Application of Integrals — Class 12 Maths Solution

We have $y = \sqrt x$ and $y = x$
Solving, we get
$x = \sqrt x$

$\Rightarrow {x^2} = x$

$\Rightarrow {x^2} - x = 0$

At $x = 0,y = 1$

Thus curves intersect at (0,0) and (1,1)
Graph of $$y = \sqrt x $$ is part of parabola lying above $x$-axis.

The graph is as shown in the adjacent figure.
From the figure, area of shaded region.

$A = \int_0^1 {\left[ {(\sqrt x - x)dx = \left[ {\frac{2}{3}{x^{3/2}} - \frac{{{x^2}}}{2}} \right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}{\rm{sq}}} \right.}$. units