The area (in sq. units) of the region bounded by \(y=-1,y=2,x=y^3\) and \(x = 0\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{17}{4}\)
The region lies between the curve \(x = y^3\) and the y-axis, from y = −1 to y = 2, so the area is found by integrating with respect to y. For \(-1 \le y \le 0\), \(y^3 \le 0\) and the region is to the left of the y-axis; there the width is \(-y^3\):
\( \displaystyle \text{Area} = \int_{-1}^{0}(-y^3)\,dy + \int_{0}^{2}y^3\,dy \)
\( \displaystyle \int_{-1}^{0}(-y^3)\,dy = \left[-\frac{y^4}{4}\right]_{-1}^{0} = \frac{1}{4} \)
\( \displaystyle \int_{0}^{2}y^3\,dy = \left[\frac{y^4}{4}\right]_{0}^{2} = 4 \)
\( \displaystyle \text{Area} = \frac{1}{4} + 4 = \frac{17}{4} \)
Why the other options are incorrect
Option (1) 4 — Incorrect. This is only the part for \(0 \le y \le 2\); the part for \(-1 \le y \le 0\) adds \(\frac{1}{4}\).
Option (2) 8 — Incorrect. The two parts give \(\frac{1}{4} + 4 = \frac{17}{4}\), not 8.
Option (4) \(\displaystyle\frac{15}{4}\) — Incorrect. This is \(\int_{-1}^{2}y^3\,dy = 4 - \frac{1}{4}\), where the part left of the y-axis is counted as negative; an area is always positive.