Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises: 18

Answer

$\frac{52}{9}$

Work Step by Step

$\int^2_{0}y^2\sqrt {1+y^3}dy = \int^2_{0}y^2 (1+y^3)^\frac{1}{2}dy$ $= [\frac{2}{3}\times\frac{1}{3y^2}\times y^2 (1+y^3)^\frac{3}{2}] ^2_{0}$ $= [\frac{2}{9}(1+y^3)^\frac{3}{2}] ^2_{0}$. Sub in the bounds and subtract the lower bound (0) from the upper bound (2): $= [\frac{2}{9}(1+(2)^3)^\frac{3}{2}] - [\frac{2}{9}(1+(0)^3)^\frac{3}{2}] $ $= (\frac{2}{9}(9)^\frac{3}{2}) - \frac{2}{9} $ $= 6 - \frac{2}{9} $ $=\frac{52}{9}$.
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