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Q.
The area of the region bounded by the parabola $y^2=2 x$ and the straight line $x-y=4$ is
Application of Integrals
Solution:
The intersecting points of the given curves are obtained by solving the equations $x-y=4$ and $y^2=2 x$ for $x$ and $y$.
We have, $y^2=8+2$ y i.e., $(y-4)(y+2)=0$
which gives $y=4,-2$ and $x=8,2$.
Thus, the points of intersection are $(8,4)$ and $(2,-2)$.
$\therefore $ Area $ =\int\limits_{-2}^4\left(4+y-\frac{1}{2} y^2\right) d y$
$ =\left|4 y+\frac{y^2}{2}-\frac{1}{6} y^3\right|_{-2}^4$
$ =18 \text { sq units }$
