Question

Find the area of the region bounded by the curve $(y-1{)}^2=4(x+1)$ and the line $y=x-1$.

Answer 1

$(y-1{)}^2=4(x+1)\Rightarrow x=\frac{(y-1{)}^2-1}{4}$

$y=x-1$ $x=y+1$

$A=\left|{\int }_{-1}^7(y+1-\frac{(y-1{)}^2}{4}+\frac{1}{4})dy\right|$

$A=|\frac{y^2}{2}+y-\frac{(y-1{)}^3}{12}+\frac{1}{4}y{|}_{-1}^7|$

$A=|\frac{49}{2}+7-\frac{216}{12}+\frac{7}{4}-\frac{1}{2}+1+\frac{B}{12}+\frac{1}{4}|$

$A=|\frac{49}{2}-\frac{216}{12}+\frac{8}{12}+\frac{38}{4}|=|\frac{49}{2}-\frac{36}{2}+\frac{2}{3}+\frac{17}{2}|$

$A=\left|\frac{147-108+4+57}{6}\right|=\left|\frac{94}{6}\right|=15.66$ sq.units