Area of bounded between $y=x-1$ and $(y-1)^2=4 (x+1)$ Find the area enclosed by curves $y=x-1$ and $(y-1)^2=4 (x+1)$.
I found point of intersection as $(0,-1)$ and $(8,7)$ and set up integral as:
$|\int_{-1}^{7} [(y+1)-\frac{(y-1)^2}{4}-1]dy|$. Am I heading in right direction?

Finding the points of intersection of the two curves: $(y-1)^2 = 4(y+2) \Rightarrow y^2-2y+1 - 4y-8 = 0\Rightarrow y^2-6y-7 = 0\Rightarrow (y+1)(y-7) = 0\Rightarrow y = -1, 7\Rightarrow A = \displaystyle \int_{-1}^7 \left(y+1- \left(\dfrac{(y-1)^2}{4}-1\right)\right)dy$. The difference between your answer and mine is the last $-1$ in your answer. I think it should be $+1$.
