 # (a) We define the improper integral (over the entire plane ℝ 2 ) I = ∬ R 2 e − ( x 2 + y 2 ) d A = ∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 ) d y d x = lim a → ∞ ∬ D a e − ( x 2 + y 2 ) d A where D a is the disk with radius a and center the origin. Show that ∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 ) d A = π (b) An equivalent definition of the improper integral in part (a) is ∬ R 2 e − ( x 2 + y 2 ) d A = lim a → ∞ ∬ S a e − ( x 2 + y 2 ) d A where S a is the square with vertices ( ±a, ±a ). Use this to show that ∫ − ∞ ∞ e − x 2 d x ∫ − ∞ ∞ e − y 2 d y = π (c) Deduce that ∫ − ∞ ∞ e − x 2 d x = π (d) By making the change of variable t = 2 x , show that ∫ − ∞ ∞ e − x 2 / 2 d x = 2 π (This is a fundamental result for probability and statistics.) ### Calculus: Early Transcendentals

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285741550 ### Calculus: Early Transcendentals

8th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781285741550

#### Solutions

Chapter
Section
Chapter 15.3, Problem 40E
Textbook Problem

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