Area (in sq. units) of the region bounded by the curves y = -1, y = 2…

The correct answer is Option (2) → \(\displaystyle\frac{17}{4}\)

The region lies between the curve \(x = y^3\) and the y-axis \(x = 0\), from \(y = -1\) to \(y = 2\), so the area is found by integrating along y:

    \( \displaystyle \text{Area} = \int_{-1}^{2} |y^3|\,dy \)

For \(-1 \le y \lt 0\), \(y^3\) is negative (the curve is to the left of the y-axis), so the area there is \(\int -y^3\,dy\):

    \( \displaystyle \int_{-1}^{0} (-y^3)\,dy = \left[-\frac{y^4}{4}\right]_{-1}^{0} = 0 + \frac{1}{4} = \frac{1}{4} \)

    \( \displaystyle \int_{0}^{2} y^3\,dy = \left[\frac{y^4}{4}\right]_{0}^{2} = \frac{16}{4} = 4 \)

Adding the two parts:

    \( \displaystyle \text{Area} = \frac{1}{4} + 4 = \frac{17}{4} \)