## Calculus: Early Transcendentals 8th Edition

$\int_{0}^{27}\sqrt[3] x~dx = \frac{243}{4}$
We can find the area under the graph: $\int_{0}^{27}\sqrt[3] x~dx$ $=\frac{3}{4}x^{4/3}~\vert_{0}^{27}$ $=\frac{3}{4}(27^{4/3} -0^{4/3})$ $=\frac{3}{4}(81 -0)$ $=\frac{243}{4}$