Question

Find the area of the region bounded by the curves x=y³ and y=x².

Answer 1

Solution

Points of intersection of the given curves are solution of equation x = x⁶  => x(x⁵ – 1) = 0  =>  x(x-1)(x⁴ + x³ + x² + x + 1) = 0  => Roots of this equation are x₁ = 0, x₂ = 1

So area to be find is the area bounded by curves y=x² and $y=\sqrt[3]{x}$  (for 0<x<1 $\sqrt[3]{x}>x^2$ )  =>

$A=\int_{0}^{1}\left(\sqrt[3]{x}-x^2\right)dx=\left(\frac{3}{4}x^{4/3}-\frac{1}{3}x^3\right)\left|\begin{matrix}1\\0\\\end{matrix}\right.$

A = 3/4 - 1/3 = 5/12 = 0.417

Answer

A = 0.417