Question

Find the area bounded by the parabola y² = 4x and the line y = 2x − 4 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.
\[\left( 2x - 4 \right)^2 = 4x\]
\[ \Rightarrow 4 x^2 + 16 - 16x = 4x\]
\[ \Rightarrow 4 x^2 - 20x + 16 = 0\]
\[ \Rightarrow x^2 - 5x + 4 = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x - 4 \right) = 0\]
\[ \Rightarrow x = 1, 4\]
\[\Rightarrow y = - 2, 4\]
Therefore, the points of intersection are C(1, −2) and A(4, 4).
Using Vertical Strips:-
The area of the required region ABCD

\[A = \int_0^4 y_2 d x - \int_1^4 y_1 d x ..........\left( \text{ Where,} y_2 = 2\sqrt{x}\text{ and }y_1 = 2x - 4 \right)\]
\[= \int_0^4 2\sqrt{x} d x - \int_1^4 \left( 2x - 4 \right) d x\]
\[ = \left[ \frac{4}{3} x^\frac{3}{2} \right]_0^4 - \left[ x^2 - 4x \right]_1^4 \]
\[ = \left[ \left\{ \frac{4}{3} \left( 4 \right)^\frac{3}{2} \right\} - \left\{ \frac{4}{3} \left( 0 \right)^\frac{3}{2} \right\} \right] - \left[ \left( 4^2 - 4 \times 4 \right) - \left( 1^2 - 4 \times 1 \right) \right]\]
\[ = \left[ \frac{32}{3} - 0 \right] - \left[ 0 - \left( 1 - 4 \right) \right]\]
\[ = \frac{32}{3} - 3\]
\[ = \frac{23}{3}\text{ square units }\]