# The area of the region bounded by the circle $x^2+y^2=1$ is $(A)\;2\pi sq.units\quad(B)\;\pi sq. units\quad(C)\;3\pi sq.units\quad(D)4\pi sq.units$

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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\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big)$$+c Step 1: Given curve is x^2+y^2=1\Rightarrow y=\sqrt{1-x^2} Clearly the given equation is the equation of the circle with centre (0,0) and radius 1. The required area is the shaded portion in the fig. Area of the shaded portion is A=4\times \int_0^1\sqrt{1-x^2}dx. We multiply by 4 because the curve is symmetrical about x-axis and y-axis. Hence the required area is 4 times area in the I quadrant. \int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big) Step 2: On integrating we get, A=4\times \begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{1-x^2}+\large\frac{1}{2}\normalsize\sin^{-1}\big(\large\frac{x}{1}\big)\end{bmatrix}_0^1 On applying the limits we get, \;\;=4\times[\large\frac{1}{2}$$\sqrt{1-1}+\frac{1}{2}\sin^{-1}\big(\frac{1}{1}\big)$$-0] \;\;=4\times [\large\frac{1}{2}$$\sin^{-1}(1)]$
But $\sin^{-1}(1)=\large\frac{\pi}{2}$
Therefore $A=4\times \large\frac{1}{2}.\frac{\pi}{2}$$=\pi$sq.units.
Hence the correct option is B.