Question

Area lying between the curves y² = 4x and y = 2x is

Options

• \[\frac{2}{3}\]
• \[\frac{1}{3}\]
• \[\frac{1}{4}\]
• \[\frac{3}{4}\]

Answer 1

\[\frac{1}{3}\]

The points of intersection of the straight line and the parabola is obtained by solving the simultaneous equations
\[y^2 = 4x\text{ and }y = 2x\]
\[ \Rightarrow \left( 2x \right)^2 = 4x\]
\[ \Rightarrow 4 x^2 = 4x\]
\[ \Rightarrow x\left( x - 1 \right) = 0\]
\[ \Rightarrow x = 0\text{ or }x = 1\]
\[ \Rightarrow y = 0\text{ or }y = 2\]
\[\text{ Thus O }\left( 0, 0 \right)\text{ and A}\left( 1, 2 \right)\text{ are the points of intersection of the parabola and straight line }\]
Shaded area is the required area . 
Using the horizontal strip method , 
\[\text{ Shaded area }= \int_0^2 \left| x_2 - x_1 \right|dy\]
\[ = \int_0^2 \left[ \left( \frac{y}{2} \right) - \left( \frac{y^2}{4} \right) \right] dy\]
\[ = \left[ \frac{1}{2}\left( \frac{y^2}{2} \right) - \frac{1}{4}\left( \frac{y^3}{3} \right) \right]_0^2 \]
\[ = \frac{1}{4}\left( 2^2 \right) - \frac{1}{12}\left( 2^3 \right) - 0\]
\[ = 1 - \frac{8}{12}\]
\[ = \frac{12 - 8}{12}\]
\[ = \frac{1}{3}\text{ sq . units }\]