Explanation Given: The provided function is, f ( y ) = y 3 + 1 , in the interval [ 1 , 2 ] . Formula used: The sum of squares of first n natural numbers is given by the formula, ∑ i = 1 n i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 The sum of first n natural numbers is given by the formula, ∑ i = 1 n i = n ( n + 1 ) 2 Sum of a constant n times is written as, ∑ i = 1 n c = n c The sum of cubes of first n natural numbers is given by the formula, ∑ i = 1 n i 3 = n 2 ( n + 1 ) 2 4 Using upper endpoints area is, A = lim n → ∞ ∑ i = 1 n f ( M i ) ( Δ x ) Here, M i is the upper endpoints. Calculation: The provided function is, f ( y ) = y 3 + 1 We will differentiate f ( y ) with respect to y and obtain f ' ( y ) , f ' ( y ) = 3 y 2 Now, equate f ' ( y ) to zero to get, 3 y 2 = 0 y = 0 So, the slope of the function w.r.t y -axis is zero at one point. Check where the curve is approach by put y = ∞ and y = − ∞ . On put y = ∞ f ( ∞ ) = ∞ 3 + 1 So, f ( ∞ ) is approach to ∞ . Now put y = − ∞ f ( − ∞ ) = − ∞ 3 + 1 That means the curve is approaching to − ∞ as y → − ∞ . Therefore, the required graph with the region is shown below, As the given function f ( y ) is continuous and non-negative in the interval [ 1 , 2 ] . So, by the partition of interval into n subintervals each of equal width. Therefore, considering the width of each rectangle to be ( Δ y ) then, The width of each rectangle is, Δ y = 2 − 1 n = 1 n The area can be calculated by the use of lower endpoints ( m i ) or upper endpoints ( M i ) . For this problem the upper endpoints are convenient. Now, find upper endpoints ( M i ) . M i = 1 + i n = i + n n So, the upper endpoint is M i = i + n n . Therefore, apply the formula to find the area ( A ) of the region, A = lim n → ∞ ∑ i = 1 n f ( M i ) ( Δ y ) Now, substitute the values to find the area of the region

Student Solutions Manual for Larson/Edwards' Calculus of a Single Variable, 11th

11th Edition

ISBN: 9781337275385

Author: Larson, Ron, Edwards, Bruce H.

Publisher: Brooks Cole

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Question

Chapter 4.2, Problem 62E

To determine

To calculate: The area of the region bounded by the function f(y)=y3+1, in the interval [1,2] and the y -axis.

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Chapter 4.2, Problem 61E

Chapter 4.2, Problem 63E

Chapter 4 Solutions

Student Solutions Manual for Larson/Edwards' Calculus of a Single Variable, 11th

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The area of the region bounded by the function f ( y ) = y 3 + 1 , in the interval [ 1 , 2 ] and the y -axis.

Solution Summary: The author explains how to calculate the area of the region bounded by the function f(y)=y

To calculate: The area of the region bounded by the function f(y)=y3+1, in the interval [1,2] and the y -axis.

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Chapter 4 Solutions