Solve this following

Question:

The area of the region bounded by $\mathrm{y}-\mathrm{x}=2$ and $\mathrm{x}^{2}=\mathrm{y}$ is equal to :-

 

  1. $\frac{16}{3}$

  2. $\frac{2}{3}$

  3. $\frac{9}{2}$

  4. $\frac{4}{3}$


Correct Option: , 3

Solution:

$y-x=2, x^{2}=y$

Now, $x^{2}=2+x$

$\Rightarrow x^{2}-x-2=0$

$\Rightarrow(x+1)(x-2)=0$

Area $=\int_{-1}^{2}\left(2+x-x^{2}\right)$

$=\left|2 x+\frac{x^{2}}{2}-\frac{x^{3}}{3}\right|_{-1}^{2}$

$=\left(4+2-\frac{8}{3}\right)-\left(-2+\frac{1}{2}+\frac{1}{3}\right)$

$=6-3+2-\frac{1}{2}=\frac{9}{2}$

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