The area of the region \( \{(x,y):x^2\le y\le x\} \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{6} \) sq.units
The region lies between the parabola \( y=x^2 \) (below) and the line y = x (above). They meet where \( x^2=x \), that is, at x = 0 and x = 1, and for 0 < x < 1 we have \( x^2\lt x \).
\( \displaystyle \text{Area}=\int_{0}^{1}(x-x^2)\,dx \)
\( \displaystyle =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{1} \)
\( \displaystyle =\frac{1}{2}-\frac{1}{3} \)
\( \displaystyle =\frac{1}{6}\ \text{sq. units} \)
Why the other options are incorrect
Option (1) 1 sq.units — Incorrect. The whole region lies inside the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and fills only a small part of it, so its area is much less than 1.
Option (2) \( \displaystyle\frac{5}{6} \) sq.units — Incorrect. \( \frac{5}{6} \) is \( \frac{1}{2}+\frac{1}{3} \); the area under the parabola must be subtracted from the area under the line, not added.
Option (3) \( \displaystyle\frac{1}{3} \) sq.units — Incorrect. \( \frac{1}{3} \) is only the area under the parabola, \( \int_{0}^{1}x^2\,dx \), not the area between the two curves.