The area (in sq. units) of the bounded region \( \{(x,y):x^{2}\le y\le x\} \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{6} \)
The region lies between the parabola \( y=x^{2} \) (below) and the line y = x (above).
They meet where
\( x^{2}=x \)
\( x(x-1)=0 \)
\( x=0,\ 1 \)
On 0 ≤ x ≤ 1 the line is above the parabola (for example, at \( x=\frac{1}{2} \): \( \frac{1}{2}\gt\frac{1}{4} \)).
\( \displaystyle \text{Area}=\int_{0}^{1}\left(x-x^{2}\right)dx \)
\( \displaystyle =\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1} \)
\( \displaystyle =\frac{1}{2}-\frac{1}{3} \)
\( \displaystyle =\frac{1}{6} \)