The area of the region bounded by the curve \( y=x^{3},y=8 \) and \( x=0 \) is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 12 Sq.units
The curve \( y=x^{3} \) meets y = 8 at x = 2. The region lies between the y-axis (x = 0), the curve and the line y = 8, so it is integrated along the y-axis with \( x=y^{1/3} \):
\( \displaystyle \text{Area}=\int_{0}^{8}y^{1/3}\,dy \)
\( \displaystyle =\left[\frac{3}{4}y^{4/3}\right]_{0}^{8} \)
\( \displaystyle =\frac{3}{4}\cdot 16 \)
\( \displaystyle =12\text{ sq. units} \)
Shortcut: Integrating along x instead, the region is the strip from the curve up to y = 8 for 0 ≤ x ≤ 2:
\( \displaystyle \int_{0}^{2}(8-x^{3})\,dx=\left[8x-\frac{x^{4}}{4}\right]_{0}^{2} \)
\( \displaystyle =16-4=12 \)