The area (in sq. units) of the region bounded by the curve \(y = 2x^3\), x-axis and ordinates \(x = -1\) and \(x = 1\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
The curve \(y = 2x^3\) is below the x-axis for \(-1 \le x \lt 0\) and above it for \(0 \lt x \le 1\). The area of the part below the axis is the integral of \(-y\):
\( \displaystyle \text{Area} = \int_{-1}^{0} (-2x^3)\,dx + \int_{0}^{1} 2x^3\,dx \)
Since \(2x^3\) is an odd function, the two parts are equal:
\( \displaystyle \text{Area} = 2\int_{0}^{1} 2x^3\,dx = 4\left[\frac{x^4}{4}\right]_{0}^{1} = 1 \)
The plain integral \(\int_{-1}^{1} 2x^3\,dx\) is 0, because the two parts cancel; an area must count both parts as positive.