The area (in square units) lying in the first | vedclass

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(A) The given circle is $x^2+y^2=4$,which has a radius $r=2$. The area is bounded by $x=0$ (the $y$-axis) and $x=2$ in the first quadrant. The area $A

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(A) The given circle is $x^2+y^2=4$,which has a radius $r=2$. The area is bounded by $x=0$ (the $y$-axis) and $x=2$ in the first quadrant. The area $A$ is given by the integral of $y$ with respect to $x$ from $0$ to $2$. Since $y^2 = 4-x^2$,we have $y = \sqrt{4-x^2}$ in the first quadrant. $A = \int_0^2 \sqrt{4-x^2} dx$ Using the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) + C$: $A = \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}(\frac{x}{2}) ight]_0^2$ $A = \left( \frac{2}{2}\sqrt{4-4} + 2\sin^{-1}(1) ight) - \left( 0 + 2\sin^{-1}(0) ight)$ $A = (0 + 2 	imes \frac{\pi}{2}) - 0 = \pi 	ext{ square units.}$

The area (in square units) lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is

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