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The area of the smaller region in first quadrant bounded by x2+y2=4x^2+y^2=4 and x=32x=\dfrac{3}{2} is:

Solution

✅ Correct Option: 2

The required area is ∫3/224−x2 dx\int_{3/2}^{2}\sqrt{4-x^2}\,dx.

Using ∫a2−x2 dx=x2a2−x2+a22sin⁡−1xa\int\sqrt{a^2-x^2}\,dx = \dfrac{x}{2}\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} with a=2a=2:

At x=2x=2 the value is π\pi; at x=3/2x=3/2 it is 378+2sin⁡−134\dfrac{3\sqrt{7}}{8} + 2\sin^{-1}\dfrac{3}{4}.

Subtracting gives π−378−2sin⁡−134\pi - \dfrac{3\sqrt{7}}{8} - 2\sin^{-1}\dfrac{3}{4}.

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