A) \[\frac{1}{9}sq\,\,unit\]
B) \[\frac{1}{3}sq\,\,unit\]
C) \[\frac{1}{\sqrt{3}}sq\,\,unit\]
D) \[\frac{2}{3}sq\,\,unit\]
Correct Answer: B
Solution :
Given curves\[y={{x}^{2}},\,\,x={{y}^{2}}\] \[\therefore \]Point of intersections \[x={{({{x}^{2}})}^{2}}\Rightarrow {{x}^{4}}-x=0\] \[\Rightarrow \] \[x({{x}^{3}}-1)=0\] \[\therefore \] \[x=0,\,\,1\] For \[x=0,\,\,y=0;\,\,x=1,\,\,y=1\] Area of shaded region \[=\int_{0}^{1}{{{x}^{1/2}}}dx-\int_{0}^{1}{{{x}^{2}}dx}\] \[=\left[ \frac{2{{x}^{3/2}}}{3} \right]_{0}^{1}-\left[ \frac{1}{3}{{x}^{3}} \right]_{0}^{1}\] \[=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}sq\,\,unit\] \[\therefore \]Required area\[=\frac{1}{3}sq\,\,unit\]You need to login to perform this action.
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