If the area (in sq. units) of the region \left \{ \left ( x,y \right ):y^{2}\leq 4x,x+y\leq 1,x\geq 0,y\geq 0 \right \} is a\sqrt{2}+b, then a-b is equal to : 

  • Option 1)

    -\frac{2}{3}

  • Option 2)

    \frac{10}{3}

  • Option 3)

    6

  • Option 4)

    \frac{8}{3}

Answers (1)

\left \{ \left ( x,y \right ):y^{2}\leqslant 4x;x+y\leqslant 1,x\geqslant 0,y\geqslant 0 \right \}

A= \int_{0}^{3-2\sqrt{2}}2\sqrt{x}dx+\frac{1}{2}\left ( 1-\left ( 3-2\sqrt{2} \right ) \right )\left ( 1-\left ( 3-2\sqrt{2} \right ) \right )

    =\frac{2\left [ x^{\frac{3}{2}} \right ]_{0}^{3-2\sqrt{2}}}{\frac{3}{2}}+\frac{1}{2}\left ( 2\sqrt{2} -2\right )\left ( 2\sqrt{2}-2 \right )

    =\frac{8\sqrt{2}}{3}+\left ( -\frac{10}{3} \right )

a=\frac{8}{3}\: \: \: ,\: \: \: b=\frac{-10}{3}

a-b=\frac{8}{3}-\left ( -\frac{10}{3} \right )=\frac{18}{3}=6


Option 1)

-\frac{2}{3}

Option 2)

\frac{10}{3}

Option 3)

6

Option 4)

\frac{8}{3}

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