Dear Student, Please find below the solution to the asked query: The required area of the region is the shaded regionAs, y2=2x⇒y=2xAnd,x2+y2=4x⇒y=4x-x2Now, area of the required region=∫02∫4x-x2-2xdx=∫024x-x2dx-∫022xdx=∫02-x2-4xdx-2∫02x12dx=∫02-x2-4x+4-4dx-22x32302=∫0222-x-22dx-222323-0=x-2222-x-22+222sin-1x-2202-222323-0=0-0-2sin-1-1-222323=2sin-11-2×223=2×π2-83=π-83Hence, the correct option is 2 Regards

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Dear Student,
Please find below the solution to the asked query:

The required area of the region is the shaded region As, y^{2} = 2 x \Rightarrow y = \sqrt{2 x} And, x^{2} + y^{2} = 4 x \Rightarrow y = \sqrt{4 x - x^{2}} Now, area of the required region = \int_{0}^{2} \int (\sqrt{4 x - x^{2}} - \sqrt{2 x}) dx = \int_{0}^{2} \sqrt{4 x - x^{2}} dx - \int_{0}^{2} \sqrt{2 x} dx = \int_{0}^{2} \sqrt{- (x^{2} - 4 x)} dx - \sqrt{2} \int_{0}^{2} x^{\frac{1}{2}} dx = \int_{0}^{2} \sqrt{- (x^{2} - 4 x + 4 - 4)} dx - \sqrt{2} {[\frac{2 x^{\frac{3}{2}}}{3}]}_{0}^{2} = \int_{0}^{2} \sqrt{2^{2} - {(x - 2)}^{2}} dx - \sqrt{2} [\frac{2 {(2)}^{\frac{3}{2}}}{3} - 0] = {[\frac{(x - 2)}{2} \sqrt{2^{2} - {(x - 2)}^{2}} + \frac{2^{2}}{2} \sin^{- 1} (\frac{x - 2}{2})]}_{0}^{2} - \sqrt{2} [\frac{2 {(2)}^{\frac{3}{2}}}{3} - 0] = [0 - 0 - 2 \sin^{- 1} (- 1)] - [\frac{2 \sqrt{2} {(2)}^{\frac{3}{2}}}{3}] = [2 \sin^{- 1} (1)] - [\frac{2 \times 2^{2}}{3}] = 2 \times \frac{π}{2} - \frac{8}{3} = π - \frac{8}{3} Hence, the correct option is (2) .

Regards

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