Questions

# The area bounded by the curves  is

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a
$\frac{2}{3}\left(\sqrt{2}+1\right)$
b
$\frac{4}{3}\left(\sqrt{2}-1\right)$
c
$2\left(\sqrt{2}-1\right)$
d
$\frac{8}{3}\left(\sqrt{2}-1\right)$
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detailed solution

Correct option is D

$\mathrm{y}=\left|{\mathrm{x}}^{2}-1\right|$

Area = ABCDEA

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

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