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Solving the above for dx we have: du = dx. When x = 1: we have u = 1.414 When x = e: we have u = 1 When u = V2 - In(x): we have x = By substitution we have: dx = V2 - In(x) du du. Letting u = x we see that the two integrals are equivalent and, thus, we have demonstrated that the area under the graphs of f and g are the same.

Solving the above for dx we have: du = dx. When x = 1: we have u = 1.414 When x = e: we have u = 1 When u = V2 - In(x): we have x = By substitution we have: dx = V2 - In(x) du du. Letting u = x we see that the two integrals are equivalent and, thus, we have demonstrated that the area under the graphs of f and g are the same.

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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Solving the above for dx we have: du = dx. When x = 1: we have u = 1.414 When x = e: we have u = 1 When u = V2 - In(x): we have x = By substitution we have: dx = V2 - In(x) du V2 du. Letting u = x we see that the two integrals are equivalent and, thus, we have demonstrated that the area under the graphs of f and g are the same.
Transcribed Image Text:Solving the above for dx we have: du = dx. When x = 1: we have u = 1.414 When x = e: we have u = 1 When u = V2 - In(x): we have x = By substitution we have: dx = V2 - In(x) du V2 du. Letting u = x we see that the two integrals are equivalent and, thus, we have demonstrated that the area under the graphs of f and g are the same.
We wish to employ a substitution to demonstrate that the area under the graphs off and g are the same. f(x) = V2 - In(x) on [1, e]; g(x) = 2e2 - x on [1, V21 By inspection we can determine the following. f(x) = V2 - In(x) is positive or nonnegative on [1, e]. g(x) = 2e2 - x* is positive or nonnegative on [1, /2). Therefore, our goal is to show that 1 | 72- In (x) dx = 2e2-2 dx. We will do this by employing a substitution in a manner which will allow us to rewrite the integral on the left side so that it is identical to the integral on the right side. Let u = V2 - In(x). -u²+2 e Then x = So by the chain rule, 1 du = dx. 2x/2 – In(x) Solving the above for dx we have: du = dx.
Transcribed Image Text:We wish to employ a substitution to demonstrate that the area under the graphs off and g are the same. f(x) = V2 - In(x) on [1, e]; g(x) = 2e2 - x on [1, V21 By inspection we can determine the following. f(x) = V2 - In(x) is positive or nonnegative on [1, e]. g(x) = 2e2 - x* is positive or nonnegative on [1, /2). Therefore, our goal is to show that 1 | 72- In (x) dx = 2e2-2 dx. We will do this by employing a substitution in a manner which will allow us to rewrite the integral on the left side so that it is identical to the integral on the right side. Let u = V2 - In(x). -u²+2 e Then x = So by the chain rule, 1 du = dx. 2x/2 – In(x) Solving the above for dx we have: du = dx.
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