The area (in square units) of the region bounded by the curve \( x=y^3 \) , \( y=-1 \) , \( y=2 \) and \( x=0 \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 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 y < 0 the curve is to the left of the y-axis (x < 0), so that part is taken with a minus sign:
\( \displaystyle \text{Area}=\int_{-1}^{2}|y^3|\,dy \)
\( \displaystyle =-\int_{-1}^{0}y^3\,dy+\int_{0}^{2}y^3\,dy \)
\( \displaystyle =-\left[\frac{y^4}{4}\right]_{-1}^{0}+\left[\frac{y^4}{4}\right]_{0}^{2} \)
\( \displaystyle =-\left(0-\frac{1}{4}\right)+\frac{16}{4} \)
\( \displaystyle =\frac{1}{4}+4 \)
\( \displaystyle =\frac{17}{4} \)
Why the other options are incorrect
Option (2) 15/4 — Incorrect. 15/4 is \( \int_{-1}^{2}y^3\,dy=4-\frac{1}{4} \); the part with x < 0 (from y = −1 to 0) is subtracted instead of added.
Option (3) 13/4 — Incorrect. The area is 17/4, not 13/4.
Option (4) 11/4 — Incorrect. The area is 17/4, not 11/4.