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Question

The area of the region $$R = \{ (x,y) / x^2 \le y \le x \} $$is 


A
16
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B
23
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C
43
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D
2
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Solution

The correct option is A $$ \dfrac {1}{6}$$
$$x : \left(x,y\right)\quad \left\{ { x }^{ 2 }\le y\le x \right\} $$
$$y=x\qquad y={x}^{2}$$
Meeting point $$x=0$$ and $$x=1$$
$$\displaystyle \therefore area=\int _{ 0 }^{ 1 }{ \left( { x }^{ 2 }-x \right)  } dx$$
$$\displaystyle \Rightarrow { \left. \dfrac { { x }^{ 3 } }{ 3 }  \right]  }_{ 0 }^{ 1 }-{ \left. \dfrac { { x }^{ 2 } }{ 2 }  \right]  }_{ 0 }^{ 1 }$$
$$\Rightarrow \left| \dfrac { 1 }{ 3 } -\dfrac { 1 }{ 2 }  \right|$$
$$\Rightarrow \dfrac { \left| 2-3 \right|  }{ 6 } \Rightarrow \dfrac { 1 }{ 6 } sq.unit$$
$$\boxed {\therefore Area\ of\ region\ ‘R’\ is\ \dfrac { 1 }{ 6 }}$$

Physics

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