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

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

 


\(y=x_R-1 \rightarrow x _R=y+1\)
 


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

horizontal cross - section: 
\(d y :-2 \rightarrow 4\)
 

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


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


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


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


\(=18\)
 

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