Question

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

Answer 1

To find the points of intersection between the parabola and the line let us substitute y = 2x − 4 in y² = 4x.
\[\left( 2x - 4 \right)^2 = 4x\]
\[ \Rightarrow 4 x^2 + 16 - 16x = 4x\]
\[ \Rightarrow 4 x^2 - 20x + 16 = 0\]
\[ \Rightarrow x^2 - 5x + 4 = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x - 4 \right) = 0\]
\[ \Rightarrow x = 1, 4\]
\[\Rightarrow y = - 2, 4\]
Therefore, the points of intersection are C(1, −2) and A(4, 4).
Using Horizontal Strips:-
The area of the required region ABCD
\[A = \int_{- 2}^4 \left( x_1 - x_2 \right) dy ...........\left(\text{where,} x_1 = \frac{y + 4}{2}\text{ and }x_2 = \frac{y^2}{4} \right)\]
\[ = \int_{- 2}^4 \left[ \left( \frac{y + 4}{2} \right) - \left( \frac{y^2}{4} \right) \right] d y\]
\[ = \left[ \frac{y^2}{4} + 2y - \frac{y^3}{12} \right]_{- 2}^4 \]
\[ = \left( \frac{4^2}{4} + 2 \times 4 - \frac{4^3}{12} \right) - \left[ \frac{\left( - 2 \right)^2}{4} + 2\left( - 2 \right) - \frac{\left( - 2 \right)^3}{12} \right]\]
\[ = \left( 4 + 8 - \frac{16}{3} \right) - \left[ 1 - 4 + \frac{2}{3} \right]\]
\[ = 12 - \frac{16}{3} + 3 - \frac{2}{3}\]
\[ = 15 - \frac{18}{3}\]
\[ = 15 - \frac{18}{3}\]
\[ = 15 - 6 = 9\text{ sq . units }\]