Discussion Forum

The area (in sq units) of the  region described by   $(x,y:y^{2}\leq2x)$ and  $(y\geq4x-1)$  is 

Answer:

Explanation:

 Given region is       $(x,y:y^{2}\leq2x)$ and  $(y\geq4x-1)$                          

    $y^{2}\leq 2x$   represents  a region inside the  parabola

           $y^{2}= 2x$                ..............(i)

 and  $y\geq 4x-1$   represents   a region to the left of the line

                           y=4x-1                    .......(ii)

 The point of intersection of the curve (i) and (ii) is 

                      $( 4x-1)^{2}=2x$

$\Rightarrow$       $16x^{2}+1-8x=2x$

$\Rightarrow$       $16x^{2}-10x+1=0$

$\Rightarrow$      $x=\frac{1}{2},\frac{1}{8}$

$\therefore$    The points where these curves interset , are    $\left(\frac{1}{2},1\right)$  and   $\left(\frac{1}{8},\frac{1}{2}\right)$

Hence, required area

                    =  $\int_{-1/2}^{1} \left( \frac{y+1}{4}-\frac{y^{2}}{2}\right)dy$

                 =   $\frac{1}{4}\left(\frac{y^{2}}{2}+y\right)_{-1/2}^1-\frac{1}{6}(y^{3})^{1}_{-1/2}$

              =  $\frac{1}{4}\left\{\left(\frac{1}{2}+1\right)-\left(\frac{1}{8}-\frac{1}{2}\right)\right\}-\frac{1}{6}\left\{1+\frac{1}{8}\right\}$

       =   $\frac{1}{4}\left\{\frac{3}{2}+\frac{3}{8}\right\}-\frac{1}{6}\left\{\frac{9}{8}\right\}$

   =  $\frac{1}{4}\times \frac{15}{8}-\frac{3}{16}=\frac{9}{32}$