The area enclosed between the graph of \(y = x^3\) and the lines \(x = 0, y = 1, y = 8\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{45}{4}\) Sq. units
The region lies between the y-axis (x = 0) and the curve, from y = 1 to y = 8, so it is found by integrating along the y-axis, with \(x = y^{1/3}\):
\( \displaystyle \text{Area} = \int_{1}^{8} y^{1/3}\,dy \)
\( \displaystyle \text{Area} = \left[\frac{3}{4}y^{4/3}\right]_{1}^{8} \)
\( \displaystyle \text{Area} = \frac{3}{4}(16 - 1) = \frac{45}{4} \)
Here \(8^{4/3} = (2^3)^{4/3} = 2^4 = 16\).