Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises - Page 555: 42

Answer

$$V = \frac{\pi }{{25}}\left( {4{e^5} + 1} \right)$$

Work Step by Step

$$\eqalign{ & {\text{Calculate the volume using the disk method about the x - Axis}} \cr & V = \int_a^b {\pi f{{\left( x \right)}^2}} dx \cr & {\text{Let }}f\left( x \right) = {x^2}\sqrt {\ln x} ,\,\,\,{\text{on the interval }}\left[ {1,e} \right] \cr & V = {\int_1^e {\pi \left[ {{x^2}\sqrt {\ln x} } \right]} ^2}dx \cr & V = \pi \int_1^e {{x^4}\ln x} dx \cr & {\text{Integrate by tables using the formula 100}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {{x^n}\ln x} dx = \frac{{{x^{n + 1}}}}{{n + 1}}\left( {\ln x - \frac{1}{{n + 1}}} \right) + C \cr & {\text{Let }}n = 4,\,\,\,{\text{then}} \cr & V = \pi \left[ {\frac{{{x^{4 + 1}}}}{{4 + 1}}\left( {\ln x - \frac{1}{{4 + 1}}} \right)} \right]_1^e \cr & V = \pi \left[ {\frac{{{x^5}}}{5}\left( {\ln x - \frac{1}{5}} \right)} \right]_1^e \cr & V = \frac{\pi }{5}\left[ {{e^5}\left( {\ln e - \frac{1}{5}} \right)} \right] - \frac{\pi }{5}\left[ {{1^5}\left( {\ln 1 - \frac{1}{5}} \right)} \right] \cr & V = \frac{\pi }{5}\left[ {{e^5}\left( {1 - \frac{1}{5}} \right)} \right] - \frac{\pi }{5}\left[ { - \frac{1}{5}} \right] \cr & V = \frac{\pi }{5}\left[ {\frac{4}{5}{e^5}} \right] + \frac{\pi }{{25}} \cr & V = \frac{{4\pi {e^5}}}{{25}} + \frac{\pi }{{25}} \cr & V = \frac{\pi }{{25}}\left( {4{e^5} + 1} \right) \cr} $$
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