Answer
$\frac{45}{28}$
Work Step by Step
$\int_{0}^{1}\sqrt[3] {1+7x}~dx$
Let $u = 1+7x$
$\frac{du}{dx} = 7$
$dx = \frac{du}{7}$
When $x = 0$, then $u = 1$
When $x = 1$, then $u = 8$
$\int_{1}^{8}\frac{1}{7}\sqrt[3] u~du$
$= \frac{1}{7}~[\frac{3}{4}(u)^{4/3}\vert_{1}^{8}~]$
$= \frac{3}{28}~[(u)^{4/3}\vert_{1}^{8}~]$
$= \frac{3}{28}~[(8)^{4/3}-(1)^{4/3}]$
$= \frac{3}{28}~(16-1)$
$= \frac{45}{28}$