The area of the region bounded by the curve y | vedclass

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(B) The given curve is $y = \sqrt{49 - x^2}$,which implies $y^2 = 49 - x^2$ or $x^2 + y^2 = 7^2$. This represents a circle centered at the origin $(0,

Vedclass · Accepted Answer

$\frac{49 \pi}{2}$ sq. units

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(B) The given curve is $y = \sqrt{49 - x^2}$,which implies $y^2 = 49 - x^2$ or $x^2 + y^2 = 7^2$. This represents a circle centered at the origin $(0, 0)$ with a radius $r = 7$. Since $y = \sqrt{49 - x^2}$ is always non-negative,it represents the upper semi-circle.The area of the region bounded by the curve and the $X$-axis is the area of the semi-circle.Area $= \int_{-7}^{7} \sqrt{49 - x^2} \, dx = 2 \int_{0}^{7} \sqrt{49 - x^2} \, dx$Using the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} ight)$:Area $= 2 \left[ \frac{x}{2} \sqrt{49 - x^2} + \frac{49}{2} \sin^{-1} \left( \frac{x}{7} ight) ight]_{0}^{7}$$= 2 \left[ \left( \frac{7}{2} \sqrt{49 - 49} + \frac{49}{2} \sin^{-1} (1) ight) - (0 + 0) ight]$$= 2 \left[ 0 + \frac{49}{2} 	imes \frac{\pi}{2} ight] = \frac{49 \pi}{2} 	ext{ sq. units}$.

The area of the region bounded by the curve $y=\sqrt{49-x^2}$ and the $X$-axis is

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