$y^2 = 4x$ and the line y = 2x – 4

Solution:

$\begin{array}{ll}{\text { The parabola } y^{2}=4 x \text { and the line } y=2 x-4 \text { intersect where }} \\ {(2 x-4)^{2}=4 x} \\ {\therefore 4 x^{2}-16 x+16=4 x} \\{\therefore 4 x^{2}-20 x+16=0} \\ {\therefore x^{2}-5 x+4=0} \\ {\therefore(x-4)(x-1)=0} \\ {\therefore x=1,4}\end{array}$

When $x=1, y=2-4=-2 ;$ and when $x=4, y=8-4=4$ . Thus, the points of intersection are $A(1,-2)$ and $B(4,4) .$

Now, consider a strip parallel to x -axis. On this strip x varies from $x=y^{2 / 4}$ to $x=(y+4) / 2 .$ The strip then moves parallel to the x -axis from $y=-2$ to $y=4 .$,

$\begin{aligned} \therefore \mathrm{A}=& \int_{-2}^{4} \int_{y^{2} / 4}^{(\mathrm{y}+4) / 2} d x d y=\int_{-2}^{4}[x]_{\frac{y^{2}}{2}}^{\frac{y+4}{2}} d y \\ &=\int_{-2}^{4}\left(\frac{y+4}{2}-\frac{y^{2}}{4}\right) d y \\ &=\frac{1}{4} \int_{-2}^{4}\left(2 y+8-y^{2}\right) d y \end{aligned}$

$\begin{aligned} &=\frac{1}{4}\left[y^{2}+8 y-\frac{y^{3}}{3}\right]_{-2}^{4} \\ &=\frac{1}{4}\left[\left(16+32-\frac{64}{3}\right)-\left(4-16+\frac{8}{3}\right)\right] \\ &=\frac{1}{4}(60-24) \\ \therefore A=9 & \end{aligned}$