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# Find the area of the region bounded by $y=\sqrt x$ and $y=x.$ • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c. Step 1: Given :y=\sqrt x and y=x. Consider y=\sqrt x Squaring on both sides we get, y^2=x Clearly this is a parabola open rightward and whose focus is (4a=1\Rightarrow a=\large\frac{1}{4}) y=x is a straight line passing through the origin which is inclined at an angle of 45^\circ to the x-axis. Step 2: The points of intersection can be obtained by equating y^2=x and y^2=x^2 x^2=x\Rightarrow (x^2-x)=0 x(x-1)=0. \Rightarrow x=0 or x=1.Hence y=0,1. Hence the points of intersection are (0,0) and (1,1). The required area is the shaded portion shown in the fig. Hence A=\int_0^1(y_2-y_1)dx. Where y_2=\sqrt{x} and y_1=x So,A=\int_0^1(\sqrt x-x)dx. \;\;\;\;=\int_0^1\sqrt{x}dx-\int_0^1 xdx. Step 3: On integrating we get, \;\;\;\;=\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_0^1-\begin{bmatrix}\large\frac{x^2}{2}\end{bmatrix}_0^1 \;\;\;\;=\large\frac{2}{3}$$\begin{bmatrix}x^{\large\frac{3}{2}}\end{bmatrix}_0^1$$-\large\frac{1}{2}\begin{bmatrix}x^2\end{bmatrix}_0^1$
$\;\;\;\;=\large\frac{2}{3}\normalsize[1^{\large\frac{3}{2}}\normalsize-0]-\large\frac{1}{2}\normalsize[1^2-0]$
$\;\;\;\;=\large\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$sq.units.
Hence the required area is $\large\frac{1}{6}$sq.units.