The area of the region enclosed by the parabola (y-2)^2=x-1, the line x-2 y+4=0 and the po…

Question

The area of the region enclosed by the parabola $(y-2)^2=x-1$, the line $x-2 y+4=0$ and the positive coordinate axes is \_\_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Solving the equations

$\begin{gathered} (y-2{)}^{2}=x-1x-2y+4=0 \ x=2(y-2) \end{gathered}$

$\begin{gathered} \frac{{x}^{2}}{4}=x-1 \ {x}^{2}-4x+4=0 \ (x-2{)}^{2}=0 \ x=2 \end{gathered}$

Exclose area (w.r.t. y-axis) $=\int _{0}^{3}xdy-\Delta$.

$\begin{gathered} =\int _{0}^{3}((y-2{)}^{2}+1)dy-\frac{1}{2}	imes 1	imes 2 \ =\int _{0}^{3}({y}^{2}-4y+5)dy-1 \ ={[\frac{{y}^{3}}{3}-2{y}^{2}+5y]}_{3}^{0}-1 \ =9-18+15-1=5 \end{gathered}$

Solutions · 1