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Let u = x + 144. Differentiate u in terms of x, du = 2 x dx. %3D Step 2 Rewrite the integration in terms of u. 2x dk = " du V x2 + 144 Step 3 Apply the power rule of integration. < 5 1 du = -1/2 Vu du 1/2 + 1 X = 5 + 1 1x 5 %3D

Let u = x + 144. Differentiate u in terms of x, du = 2 x dx. %3D Step 2 Rewrite the integration in terms of u. 2x dk = " du V x2 + 144 Step 3 Apply the power rule of integration. < 5 1 du = -1/2 Vu du 1/2 + 1 X = 5 + 1 1x 5 %3D

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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Let u = x + 144. Differentiate u in terms of x, du = x dx. Step 2 Rewrite the integration in terms of u. 2x x = 5 dx = du + 144 Step 3 Apply the power rule of integration. 'x = 5 du = - 1/2 du 1/2 + 1 1x = 5 + 1lo 1x = 5 %3D Submit Skip (you cannot come back) Viewing Saved Work Revert to Last Response Submit Answer acer FULL HD 1080►
Transcribed Image Text:Let u = x + 144. Differentiate u in terms of x, du = x dx. Step 2 Rewrite the integration in terms of u. 2x x = 5 dx = du + 144 Step 3 Apply the power rule of integration. 'x = 5 du = - 1/2 du 1/2 + 1 1x = 5 + 1lo 1x = 5 %3D Submit Skip (you cannot come back) Viewing Saved Work Revert to Last Response Submit Answer acer FULL HD 1080►
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