Area of the region bounded by the curves \( x=y^3,\ y=-1,y=2 \) and \( y \)-axis, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{17}{4} \) square units
The region lies between the y-axis and the curve \( x=y^3 \), from y = −1 to y = 2. For y < 0 the curve is on the left of the y-axis, so the two parts are found separately:
\( \displaystyle \int_{-1}^{0}\left|y^3\right|dy=\int_{-1}^{0}(-y^3)\,dy=\frac{1}{4} \)
\( \displaystyle \int_0^2y^3\,dy=\frac{16}{4}=4 \)
\( \displaystyle \text{Area}=\frac{1}{4}+4=\frac{17}{4}\ \text{square units} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{11}{4} \) square units — Incorrect. The area is \( \frac{17}{4} \), not \( \frac{11}{4} \).
Option (2) \( \displaystyle\frac{15}{4} \) square units — Incorrect. This is \( \int_{-1}^{2}y^3\,dy \); the part on the left of the y-axis comes out negative and cancels part of the area.
Option (4) \( \displaystyle\frac{19}{4} \) square units — Incorrect. The area is \( \frac{17}{4} \), not \( \frac{19}{4} \).