University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 6


$$\int^e_1 x^3\ln xdx=\frac{3e^4+1}{16}$$

Work Step by Step

$$A=\int^e_1 x^3\ln xdx$$ Set $u=\ln x$ and $dv=x^3dx$ Then we would have $du=\frac{dx}{x}$ and $v=\frac{x^4}{4}$ Using the formula $\int^b_a udv= uv]^b_a-\int^b_a vdu$: $$A=\frac{x^4\ln x}{4}\Big]^e_1-\int^e_1\Big(\frac{x^4}{4}\times\frac{dx}{x}\Big)$$ $$A=\frac{1}{4}(e^4\ln e-1^4\ln1)-\frac{1}{4}\int^e_1x^3dx$$ $$A=\frac{1}{4}(e^4-0)-\frac{1}{4}\Big(\frac{x^4}{4}\Big)\Big]^e_1$$ $$A=\frac{e^4}{4}-\frac{1}{16}(e^4-1^4)$$ $$A=\frac{4e^4-(e^4-1)}{16}$$ $$A=\frac{3e^4+1}{16}$$
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