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The area (in sq. units) of the region

\left \{ \left ( x,y \right ) \: :x\geqslant 0,x+y\leq 3,x^{2}\leq 4y\: \: and\: \: y\leq 1+\sqrt{x}\right \}

 is :

 

 

  • Option 1)

    \frac{3}{2}

  • Option 2)

    \frac{7}{3}

  • Option 3)

    \frac{5}{2}

  • Option 4)

    \frac{59}{12}

 

Answers (1)

As learnt in

Area along x axis -

Let y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x) be two curve then area bounded between the curves and the lines

x = a and x = b is

\left | \int_{a}^{b} \Delta y\, dx\right |= \left | \int_{a}^{b}\left ( y_{2}-y_{1} \right ) dx\right |

 

- wherein

Where \Delta y= f_{2}\left ( x \right )-f_{1}(x)

 

 

Point of intersection of 

x+y=3; y=1+\sqrt{x}; x^{2}=4y

Area =\int_{0}^{1}(1+\sqrt{x}) \: dx+\int_{1}^{2}\left ( 3-x \right )dx-\int_{0}^2{}\left ( \frac{x^{2}}{4} \right )dx

=\left [ x+\frac{2}{3}{x^{\frac{3}{2}}} \right ]^{1}_{0}+\left [ 3x-\frac{x^{2}}{2} \right ]^{2}_{1}-\left [ \frac{x^{3}}{12} \right ]^{2}_{0}

=\left [ 1+\frac{2}{3} \right ]+3-\frac{3}{2}-\frac{8}{12}

=\frac{5}{3}+\frac{3}{2}-\frac{2}{3}

=1+\frac{3}{2}

=\frac{5}{2}


Option 1)

\frac{3}{2}

Incorrect option    

Option 2)

\frac{7}{3}

Incorrect option    

Option 3)

\frac{5}{2}

Correct option

Option 4)

\frac{59}{12}

Incorrect option    

Posted by

Sabhrant Ambastha

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