Find the area of the region bounded by the two parabolas y = x^2 and…

The correct answer is Option (2) → \(\displaystyle\frac{1}{3}\)

The parabolas meet where \(y = x^2\) and \(y^2 = x\) hold together:

    \(x^4 = x\)
    \(x(x^3 - 1) = 0\)
    \(x = 0 \text{ or } x = 1\)

For 0 < x < 1 the upper curve is \(y = \sqrt{x}\) and the lower curve is \(y = x^2\). So the area between the curves is

    \( \displaystyle A = \int_{0}^{1}\left(\sqrt{x} - x^2\right)dx \)

    \( \displaystyle = \left[\frac{2}{3}x^{3/2} - \frac{x^3}{3}\right]_{0}^{1} \)

    \( \displaystyle = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \)

Shortcut: The parabolas \(y^2 = 4ax\) and \(x^2 = 4by\) enclose an area \(\frac{16ab}{3}\). Here 4a = 1 and 4b = 1, so

    \( \displaystyle A = \frac{16}{3}\times\frac{1}{4}\times\frac{1}{4} = \frac{1}{3} \)