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Use the value of the first integral I to evaluate the two given integrals. a. 3 || = √(x³ - 6x) dx = - 3 0 0 (12x-2x³) dx - 27 4 0 - b. •. S (6x − x³) dx 3 3 a. | (12x − 2x³) dx = | (੧) 0 0 (Simplify your answer.) b. S (6x-x³) dx = | (Simplify your answer.) 3

Use the value of the first integral I to evaluate the two given integrals. a. 3 || = √(x³ - 6x) dx = - 3 0 0 (12x-2x³) dx - 27 4 0 - b. •. S (6x − x³) dx 3 3 a. | (12x − 2x³) dx = | (੧) 0 0 (Simplify your answer.) b. S (6x-x³) dx = | (Simplify your answer.) 3

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Use the value of the first integral I to evaluate the two given integrals. a. 3 || = √(x³ - 6x) dx = - 3 0 0 (12x-2x³) dx - 27 4 0 - b. •. S (6x − x³) dx 3 3 a. | (12x − 2x³) dx = | (੧) 0 0 (Simplify your answer.) b. S (6x-x³) dx = | (Simplify your answer.) 3
Transcribed Image Text:Use the value of the first integral I to evaluate the two given integrals. a. 3 || = √(x³ - 6x) dx = - 3 0 0 (12x-2x³) dx - 27 4 0 - b. •. S (6x − x³) dx 3 3 a. | (12x − 2x³) dx = | (੧) 0 0 (Simplify your answer.) b. S (6x-x³) dx = | (Simplify your answer.) 3
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