Question

Calculate the area bounded by y = x³, x = 0 and y = 4.

Answer 1

Tracing the graph of $y=x^3$ shows us that the area is bounded by $x=0$ on the left, $y=4$ from above and $y=x^3$ from below, with two latter encountering at $x=4^{1/3}$. Therefore, the area is given by :

$S=\int_0^{\sqrt[3]4}(4-x^3)dx$ this integral can be easily calculated :

$S=4\cdot\sqrt[3]{4} - \frac{1}{4}x^4|^{\sqrt[3]{4}}_{0}$

$S=4 \sqrt[3]{4}-\sqrt[3]{4}=3\cdot \sqrt[3]{4}$