Question

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

Answer 1

$(y-1{)}^2=4(x+1)$ ,$y=1$ , $x=1$

$A=(-1,1)$

$B=(-0.75,0)$

$C=(-1,0)$

As under ABC $={\int }_0^1\left[\frac{1}{4}\right(y-1{)}^2-1]dy$

$=\frac{1}{12}\left[\right(y-1{)}^2{]}_0^1-[y{]}_0^1$

$=\frac{-1}{12}-1=\frac{13}{12}$

Rectangle with sides 2 and 1 units

${\int }_0^{-1}\left[\frac{1}{4}\right(y-1{)}^2-1]dy$

$\Rightarrow$ $2-\frac{13}{12}$

$=\frac{11}{12}units+{\int }_0^{-1.828}\left[\frac{1}{4}\right(y-1{)}^2-1]dy$ (area under BDE)

$\Rightarrow$ $\frac{11}{12}+\frac{2^{4.5}-43}{12}units$