Q.

The area bounded by the curves $y = |x^{2} - 1| and y = 1$ is

Moderate

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By Expert Faculty of Sri Chaitanya

$\frac{2}{3} (\sqrt{2} + 1)$

$\frac{4}{3} (\sqrt{2} - 1)$

$2 (\sqrt{2} - 1)$

$\frac{8}{3} (\sqrt{2} - 1)$

answer is D.

(Detailed Solution Below)

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Detailed Solution

$y = |x^{2} - 1|$

Area = ABCDEA

$\begin{array}{l} = 2 (\int_{0}^{1} (1 - (1 - x^{2})) dx + \int_{1}^{\sqrt{2}} (1 - (x^{2} - 1)) dx) \\ = 2 (\int_{0}^{1} x^{2} dx + \int_{1}^{\sqrt{2}} (2 - x^{2}) dx) \\ = 2 ({(\frac{x^{3}}{3})}_{0}^{1} + {(2 x - \frac{x^{3}}{3})}_{1}^{\sqrt{2}}) \\ 2 (\frac{1}{3} + ((2 \sqrt{2} - \frac{2 \sqrt{2}}{3}) - (2 - \frac{1}{3}))) = 2 (\frac{1}{3} - \frac{5}{3} + \frac{4 \sqrt{2}}{3}) \\ = \frac{8}{3} (\sqrt{2} - 1) \end{array}$

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