Explanation Formula used: Integration by parts: “If u and v are differentiable functions, then ∫ u d v = u v − ∫ v d u .” Calculation: The given integral is, ∫ 0 5 1000 t e − 0.04 t d t . Solve the above integral with the help of integration by parts by letting u = 1000 t and d v = e − 0.04 t d t . Differentiate u = 1000 t as follows. d u d t = 1000 ( 1 ) d u d t = 1000 d u = 1000 d t Integrate d v = e − 0.04 t d t as follows. ∫ d v = ∫ e − 0.04 t d t v = e − 0.04 t − 0.04 Substitute the corresponding values in ∫ u d v = u v − ∫ v d u , ∫ 1000 t e − 0.04 t d t = 1000 t [ e − 0.04 t − 0.04 ] − ∫ [ e − 0.04 t − 0.04 ] 1000 d t = 1000 − 0.04 t e − 0.04 t + 1000 0.04 [ e − 0.04 t − 0

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 8.3, Problem 4WE

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Chapter 8.3, Problem 3WE

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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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The value of ∫ 0 5 1000 t e − 0.04 t d t .

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Chapter 8 Solutions