Question

Find the area enclosed by the parabolas y = 5x² and y = 2x² + 9.

Answer 1

\[y = 5 x^2\text{ represents a parabola with vertex at O }\left( 0, 0 \right)\text{ and opening upwards , symmetrical about + ve }y \text{ axis }\]
\[y = 2 x^2 + 9\text{ represents the wider parabola , with vertex at C }\left( 0, 9 \right) \]
To find point of intersection , solve the two equations
\[5 x^2 = 2 x^2 + 9 \]
\[ \Rightarrow 3 x^2 = 9\]
\[ \Rightarrow x = \pm \sqrt{3}\]
\[ \Rightarrow y = 15\]
\[\text{ Thus A }\left( \sqrt{3}, 15 \right)\text{ and A'}\left( - \sqrt{3}, 15 \right)\text{ are points of intersection of the two parabolas . }\]
\[\text{ Shaded area A'OA }= 2 \times\text{ area }\left( OCAO \right)\]
\[\text{ Consider a vertical stip of length }= \left| y_2 - y_1 \right|\text{ and width }= dx \]
\[\text{ Area of approximating rectangle }= \left| y_2 - y_1 \right|dx \]
\[\text{ The approximating rectangle moves from }x = 0 \text{ to }x = \sqrt{3}\]
\[ \therefore \text{ Area }\left( OCAO \right) = \int_0^\sqrt{3} \left| y_2 - y_1 \right|dx = \int_0^\sqrt{3} \left( y_2 - y_1 \right)dx .............\left\{ \because \left| y_2 - y_1 \right| = y_2 - y_1\text{ as }y_2 > y_1 \right\}\]
\[ = \int_0^\sqrt{3} \left( 2 x^2 + 9 - 5 x^2 \right)dx \]
\[ = \int_0^\sqrt{3} \left( 9 - 3 x^2 \right)dx\]
\[ = \left[ \left( 9x - 3\frac{x^3}{3} \right) \right]_0^\sqrt{3} \]
\[ = 9\sqrt{3} - 3\sqrt{3}\]
\[ = 6\sqrt{3}\text{ sq units }\]
\[ \therefore\text{ Shaded area B'A'AB }= 2 \text{ area OCAO }= 2 \times 6\sqrt{3} = 12\sqrt{3}\text{ sq units }\]
\[\text{ Thus area enclosed by two parabolas }= 12\sqrt{3}\text{ sq units }\]