Explanation Given information: 1) The interval 0 ≤ x ≤ 1 is divided into 100 subintervals of width Δ x = 0.01 . 2) The sum is [ 2 ( 0.01 ) + ( 0.01 ) 3 ] Δ x + [ 2 ( 0.02 ) + ( 0.02 ) 3 ] Δ x + ... + [ 2 ( 1.0 ) + ( 1.0 ) 3 ] 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 ) + ..

MyLab Math with Pearson eText -- 24 Month Access -- for Calculus & Its Applications

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ISBN: 9780137590469

Author: Larry Goldstein / David Lay

Publisher: Pearson Education (US)

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Question

Chapter 6.5, Problem 42E

To determine

To prove: The sum [2(0.01)+(0.01)3]Δx+[2(0.02)+(0.02)3]Δx+...+[2(1.0)+(1.0)3]Δx is close to 54, where the interval 0≤x≤1 is divided into 100 subintervals of width Δx=0.01.

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Chapter 6.5, Problem 41E

Chapter 6.5, Problem 43E

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Chapter 6 Solutions

MyLab Math with Pearson eText -- 24 Month Access -- for Calculus & Its Applications

Ch. 6.1 - Determine the following: a. t7/2dt b....Ch. 6.1 - Find a function f(t) that satisfies f(t)=3t+5 and...Ch. 6.1 - Find all antiderivatives of each following...Ch. 6.1 - Find all antiderivatives of each following...Ch. 6.1 - Find all antiderivatives of each following...Ch. 6.1 - Find all antiderivatives of each following...Ch. 6.1 - Find all antiderivatives of each following...Ch. 6.1 - Find all antiderivatives of each following...Ch. 6.1 - Determine the following: 4x3dx Ch. 6.1 - Determine the following: 13xdx

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To prove: The sum [ 2 ( 0.01 ) + ( 0.01 ) 3 ] Δ x + [ 2 ( 0.02 ) + ( 0.02 ) 3 ] Δ x + ... + [ 2 ( 1.0 ) + ( 1.0 ) 3 ] Δ x is close to 5 4 , 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: The sum [2(0.01)+(0.01)3]Δx+[2(0.02)+(0.02)3]Δx+...+[2(1.0)+(1.0)3]Δx is close to 54, where the interval 0≤x≤1 is divided into 100 subintervals of width Δx=0.01.

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Chapter 6 Solutions