Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 13

Answer

$ e^x(x^3 -3 x^2 + 6x - 6) $

Work Step by Step

$ \int{ x^3 e^x } dx $ $ x^3 e^x - \int{ 3x^2 e^x} dx $ Parts: $ u=x^3, dv=e^x$ $ x^3 e^x -3 \int{ x^2 e^x} dx $ $ x^3 e^x -3 (x^2 e^x - \int{2x e^x}dx ) $ Parts: $ u=x^2, dv=e^x$ $ x^3 e^x -3 x^2 e^x + 6\int{ x e^x} dx $ $ x^3 e^x -3 x^2 e^x + 6(xe^x -\int{ e^x} dx) $ Parts: $u = x, dv=e^x$ $ x^3 e^x -3 x^2 e^x + 6xe^x - 6e^x $ $ e^x(x^3 -3 x^2 + 6x - 6) $
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