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

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KollegeApply · Accepted Answer

$\frac{1}{3}$ 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( y/2 \right) - \left( y^$\frac{2}{4}$ \right) \right] dy = \left[ $\frac{1}{2}$\left( y^$\frac{2}{2}$ \right) - $\frac{1}{4}$\left( y^$\frac{3}{3}$ \right) \right]_0^2 = $\frac{1}{4}$\left( 2^2 \right) - $\frac{1}{12}$\left( 2^3 \right) - 0 = 1 - $\frac{8}{12}$ = 12 - $\frac{8}{12}$ = $\frac{1}{3}$\text{ sq . units } [Solution diagram: 44146_solution_1.png]

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