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 $
