## Calculus 8th Edition

Published by Cengage

# Chapter 4 - Integrals - 4.4 Indefinite Integrals and the Net Change Theorem - 4.4 Exercises - Page 337: 30

#### Answer

$- (2 \frac{1}{4})$

#### Work Step by Step

We can rewrite $\sqrt[3] w$ as $w^{\frac{1}{3}}$ to make the terms easier to integrate. Thus, we can rewrite the integral as $\int_1^8 2w^{\frac {-1}{3}} - w^{\frac{1}{3}}dw$ Evaluating the integral gives us $\frac{2w^{\frac{2}{3}}} {\frac {2}{3}} - \frac{w^{\frac{4}{3}}} {\frac {4}{3}}$ $= 3w^{\frac{2}{3}} - \frac{3w^{\frac{4}{3}}} {4}$ We now substitute $8$ and $1$ and get $(3(8)^{\frac{2}{3}} - \frac{3(8)^{\frac{4}{3}}} {4}) - (3 - \frac{3} {4})$ , which can be simplified to $(3(4) - \frac{3(16)}{4}) - (2 \frac{1}{4})$ $= (12 - 12) - (2 \frac{1}{4}) = - (2 \frac{1}{4})$ *Tip: Recall that $8^{\frac{2}{3}} = 2^2 = 4$ and that $8^{\frac{4}{3}} = 2^4 = 16$

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