The area enclosed by the curve y=4−x2,y≥2sin(22xπ) , and the x−aξs is divided by the y−aξs in the ratio. π2+8π2−8 (b) π2+4π2−4 π−4π−4 (d) 2π+π2−82π2

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The area enclosed by the curve {y}={{{4}-{x}^{{2}}}},{y}{{{2}}

Question

The area enclosed by the curve $y = 4 - x^{2}, y \geq 2 sin (\frac{x π}{2 2})$ , and the $x - a ξ s$ is divided by the $y - a ξ s$ in the ratio. $\frac{π ^{2} - 8}{π ^{2} + 8}$ (b) $\frac{π ^{2} - 4}{π ^{2} + 4}$ $\frac{π - 4}{π - 4}$ (d) $\frac{2 π ^{2}}{2 π + π ^{2} - 8}$

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The area enclosed by the curve

y = 4 - x^{2}, y \geq 2 sin (\frac{x π}{2 2})

, and the

x - a ξ s

is divided by the

y - a ξ s

in the ratio.

\frac{π ^{2} - 8}{π ^{2} + 8}

(b)

\frac{π ^{2} - 4}{π ^{2} + 4}

\frac{π - 4}{π - 4}

(d)

\frac{2 π ^{2}}{2 π + π ^{2} - 8}

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Question Text

The area enclosed by the curve

y = 4 - x^{2}, y \geq 2 sin (\frac{x π}{2 2})

, and the

x - a ξ s

is divided by the

y - a ξ s

in the ratio.

\frac{π ^{2} - 8}{π ^{2} + 8}

(b)

\frac{π ^{2} - 4}{π ^{2} + 4}

\frac{π - 4}{π - 4}

(d)

\frac{2 π ^{2}}{2 π + π ^{2} - 8}