.....+[3e−1]Δx= 3(1 − e−1), where the interval 0≤x≤1 is divided into 100 subintervals of width Δx=0.01.

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Chapter 6 Solutions