### Area Under Curves Question 10

#### Question 10 - 2024 (30 Jan Shift 2)

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

## Show Answer

#### Answer (5)

#### Solution

Solving the equations

$(y-2)^{2}=x-1$ and $x-2 y+4=0$

$ x=2(y-2) $

$\frac{x^{2}}{4}=x-1$

$x^{2}-4 x+4=0$

$(x-2)^{2}=0$

$x=2$

Exclose area (w.r.t. y-axis) $=\int _0^{3} x d y-$ Area of $\Delta$.

$ \begin{aligned} & =\int _0^{3}\left((y-2)^{2}+1\right) d y-\frac{1}{2} \times 1 \times 2 \\ & =\int _0^{3}\left(y^{2}-4 y+5\right) d y-1 \\ & =\left[\frac{y^{3}}{3}-2 y^{2}+5 y\right] _0^{3}-1 \\ & =9-18+15-1=5 \end{aligned} $