The area (in square units) of the region bounded by curves \(y=x\) and \(y=x^3\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1/2
Points of intersection:
\(x = x^3\)
\(x(x-1)(x+1) = 0\)
\(x = -1,\ 0,\ 1\)
The curves enclose two regions: on (0, 1) the line is above the cubic (for example at x = 1/2, 1/2 > 1/8), and on (−1, 0) the cubic is above the line. Both functions are odd, so the two regions are symmetric about the origin and have equal areas:
\( \displaystyle A = 2\int_{0}^{1}\left(x - x^3\right)dx \)
\( \displaystyle = 2\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1} \)
\( \displaystyle = 2\left(\frac{1}{2} - \frac{1}{4}\right) \)
\( \displaystyle = \frac{1}{2} \)
Why the other options are incorrect
Option (1) 0 — Incorrect. This is \(\int_{-1}^{1}(x - x^3)\,dx\): the part on (−1, 0) is negative and cancels the part on (0, 1). An area counts both regions as positive.
Option (3) 1/4 — Incorrect. This is the area of only one of the two regions.
Option (4) 4 — Incorrect. The total area of the two regions is 1/2, not 4.