Question

Find the area of the region between the parabola x = 4y − y² and the line x = 2y − 3.

Answer 1

To find the point of intersection of the parabola x = 4y − y² and the line x = 2y − 3
Let us substitute x = 2y − 3 in the equation of the parabola.
$$\begin{gathered} 2y-3=4y-y^2 \\ \Rightarrow y^2-2y-3=0 \\ \Rightarrow \left(y+1\right)\left(y-3\right) \\ \Rightarrow y=-1,3 \end{gathered}$$
Therefore, the points of intersection are D(−1, −5) and A(3, 3).
The area of the required region ABCDOA,
$$\begin{gathered} A={\int }_{-1}^3\left|x_1-x_2\right|\mathit{d}y\mathit{}\mathit{}\mathit{}\mathrm{where},x_1=4y-y^2\mathrm{and}{x}_2=2y-3 \\ ={\int }_{-1}^3\left(x_1-x_2\right)\mathit{d}y\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }{\mathit{x}}_{\mathit{1}}\mathit{>}{\mathit{x}}_{\mathit{2}}\right) \\ ={\int }_{-1}^3\left[\left(4y-y^2\right)-\left(2y-3\right)\right]\mathit{d}y \\ ={\int }_{-1}^3\left(4y-y^2-2y+3\right)\mathit{d}y \\ ={\int }_{-1}^3\left(-y^2+2y+3\right)\mathit{d}y \\ ={\left[-\frac{y^3}{3}+\frac{2y^2}{2}+3y\right]}_{-1}^3 \\ ={\left[-\frac{y^3}{3}+y^2+3y\right]}_{-1}^3 \\ =-\frac{3^3}{3}+3^2+9-\left(\frac{1}{3}+1-3\right) \\ =-3^2+3^2+9-\frac{1}{3}-1+3 \\ =11-\frac{1}{3} \\ =\frac{32}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$