Question

Find the area bounded by the parabola y² = 4x and the line y = 2x − 4.
(i) By using horizontal strips
(ii) By using vertical strips.

Answer 1

To find the points of intersection between the parabola and the line let us substitute y = 2x − 4 in y² = 4x.
$$\begin{gathered} {\left(2x-4\right)}^2=4x \\ \Rightarrow 4x^2+16-16x=4x \\ \Rightarrow 4x^2-20x+16=0 \\ \Rightarrow x^2-5x+4=0 \\ \Rightarrow \left(x-1\right)\left(x-4\right)=0 \\ \Rightarrow x=1,4 \end{gathered}$$
$\Rightarrow y=-2,4$
Therefore, the points of intersection are C(1, −2) and A(4, 4).

(i) Using Horizontal Strips:
The area of the required region ABCD
$$\begin{gathered} A={\int }_{-2}^4\left(x_1-x_2\right)\mathit{d}y\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathrm{where},x_1=\frac{y+4}{2}\mathrm{and}{x}_2=\frac{y^2}{4}\right) \\ ={\int }_{-2}^4\left[\left(\frac{y+4}{2}\right)-\left(\frac{y^2}{4}\right)\right]\mathit{d}y \\ ={\left[\frac{y^2}{4}+2y-\frac{y^3}{12}\right]}_{-2}^4 \\ =\left(\frac{4^2}{4}+2\times 4-\frac{4^3}{12}\right)-\left[\frac{{\left(-2\right)}^2}{4}+2\left(-2\right)-\frac{{\left(-2\right)}^3}{12}\right] \\ =\left(4+8-\frac{16}{3}\right)-\left[1-4+\frac{2}{3}\right] \\ =12-\frac{16}{3}+3-\frac{2}{3} \\ =15-\frac{18}{3} \\ =15-\frac{18}{3} \\ =15-6=9\mathrm{sq}.\mathrm{units} \end{gathered}$$


(ii) Using Vertical Strips:
The area of the required region ABCD
$A={\int }_0^4{y}_{2}dx-{\int }_1^4{y}_{1}dx\left(\mathrm{Where},y_2=2\sqrt{x}\mathrm{and}{y}_1=2x-4\right)$
$$\begin{gathered} ={\int }_0^{4}2\sqrt{x}dx-{\int }_1^4\left(2x-4\right)dx \\ ={\left[\frac{4}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^4-{\left[x^2-4x\right]}_1^4 \\ =\left[\left\{\frac{4}{3}{\left(4\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}-\left\{\frac{4}{3}{\left(0\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}\right]-\left[\left(4^2-4\times 4\right)-\left(1^2-4\times 1\right)\right] \\ =\left[\frac{32}{3}-0\right]-\left[0-\left(1-4\right)\right] \\ =\frac{32}{3}-3 \\ =\frac{23}{3}\mathrm{square}\mathrm{units} \end{gathered}$$