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The area (in sq. units) of the region \left \{ x\equiv \mathbf{R}\: :\: x\geq 0,\: y\geq 0,\: y\geq x-2\;\: and\; y\leq \sqrt{x} \right \} is :

  • Option 1)

    \frac{13}{3}

     

     

     

  • Option 2)

    \frac{8}{3}

  • Option 3)

    \frac{10}{3}

  • Option 4)

    \frac{5}{3}

 

Answers (2)

best_answer

As we learned 

 

Geometrical integration of a definite integral -

An algebraic sum of the area of the figure bounded by the curve y = f(x), the x axis and the striaght lines x=a and x=b. The areas above x axis are taken as positive and the areas below x axis are taken as negative.  
 

- wherein

Where a< b

Hence

\int_{a}^{b}f\left ( x \right )dx=

ar\left ( PAQP \right )-ar\left ( QTRQ \right )+ar\left ( RBSR \right )

 

 

I= \int_{0}^{4}\sqrt{x}dx=\frac{2}{3}\left [ x^{\frac{3}{2}}\right ]_{0}^{4}=\frac{16}{3}

Also shaded region    \frac{16}{3}-\frac{1}{2}\times 2\times 2=\frac{16}{3}-2=\frac{10}{3}

 


Option 1)

\frac{13}{3}

 

 

 

Option 2)

\frac{8}{3}

Option 3)

\frac{10}{3}

Option 4)

\frac{5}{3}

Posted by

Himanshu

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