The area bounded by two branches of the curve $(y - x)^{2} = x^{3} & x = 1$ equals

3 / 5

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5 / 4

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6 / 5

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4 / 5

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Solution

The correct option is B $4 / 5$
$(y - x)^{2} = x^{3}$
$y = x^{3 / 2} + x$ and $y = - x^{3 / 2} + x$
Area 1 $=_{0}^{1} \int (x^{3 / 2} + x) d x = \frac{2 x^{5 / 2}}{5} + \frac{x^{2}}{2} |_{0}^{1} = 2 / 5 + 1 / 2$
Area 2 $=_{0}^{1} \int (- x^{3 / 2} + x) d x = \frac{- 2 x^{5 / 2}}{5} + \frac{x^{2}}{2} |_{0}^{1} = - 2 / 5 + 1 / 2$
Total Area $=$ Area 1 - Area 2 $=$ $\frac{4}{5}$

Suggest Corrections

Similar questions

{(y - x)}^{2} = x^{3}

x = 1

y = \sqrt{5 - x^{2}}

y = | x - 1 |

y = \sqrt{5 - x^{2}}

y = | x - 1 |

\frac{x^{2}}{25} + \frac{y^{2}}{36} \leq 1 \leq \frac{x}{5} + \frac{y}{6}

y = 1 + \frac{8}{x^{2}}

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