Explanation Definition used 1: “Let f be the function defined on a closed interval [ a , b ] , which is divided into n subintervals of equal length Δ x . If x k * is a point in the k th subinterval [ x k − 1 , x k ] , for k = 1 , 2 , .... n , then the Riemann sum for f on [ a , b ] is ∑ k = 1 n f ( x k * ) Δ x . That is, 1. Right Riemann sum if x k * = a + k Δ x ”. Definition used 2: The limit definition of the definite integral is given by ∫ a b f ( x ) d x = lim n → ∞ ∑ k = 1 n f ( x k * ) Δ x . Theorem: Fundamental theorem of calculus : “If f is continuous on [ a , b ] and F is any anti-derivative of f on [ a , b ] , then ∫ a b f ( x ) d x = F ( b ) − F ( a ) ”. Given: The definite integral ∫ 0 4 ( x 3 − x ) d x . Calculation: According to the given data, a = 0 and b = 4 . Divide the interval [ 0 , 4 ] into n equal subintervals. Calculate the length of the each subinterval as follows. Δ x = b − a n = 4 − 0 n = 4 n Therefore, the length of each sub interval is, Δ x = 4 n . By the definition, Right Riemann sum if x k * = a + k Δ x ”. That is, x k * = 0 + k 4 n x k * = 4 k n Consider ∑ k = 1 n f ( x k * ) Δ x (1) By the above information, the equation (1) can be calculated as follows. ∑ k = 1 n f ( x k * ) Δ x = ∑ k = 1 n ( ( 4 k n ) 3 − 4 k n ) 4 n = ∑ k = 1 n ( 256 k 3 n 4 − 16 k n 2 ) = 256 n 4 ∑ k = 1 n k 3 − 16 n 2 ∑ k = 1 n k

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

3rd Edition

ISBN: 9780134996103

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

Publisher: PEARSON

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Chapter 5, Problem 13RE

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To evaluate: The value of the right Riemann sum by using the limit definition of the definite integral and check the required result by using Fundamental theorem of calculus.

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Chapter 5, Problem 12RE

Chapter 5, Problem 14RE

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(Drag and drop the missing words) An is the of a definite integral as an endpoint of the interval of integration approaches either a specified real number or positive or negative . Although the integral da is , the integral da is improper. In this case, we define the improper integral as a limit dr de lim In general, an improper integral if the limit defining it Thus for example dx does not .It is possible for an improper integral to to infinity. For instance, 00 dr dr = lim C00 is However, there are many improper integrals which diverge in no particular direction, such as z sin zdr oscillates infinity diverge divergent Riemann integral infinite integral exists converge definite integral improper integral limit converges not improper zero 一”

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Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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 value of the right Riemann sum by using the limit definition of the definite integral and check the required result by using Fundamental theorem of calculus.

Solution Summary: The author evaluates the right Riemann sum by using the limit definition of the definite integral and checks the required result using Fundamental theorem of calculus.

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To evaluate: The value of the right Riemann sum by using the limit definition of the definite integral and check the required result by using Fundamental theorem of calculus.

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Chapter 5 Solutions