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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region bounded by $y=\sqrt x$ and $y=x.$

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Toolbox:
  • 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$
On applying limits we get,
$\;\;\;\;=\large\frac{2}{3}\normalsize[1^{\large\frac{3}{2}}\normalsize-0]-\large\frac{1}{2}\normalsize[1^2-0]$
$\;\;\;\;=\large\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$sq.units.
Hence the required area is $\large\frac{1}{6}$sq.units.
answered May 8, 2013 by sreemathi.v
 

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