Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 43

$$\frac{243}{4}$$

Given $$y=\sqrt[3]{x}, \ \ 0 \leqslant x \leqslant 27$$ From the graph of the region, we can observe that area of the bounded region approximately equal to $\frac{3}{4}$ area of the rectangle with width 3 and length 27 $$\text{Area} \approx \frac{3}{4} (3)(27)=60.75 $$ Now we use integration \begin{aligned} \text{Area}&= \int_0^{27}\sqrt[3]{x}dx\\ &= \frac{3}{4}x^{4/3}\bigg|_{0}^{27} \\ &=\frac{3}{4}(27)^{4/3}\\ &= \frac{243}{4}=60.75 \end{aligned}