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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Choose the correct answer in area lying in the first quadrant and bounded by the circle \(x^2 + y^2 = 4\) and the lines \(x = 0\) and \(x = 2\) is

$(a)\;\pi\qquad(b)\;2 \pi\qquad(c)\;3 \pi\qquad(d)\;4 \pi$

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1 Answer

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Toolbox:
  • The area bounded by the curve f(x),x-axis and the ordinate x=a,x=b is given by\[A=\int_a^by\;dx=\int_a^{-b}f(x)dx.\]
Here the area of the region bounded by the line x=0 and x=2 and the circle $x^2+y^2=4$ is the shaded portion as shown in the fig:
Hence $A=\int_0^2y dx$.
Here y=$\sqrt{4-x^2}dx$
$A=\int_0^2\sqrt {4-x^2}dx.$
On integrating we get,
$A=\begin{bmatrix}\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\big(\frac{x}{2}\big)\end{bmatrix}_0^2$
On applying limits we get,
$A=\begin{bmatrix}\frac{2}{2}\sqrt{4-4}+\frac{4}{2}\sin^{-1}\big(\frac{2}{2}\big)\end{bmatrix}$
$\;\;\;=0+2.\frac{\pi}{2}$
$\;\;\;=\pi\; sq.units.$
Hence A is the correct answer.
answered Dec 20, 2013 by yamini.v
 

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