Answer
a. $F(x)=\displaystyle \frac{3}{4}x^{4/3}-12$
b. $F'(x)=\sqrt[3]{x}$
Work Step by Step
a.
$F(x)= \displaystyle \int_{8}^{x}\sqrt[3]{t}dt$=$\displaystyle \int_{8}^{x}t^{1/3}dt$
$=[\displaystyle \frac{3}{4}t^{4/3}]_{8}^{x}$
$=\displaystyle \frac{3}{4}(x^{4/3}-8^{4/3})$
$=\displaystyle \frac{3}{4}(x^{4/3}-(2^{3})^{4/3})$
$=\displaystyle \frac{3}{4}(x^{4/3}-16)$
$F(x)=\displaystyle \frac{3}{4}x^{4/3}-12$
b.
$\displaystyle \frac{d}{dx}[\frac{3}{4}x^{4/3}-12]=x^{1/3}-0=\sqrt[3]{x}$
