The area (in sq. units) of the region (x, y): y^2 ≤ x ≤ 2y is

The correct answer is Option (1) → \(\displaystyle\frac{4}{3}\)

For each y, x runs from the parabola \(x = y^{2}\) to the line x = 2y. This is possible only when

    \(y^{2} \le 2y\)
    \(y(y-2) \le 0\)
    \(0 \le y \le 2\)

So the region is found by integrating in y, with width \(2y-y^{2}\):

    \( \displaystyle A = \int_{0}^{2}(2y-y^{2})\,dy \)

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

    \( \displaystyle = 4-\frac{8}{3} = \frac{4}{3} \)