The area of the region bounded by the curves \(y = x\) and \(y = x^3\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{2}\) sq. units
The curves meet where
\( x^3 = x \)
\( x(x-1)(x+1) = 0,\quad x = -1, 0, 1 \)
So the region bounded by the curves has two parts: one for \(0 \le x \le 1\), where \(x \ge x^3\) (at \(x = \frac{1}{2}\): \(\frac{1}{2} \gt \frac{1}{8}\)), and one for \(-1 \le x \le 0\), where \(x^3 \ge x\).
\( \displaystyle \int_{0}^{1} (x - x^3)\,dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1} \)
\( \displaystyle = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \)
Both curves are odd functions, so the region is symmetric about the origin and the part on \([-1, 0]\) also has area \(\frac{1}{4}\).
\( \displaystyle \text{Total area} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \) sq. units
Why the other options are incorrect
Option (1) 1 sq. units — Incorrect. The two parts together give \(\frac{1}{2}\), not 1.
Option (2) \(\displaystyle\frac{3}{4}\) sq. units — Incorrect. The two parts together give \(\frac{1}{2}\), not \(\frac{3}{4}\).
Option (4) \(\displaystyle\frac{1}{4}\) sq. units — Incorrect. This is the area of only the part for \(0 \le x \le 1\). The part for \(-1 \le x \le 0\) is also bounded by the two curves.
Tip: Integrating \(x - x^3\) straight from −1 to 1 gives 0, because the two parts cancel. Find where the curves cross and add the areas of the parts.