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The area of the region bounded by parabola $y^2=x$ and the straight line $2y=x$ is \begin{array}{1 1}(A)\;\frac{4}{3} sq.units & (B)\;1 sq.units\\(C)\;\frac{2}{3} sq.units & (D)\;\frac{1}{3}sq.units \end{array}

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  • 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 curve $y^2=x$ and the line $2y=x$.
Required area is the shaded portion shown in the fig.
Step 2:
Next let us find the point of intersection.
Clearly one point is (0,0)
$y^2=x$ and $4y^2=x^2\Rightarrow y^2=\large\frac{x^2}{4}$
$x=\large\frac{x^2}{4}$$\Rightarrow 4x=x^2$
$\Rightarrow x^2-4x=0$
$x(x-4)=0\Rightarrow x=0$ or $x=4$
Hence Area of the shaded portion is $A=\int_0^4(y_2-y_1)dx.$
Where $y_2=\sqrt x$ and $y_1=\large\frac{x}{2}$
$A=\int_0^4(\sqrt x-\large\frac{x}{2})$
Step 3:
On integrating we get,
$A=\int_0^4\sqrt xdx-\large\frac{1}{2}$$\int_0^4 xdx$
On applying limits we get,
$\;\;\;=\large\frac{2}{3}$$[8]-\large\frac{1}{4}$$\times 16$
Hence A is the correct option.
answered May 7, 2013 by sreemathi.v

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