Explanation Given information: ∫ 1 ∞ ln y d y y 3 Formula: ∫ a b x a d x = [ x n + 1 n + 1 ] a b ∫ a b e x d x = [ e x ] a b ∫ u ( t ) v ′ ( t ) d t = u ( t ) v ( t ) − ∫ v ( t ) u ′ ( t ) d t (1) Calculation: Evaluate the integral given below. ∫ 1 ∞ ln y d y y 3 (2) Assume the variable x as ln y . x = ln y Differentiate the above equation with respect to y. d x = 1 y d y y d x = d y When the value of y is 1, the value of x is 0. When the value of y is ∞ , the value of x is ∞ . Substitute y d x for d y and x for ln y by changing the limits in the equation (2). ∫ 1 ∞ ln y d y y 3 = ∫ 0 ∞ x y d y y 3 ∫ 1 ∞ ln y d y y 3 = ∫ 0 ∞ x d y y 3 (3) Assume the variable y as e x . y = e x Differentiate the above equation with respect to x. d y d x = e x d y = e x d x Substitute e x d x for d y and e x for y in equation (3). ∫ 1 ∞ ln y d y y 3 = ∫ 0 ∞ x e x d x ( e x ) 3 = ∫ 0 ∞ x e x d x e 3 x = ∫ 0 ∞ x d x e 2 x ∫ 1 ∞ ln y d y y 3 = ∫ 0 ∞ x e − 2 x d x (4) Assume the variable u as − 2 x . u = − 2 x Differentiate the equation with respect to x . d u d x = − 2 d x = − d u 2 When the value of x is 0, the value of u is 0. When the value of x is ∞ , the value of u is ∞ . Substitute − d u 2 for dx , − u 2 for x and u for - 2 x without changing the limits in the equation (4). ∫ 1 ∞ ln y d y y 3 = − ∫ 0 ∞ ( − u 2 ) e u d u 2 ∫ 1 ∞ ln y d y y 3 = 1 4 ∫ 0 ∞ u e u d u (5) Consider the term ∫ 0 ∞ u e u d u from the equation (5) and change the variable u as t

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 8, Problem 111PE

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