Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 49

$\int_{0}^{27}\sqrt[3] x~dx = \frac{243}{4}$

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