Explanation Given information: The integral function is ∫ 0 ∞ x 2 e − x 2 d x . Calculation: Show the integral function as follows: ∫ 0 ∞ x 2 e − x 2 d x (1) Show the definition of an Improper Integral of Type 1 : (a) If ∫ a t f ( x ) d x exists for every number t ≥ a , then ∫ a ∞ f ( x ) d x = lim t → ∞ ∫ a t f ( x ) d x Provided this limit exists (as a finite number). (b) If ∫ t b f ( x ) d x exists for every number t ≤ b , then ∫ − ∞ b f ( x ) d x = lim t → − ∞ ∫ t b f ( x ) d x Provided this limit exists (as a finite number). The improper integrals ∫ a ∞ f ( x ) d x and ∫ − ∞ b f ( x ) d x are called convergent if the corresponding limit exists. The improper integrals ∫ a ∞ f ( x ) d x and ∫ − ∞ b f ( x ) d x are called divergent if the corresponding limit does not exist. (c) If both the integral function ∫ a ∞ f ( x ) d x and ∫ − ∞ a f ( x ) d x are convergent, then ∫ − ∞ ∞ f ( x ) d x = ∫ − ∞ a f ( x ) d x + ∫ a ∞ f ( x ) d x Here, a is any real number. Apply Part (a) of the definition of improper integral in Equation (1). ∫ 0 ∞ x 2 e − x 2 d x = lim t → ∞ ∫ 0 t x 2 e − x 2 d x (2) Consider the value of the function u = x (3) Differentiate both sides of Equation (3) with respect to x . d u d x = d d x ( x ) = 1 d u = d x Consider the value of the function d v = x e − x 2 d x (4) Integrate Equation (4). ∫ d v = ∫ x e − x 2 d x v = ∫ x e − x 2 d x (5) Consider t = − x 2 (6) Differentiate both sides of Equation (6) with respect to x . d t d x = d d x ( − x 2 ) = − 2 x − d t 2 = x d x Substitute − d t 2 for x d x and t for x 2 in Equation (5). v = ∫ x e − x 2 d x = ∫ e t ( − d t 2 ) = − 1 2 ∫ e t d t = − 1 2 e t (7) Substitute − x 2 for t in Equation (7). v = − 1 2 e − x 2 Show the formula of integration by parts. ∫ a b u v = [ u v ] a b − ∫ a b v d u (8) Substitute x for u , − 1 2 e − x 2 for v , d x for d u , x e − x 2 d x for d v , 0 for a and t for b in Equation (6). ∫ 0 t x e − x 2 d x = [ x × ( − 1 2 e − x 2 ) ] 0 t − ∫ 0 t ( − 1 2 e − x 2 ) d x = [ − 1 2 x e − x 2 ] 0 t + 1 2 ∫ 0 t e − x 2 d x = [ − 1 2 t e − t 2 + 0 ] + 1 2 ∫ 0 t e − x 2 d x = − 1 2 t e − t 2 + 1 2 ∫ 0 t e − x 2 d x (9) Rearrange Equation (2)

Single Variable Calculus: Early Transcendentals

8th Edition

ISBN: 9781305270336

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 7.8, Problem 77E

To determine

To show: the integral function ∫0∞x2e−x2dx=12∫0∞e−x2dx.

To evaluate: The integral function, if it is convergent.

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Chapter 7.8, Problem 76E

Chapter 7.8, Problem 78E

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Chapter 7 Solutions

Single Variable Calculus: Early Transcendentals

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the integral function ∫ 0 ∞ x 2 e − x 2 d x = 1 2 ∫ 0 ∞ e − x 2 d x . To evaluate: The integral function, if it is convergent.

Solution Summary: The author evaluates the integral function, if it is convergent, and shows the definition of an Improper Integral of Type 1.

Videos

To show: the integral function ∫0∞x2e−x2dx=12∫0∞e−x2dx.

To evaluate: The integral function, if it is convergent.

Want to see the full answer?

Chapter 7 Solutions

the integral function ∫ 0 ∞ x 2 e − x 2 d x = 1 2 ∫ 0 ∞ e − x 2 d x . To evaluate: The integral function, if it is convergent.

Solution Summary: The author evaluates the integral function, if it is convergent, and shows the definition of an Improper Integral of Type 1.

Videos

To show: the integral function ∫0∞x2e−x2dx=12∫0∞e−x2dx. To evaluate: The integral function, if it is convergent.

Want to see the full answer?

Chapter 7 Solutions

To show: the integral function ∫0∞x2e−x2dx=12∫0∞e−x2dx.

To evaluate: The integral function, if it is convergent.