To find the area of the region defined by $\{(x, y) : x^2 \le y \le x\}$, we need to calculate the definite integral of the difference between the upper curve ($y=x$) and the lower curve ($y=x^2$) between their points of intersection.
Set the equations equal to find the intersection points:
$x^2 = x$
$x^2 - x = 0$
$x(x - 1) = 0$
The points of intersection are at $x=0$ and $x=1$.
In the interval $[0, 1]$, the line $y=x$ is above the parabola $y=x^2$. We can verify this by testing a point, e.g., $x=0.5$: $y=0.5$ for the line and $y=(0.5)^2=0.25$ for the parabola. Since $0.5 > 0.25$, $y=x$ is the upper boundary.
The area $A$ is given by the integral:
$A = \int_{0}^{1} (\text{upper curve} - \text{lower curve}) \, dx$
$A = \int_{0}^{1} (x - x^2) \, dx$
Evaluate the definite integral:
$A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}$
$A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right)$
$A = \left( \frac{1}{2} - \frac{1}{3} \right) - (0 - 0)$
$A = \frac{3 - 2}{6}$
$A = \frac{1}{6}$
The area of the region is $\frac{1}{6}$ square units.
| List-I | List-II |
| Electronic Configuration | First Ionisation energy (kJ mol$^{-1}$) |
| (A). ns$^2$ | (I). 2100 |
| (B). ns$^2$np$^1$ | (II). 1400 |
| (C). ns$^2$np$^3$ | (III). 800 |
| (D). ns$^2$np$^6$ | (IV). 900 |
| List-I | List-II |
| Spectroscopy | Property |
| (A). Raman | (I). Polarizability |
| (B). FTIR | (II). Dipole Moment |
| (C). UV-Visible | (III). Absorbance |
| (D). NMR | (IV). Spin |