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• Notes

Find the area enclosed by the line
$y=x-1 \text { and the parabolas } y^{2}=2 x+6$

$2 x+6=(x-1)^{2}$

$2 x+6=x^{2}-2 x+1$

$x^{2}-2 x+1-2 x-6=0$

$x^{2}-4 x-5=0$

$(x-5)(x+1)=0$

إما
$x-5=0 \rightarrow x=5 \rightarrow y=4 \rightarrow(5,4)$

أو
$x+1=0 \longrightarrow x=-1 \rightarrow y=-2 \rightarrow(-1,-2)$

$$$\begin{array}{l}{\frac{x}{0}+\frac{y}{-1}} \\ {1} & {0} \\ {2} & {1} \\ {3} & {2} \\ {-1} & {-2}\end{array}$$$

$y=x_R-1 \rightarrow x _R=y+1$

$y^{2}=2 x_L+6 \rightarrow 2 x_{L}=y^{2}-6 \rightarrow x_{L}=\frac{1}{2} y^{2}-3$

horizontal cross - section:
$d y :-2 \rightarrow 4$

Area = $\int_{-2}^{4}(x_{right} - x_{left})dy$

$=\int_{-2}^{4}\left[y+1-\left(\frac{1}{2} y^{2}-3\right)\right] d y=\int_{-2}^{4}\left(y+1-\frac{1}{2} y^{2}+3\right) d y$

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

$=\left[\left(\frac{(4)^{2}}{2}-\frac{(4)}{6}+4(4)\right)-\left(\frac{(-2)^{2}}{2}-\frac{(-2)^{3}}{6}+4+2\right)\right]$

$=18$