Part (a): Explanation Given Information: The integral is shown as follows: ∫ 0 ∞ x n e − x d x . Here, n is an arbitrary positive integer. Calculation: Show the integral as shown below: I = ∫ 0 ∞ x n e − x d x (1) Calculate the value of the integral for n = 0 . Substitute 0 for n in Equation (1). I = ∫ 0 ∞ x 0 e − x d x = ∫ 0 ∞ e − x d x = [ e − x − 1 ] 0 ∞ = [ − e − ∞ + e − 0 ] I = [ 0 + 1 ] = 1 = 0 ! Thus, the value of the integral ∫ 0 ∞ x n e − x d x for n = 0 is 1. Calculate the value of the integral for n = 1 . Substitute 1 for n in Equation (1). I = ∫ 0 ∞ x 1 e − x d x = ∫ 0 ∞ x e − x d x (2) Consider the value of the function u = x (3) Differentiate Equation (3). d u = d x Consider the value of the function d v = e − x d x (4) Integrate Equation (4). ∫ d v = ∫ e − x d x v = − e − x Show the formula of integration by parts. ∫ 0 ∞ u d v = ( u v ) 0 ∞ − ∫ 0 ∞ v d u (5) Substitute x for u , e − x d x for d v , − e − x for v and dx for dv in Equation (5). ∫ 0 ∞ x e − x d x = [ − x e − x ] 0 ∞ − ∫ 0 ∞ ( − e − x ) d x = lim t → ∞ [ − x e − x ] 0 t − ∫ 0 ∞ ( − e − x ) d x = lim t → ∞ [ − t e − t + 0 ] + ∫ 0 ∞ ( e − x ) d x = lim t → ∞ [ − t e − t + 0 ] + ( − e − x ) 0 ∞ ∫ 0 ∞ x e − x d x = lim t → ∞ [ − t e − t ] + ( − e − ∞ + e − 0 ) = lim t → ∞ [ − t e − t ] + ( 0 + 1 ) = lim t → ∞ [ − t e − t ] + 1 = lim t → ∞ [ − t e t ] + 1 (6) Apply L’Hospital’s Rule in lim t → ∞ [ − t e t ] part of Equation (6). Differentiate numerator and denominator of lim t → ∞ [ − t e − t ] of Equation (6). ∫ 0 ∞ x e − x d x = lim t → ∞ [ − d d t ( t ) d d t ( e t ) ] + 1 = lim t → ∞ [ − 1 e t ] + 1 = lim t → ∞ [ − e − t ] + 1 = − e − ∞ + 1 ∫ 0 ∞ x e − x d x = 0 + 1 = 1 ! Thus, the value of the integral ∫ 0 ∞ x n e − x d x for n = 1 is 1. Calculate the value of the integral for n = 2 . Substitute 2 for n in Equation (1). I = ∫ 0 ∞ x 2 e − x d x = ∫ 0 ∞ x 2 e − x d x (7) Show the formula of integration . ∫ u n e a u d u = 1 a u n e a u − n a ∫ u n − 1 e a u d u (8) Apply the formula of integration along with limits [ 0 , ∞ ] in Equation (8). Substitute 2 for n , x for u , dx for du , and − 1 for a . ∫ 0 ∞ x 2 e − x d x = [ 1 ( − 1 ) x 2 e − x ] 0 ∞ − 2 ( − 1 ) ∫ 0 ∞ x 2 − 1 e − x d x = [ − x 2 e − x ] 0 ∞ + 2 ∫ 0 ∞ x e − x d x = lim t → ∞ [ − x 2 e − x ] 0 t + 2 ∫ 0 ∞ x e − x d x = lim t → ∞ [ − t 2 e − t + 0 ] + 2 ∫ 0 ∞ x e − x d x = lim t → ∞ [ − t 2 e t ] + 2 ∫ 0 ∞ x e − x d x (9) Substitute 1 for ∫ 0 ∞ x e − x d x in Equation (9). ∫ 0 ∞ x 2 e − x d x = lim t → ∞ [ − t 2 e t ] + 2 ( 1 ) = lim t → ∞ [ − t 2 e t ] + 2 (10) Apply L’Hospital’s Rule in lim t → ∞ [ − t 2 e t ] part of Equation (10) Part (b): To determine To guess: the value of the integral ∫ 0 ∞ x n e − x d x with n as an arbitrary positive integer. Part (c): To determine To prove: the guess with mathematical induction.

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Author: Stewart

Publisher: VST

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Question

Chapter 7.8, Problem 60E

Part (a):

To determine

To Evaluate: The integral ∫0∞xne−xdx for n=0, 1, 2, and 3.

Part (b):

To determine

To guess: the value of the integral ∫0∞xne−xdx with n as an arbitrary positive integer.

Part (c):

To determine

To prove: the guess with mathematical induction.

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Chapter 7.8, Problem 59E

Chapter 7.8, Problem 61E

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Chapter 7 Solutions

EBK SINGLE VARIABLE CALCULUS: EARLY TRA

Ch. 7.1 - Evaluate the integral using integration by parts...Ch. 7.1 - Prob. 2E Ch. 7.1 - Evaluate the integral. 3. xcos5xdx Ch. 7.1 - Evaluate the integral. 4. ye0.2ydy Ch. 7.1 - Evaluate the integral. 5. te3tdt Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Evaluate the integral. 8. t2sintdt Ch. 7.1 - Evaluate the integral. 9. cos1xdx Ch. 7.1 - Prob. 10E

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The integral ∫ 0 ∞ x n e − x d x for n = 0 , 1, 2, and 3.

Solution Summary: The author evaluates the value of the integral for n=0, 1, 2, and 3.

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To Evaluate: The integral ∫0∞xne−xdx for n=0, 1, 2, and 3.

To guess: the value of the integral ∫0∞xne−xdx with n as an arbitrary positive integer.

To prove: the guess with mathematical induction.

Want to see the full answer?

Chapter 7 Solutions