Explanation Given: The statement is “we would need to apply the method of integration by parts twice to determine ∫ x 3 e − x 2 d x .” Formula used: “If u and v are differentiable functions, then ∫ u d v = u v − ∫ v d u .” Calculation: The given integral is, ∫ x 3 e − x 2 d x . The integrand f ( x ) = x 3 e − x 2 is in product of two functions. Let u = x 2 and v = x e − x 2 . Here, two functions are not related in terms of derivative one another. Hence, the substitution method is not applicable. In that situation only integration by parts is helpful once. The evaluation of the above integral ∫ x 3 e − x 2 d x is obtain through integration by parts as follows, Solve the above integral with the help of integration by parts by letting u = x 2 and d v = x e − x 2 d x . Differentiate u = x 2 as follows. d u d x = 2 x d u = 2 x d x Integrate d v = x e − x 2 d x as follows

Calculus with Applications (11th Edition)

11th Edition

ISBN: 9780321979421

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 8 Solutions

Calculus with Applications (11th Edition)

Ch. 8.1 - YOUR TURN 1 Find Ch. 8.1 - YOUR TURN 2 Find Ch. 8.1 - YOUR TURN 3 Find Ch. 8.1 - YOUR TURN 4 Find . Ch. 8.1 - YOUR TURN 5 Find Ch. 8.1 - Prob. 1WE Ch. 8.1 - Find the following. W2. Ch. 8.1 - Find the following. W3. Ch. 8.1 - Find the following. W4. Ch. 8.1 - Use integration by parts to find the integrals in...

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