The area (in square units) of the region bounded by the curves $x = y^{2}$ and $x = 3 - 2 y^{2}$ is

\frac{3}{2}

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Solution

The correct option is D 4
Required area $= 2 \int_{0}^{1} (3 - 2 y^{2} - y^{2}) d y = 6 \int_{0}^{1} (1 - y^{2}) d y = 6 {(y - \frac{y^{3}}{3})}_{0}^{1} = 4$
Since given curves are symmetrical about x-axis

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x = - 2 y^{2}

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y = | x - 1 |

y = 3 - | x |

x = y^{2}

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