JEE MAIN - Mathematics (2022 - 26th July Evening Shift - No. 14)

The area bounded by the curves $$y=\left|x^{2}-1\right|$$ and $$y=1$$ is
$$\frac{2}{3}(\sqrt{2}+1)$$
$$\frac{4}{3}(\sqrt{2}-1)$$
$$2(\sqrt{2}-1)$$
$$\frac{8}{3}(\sqrt{2}-1)$$

Explanation

JEE Main 2022 (Online) 26th July Evening Shift Mathematics - Area Under The Curves Question 64 English Explanation

Area $$ = 2\int\limits_0^{\sqrt 2 } {(1 - |{x^2} - 1|)dx} $$

$$2\left[ {\int\limits_0^1 {\left( {1 - (1 - {x^2})} \right)dx + \int\limits_1^{\sqrt 2 } {\left( {2 - {x^2}} \right)dx} } } \right]$$

$$ = 2\left[ {\left. {\left[ {{{{x^3}} \over 3}} \right]_0^1 + \left[ {2x - {{{x^3}} \over 3}} \right]_1^{\sqrt 2 }} \right]} \right]$$

$$ = 2\left( {{{4\sqrt 2 - 4} \over 3}} \right) = {8 \over 3}\left( {\sqrt 2 - 1} \right)$$

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