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Math Calculus University Calculus: Early Transcendentals, Single Variable (3rd Edition)

a. To find: The area under the graph of function by using a lower sum with two rectangles of equal width. The approximate area under the graph of function by using a lower sum with two rectangles of equal width is 0.125 sq. units. Given information: f x = x 2 between x = 0 and x = 1 . Concept used: The formula for a lower Riemann sum is ∑ i = 1 n Δ f x i , where Δ x = b − a n [ n = Number of rectangles, a = lower limit, b = upper limit ] is the width of each rectangle and f x i gives the height of each rectangle. Explanation: Here, the width of each rectangle will be given by: Δ x = 1 − 0 2 = 1 2 Thus, x 1 = 0 , x 2 = 1 2 . ∑ i = 1 n Δ x f x i = 1 2 0 2 + 1 2 2 = 1 2 3 = 1 8 = 0.125 Hence, the area under the graph of function by using a lower sum with two rectangles of equal width is 0.125 sq. units. b. To find: The area under the graph of function by using a lower sum with four rectangles of equal width. The approximate area under the graph of function by using a lower sum with four rectangles of equal width is 0.21875sq. units. Given information: f x = x 2 between x = 0 and x = 1 . Concept used: The formula for a lower Riemann sum is ∑ i = 1 n Δ f x i , where Δ x = b − a n [ n = Number of rectangles, a = lower limit, b = upper limit ] is the width of each rectangle and f x i gives the height of each rectangle. Explanation: Here, the width of each rectangle will be given by: Δ x = 1 − 0 4 = 1 4 Thus, x 1 = 0 , x 2 = 1 4 , x 3 = 1 2 , and x 4 = 3 4 . ∑ i = 1 n Δ x f x i = 1 4 0 2 + 1 4 2 + 1 2 2 + 3 4 2 = 1 4 1 16 + 1 4 + 9 16 = 1 4 14 16 = 7 32 = 0.21875 Hence, the area under the graph of function by using a lower sum with four rectangles of equal width is 0.21875 sq. units. c. To find: The area under the graph of function by using an upper sum with two rectangles of equal width. The approximate area under the graph of function by using an upper sum with two rectangles of equal width is 0.21875sq. units. Given information: f x = x 2 between x = 0 and x = 1 . Concept used: The formula for an upper Riemann sum is ∑ i = 1 n Δ f x i , where Δ x = b − a n [ n = Number of rectangles, a = lower limit, b = upper limit ] is the width of each rectangle and f x i gives the height of each rectangle. Explanation: Here, the width of each rectangle will be given by: Δ x = 1 − 0 2 = 1 2 Thus, x 1 = 1 2 and x 2 = 1 . ∑ i = 1 n Δ x f x i = 1 2 1 2 2 + 1 2 = 1 2 1 4 + 1 = 1 2 5 4 = 5 8 = 0.625 Hence, the area under the graph of function by using an upper sum with two rectangles of equal width is 0.625 sq. units. b. To find: The area under the graph of function by using a lower sum with four rectangles of equal width. The approximate area under the graph of function by using an upper sum with four rectangles of equal width is 0.21875sq. units. Given information: f x = x 2 between x = 0 and x = 1 . Concept used: The formula for an upper Riemann sum is ∑ i = 1 n Δ f x i , where Δ x = b − a n [ n = Number of rectangles, a = lower limit, b = upper limit ] is the width of each rectangle and f x i gives the height of each rectangle. Explanation: Here, the width of each rectangle will be given by: Δ x = 1 − 0 4 = 1 4 Thus, x 1 = 1 4 , x 2 = 1 2 , x 3 = 3 4 , and x 4 = 1 . ∑ i = 1 n Δ x f x i = 1 4 1 4 2 + 1 2 2 + 3 4 2 + 1 2 = 1 4 1 16 + 1 4 + 9 16 + 1 = 1 4 30 16 = 15 32 = 0.46875 Hence, the area under the graph of function by using an upper sum with four rectangles of equal width is 0.46875 sq. units.

a. To find: The area under the graph of function by using a lower sum with two rectangles of equal width. The approximate area under the graph of function by using a lower sum with two rectangles of equal width is 0.125 sq. units. Given information: f x = x 2 between x = 0 and x = 1 . Concept used: The formula for a lower Riemann sum is ∑ i = 1 n Δ f x i , where Δ x = b − a n [ n = Number of rectangles, a = lower limit, b = upper limit ] is the width of each rectangle and f x i gives the height of each rectangle. Explanation: Here, the width of each rectangle will be given by: Δ x = 1 − 0 2 = 1 2 Thus, x 1 = 0 , x 2 = 1 2 . ∑ i = 1 n Δ x f x i = 1 2 0 2 + 1 2 2 = 1 2 3 = 1 8 = 0.125 Hence, the area under the graph of function by using a lower sum with two rectangles of equal width is 0.125 sq. units. b. To find: The area under the graph of function by using a lower sum with four rectangles of equal width. The approximate area under the graph of function by using a lower sum with four rectangles of equal width is 0.21875sq. units. Given information: f x = x 2 between x = 0 and x = 1 . Concept used: The formula for a lower Riemann sum is ∑ i = 1 n Δ f x i , where Δ x = b − a n [ n = Number of rectangles, a = lower limit, b = upper limit ] is the width of each rectangle and f x i gives the height of each rectangle. Explanation: Here, the width of each rectangle will be given by: Δ x = 1 − 0 4 = 1 4 Thus, x 1 = 0 , x 2 = 1 4 , x 3 = 1 2 , and x 4 = 3 4 . ∑ i = 1 n Δ x f x i = 1 4 0 2 + 1 4 2 + 1 2 2 + 3 4 2 = 1 4 1 16 + 1 4 + 9 16 = 1 4 14 16 = 7 32 = 0.21875 Hence, the area under the graph of function by using a lower sum with four rectangles of equal width is 0.21875 sq. units. c. To find: The area under the graph of function by using an upper sum with two rectangles of equal width. The approximate area under the graph of function by using an upper sum with two rectangles of equal width is 0.21875sq. units. Given information: f x = x 2 between x = 0 and x = 1 . Concept used: The formula for an upper Riemann sum is ∑ i = 1 n Δ f x i , where Δ x = b − a n [ n = Number of rectangles, a = lower limit, b = upper limit ] is the width of each rectangle and f x i gives the height of each rectangle. Explanation: Here, the width of each rectangle will be given by: Δ x = 1 − 0 2 = 1 2 Thus, x 1 = 1 2 and x 2 = 1 . ∑ i = 1 n Δ x f x i = 1 2 1 2 2 + 1 2 = 1 2 1 4 + 1 = 1 2 5 4 = 5 8 = 0.625 Hence, the area under the graph of function by using an upper sum with two rectangles of equal width is 0.625 sq. units. b. To find: The area under the graph of function by using a lower sum with four rectangles of equal width. The approximate area under the graph of function by using an upper sum with four rectangles of equal width is 0.21875sq. units. Given information: f x = x 2 between x = 0 and x = 1 . Concept used: The formula for an upper Riemann sum is ∑ i = 1 n Δ f x i , where Δ x = b − a n [ n = Number of rectangles, a = lower limit, b = upper limit ] is the width of each rectangle and f x i gives the height of each rectangle. Explanation: Here, the width of each rectangle will be given by: Δ x = 1 − 0 4 = 1 4 Thus, x 1 = 1 4 , x 2 = 1 2 , x 3 = 3 4 , and x 4 = 1 . ∑ i = 1 n Δ x f x i = 1 4 1 4 2 + 1 2 2 + 3 4 2 + 1 2 = 1 4 1 16 + 1 4 + 9 16 + 1 = 1 4 30 16 = 15 32 = 0.46875 Hence, the area under the graph of function by using an upper sum with four rectangles of equal width is 0.46875 sq. units.

University Calculus: Early Transcendentals, Single Variable (3rd Edition)

University Calculus: Early Transcendentals, Single Variable (3rd Edition)

3rd Edition

ISBN: 9780321999634

Author: Joel R. Hass, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 4, Problem 35AAE

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University Calculus: Early Transcendentals, Single Variable (3rd Edition)

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