Find the area of the loop of the curve $x = t^{2} - 1, y = t^{3} - t$

\frac{8}{15} s q . u n i t s

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\frac{7}{15} s q . u n i t s

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Solution

The correct option is A $\frac{8}{15} s q . u n i t s$
$x = t^{2} - 1$
$t = \pm \sqrt{x + 1}$
$y = \pm x \sqrt{x + 1}$
Area under one loop = $\int_{- 1}^{0} y d x$
Integrate by parts
$= 2 \int_{- 1}^{0} (x \sqrt{x + 1}) d x$
$= \frac{8}{15}$

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