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 :

π

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\frac{π}{2}

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\frac{π}{3}

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\frac{π}{4}

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Solution

The correct option is A $π$
Required area = Area OAB = $\int_{0}^{2} y d x$

$= \int_{0}^{2} \sqrt{4 - x^{2}} d x$

$= {[\frac{x}{2} \sqrt{4 - x^{2}} + \frac{4}{2} {sin}^{- 1} \frac{x}{2}]}_{0}^{- 1}$

$= 2 \times \frac{π}{2} = π u n i t s$

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Similar questions

Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is

A. π

^{'} a^{'}

^{'} b^{'}

f (x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

x = \frac{π}{4}, \frac{π}{2}

\frac{π}{2}

\frac{π}{3}

\frac{π}{4}

x^{2} + y^{2} = 4

x = 0

x = 2

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