The area ( in square units) of the region bounded by the curves $y + 2 x^{2} = 0$ and $y + 3 x^{2} = 1$ , is equal to:

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Solution

$y + 2 x^{2} = 0$

$y + 3 x^{2} = 1$

To find $P$ and $Q$

$1 - 3 x^{2} = - 2 x^{2}$

$1 = 3 x^{2} - 2 x^{2}$

$x^{2} = 1$

$x = \pm 1$

$y = - 2 x^{2}$

$y = - 2$

$P = (- 1, - 2), Q = (1, - 2)$

$A = 2 \int_{0}^{1} 1 - 3 x^{2} + {2 x}^{2} d x = 2 \int_{0}^{1} 1 - x^{2} d x$

$A = 2 {[x - \frac{x^{3}}{3}]}_{0}^{1} A = 2 [1 - \frac{1}{3}] = \frac{4}{3}$

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