The area (in square units) of the region bounded by the curves $2 x = y^{2} - 1$ and $x = 0$ is

\frac{1}{3}

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\frac{2}{3}

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Solution

The correct option is B $\frac{2}{3}$
$2 x = y^{2} - 1$ and $x = 0$
$2 0 \int - 1 / 2 (\sqrt{2 x + 1}) d x$
$(2 x + 1)^{3 / 2} + 2 / 3 |_{- 1 / 2}^{0}$
$= 2 / 3 s q u n i t s$

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Similar questions

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

2 x = y^{2} - 1

x = 0

y = | x - 1 |

y = 3 - | x |

y + 2 x^{2} = 0 and y + 3 x^{2} = 1

x = - 1, x = 2, y = x^{2} + 1

y = 2 x - 2

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