Question

Find the area enclosed by the parabolas y = 4x − x² and y = x² − x.

Answer 1

We have, 
\[y = 4x - x^2\]  and \[y = x^2 - x\]
The points of intersection of two curves is obtained by solving the simultaneous equations
\[\therefore x^2 - x = 4x - x^2 \]
\[ \Rightarrow 2 x^2 - 5x = 0 \]
\[ \Rightarrow x = 0\text{ or }x = \frac{5}{2}\]
\[ \Rightarrow y = 0\text{ or }y = \frac{15}{4}\]
\[ \Rightarrow O\left( 0, 0 \right)\text{ and }D \left( \frac{5}{2} , \frac{15}{4} \right)\text{ are points of intersection of two parabolas .} \]
\[ \text{ In the shaded area CBDC , consider P }(x, y_2 )\text{ on }y = 4x - x^2\text{ and Q }(x, y_1 )\text{ on }y = x^2 - x\]
\[\text{ Area }\left( OBDCO \right) =\text{ area }\left( OBCO \right) +\text{ area }\left( CBDC \right)\]
\[ = \int_0^1 \left| y \right| dx + \int_1^\frac{5}{2} \left| y_2 - y_1 \right| dx\]
\[ = \int_0^1 y dx + \int_1^\frac{5}{2} \left( y_2 - y_1 \right) dx ............\left\{ \because y > 0 \Rightarrow \left| y \right| = y\text{ and } \left| y_2 - y_1 \right| \Rightarrow y_2 - y_1\text{ as }y_2 > y_1 \right\} \]
\[ = \int_0^1 \left( 4x - x^2 \right)dx + \int_1^\frac{5}{2} \left\{ \left( 4x - x^2 \right) - \left( x^2 - x \right) \right\}dx\]
\[ = \left[ \frac{4 x^2}{2} - \frac{x^3}{3} \right]_0^1 + \int_1^\frac{5}{2} \left( 5x - 2 x^2 \right)dx\]
\[ = \left[ 2 x^2 - \frac{x^3}{3} \right]_0^1 + \left[ \frac{5 x^2}{2} - \frac{2 x^3}{3} \right]_1^\frac{5}{2} \]
\[ = \left( 2 - \frac{1}{3} \right) + \left[ \frac{5}{2} \left( \frac{5}{2} \right)^2 - \frac{2}{3} \left( \frac{5}{2} \right)^3 - \frac{5}{2} + \frac{2}{3} \right]\]
\[ = \left( \frac{5}{3} \right) + \left[ \left( \frac{5}{2} \right)^3 \left( 1 - \frac{2}{3} \right) - \frac{11}{6} \right]\]
\[ = \frac{5}{3} + \left( \frac{5}{2} \right)^3 \frac{1}{3} - \frac{11}{6}\]
\[ = \frac{10 - 11}{6} + \frac{125}{24} \]
\[ = \frac{121}{24}\text{ sq units }. ...... . \left( 1 \right)\]
\[\text{ Area }\left( OCV'O \right) = \int_0^1 \left| y \right| dx = \int_0^1 - y dx ............\left\{ \because y < 0 \Rightarrow \left| y \right| = - y \right\}\]
\[ = \int_0^1 - \left( x^2 - x \right)dx\]
\[ = \int_0^1 \left( x - x^2 \right) dx\]
\[ = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 \]
\[ = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\text{ sq units }. ...... . \left( 2 \right)\]
\[\text{ From }\left( 1 \right)\text{ and }\left( 2 \right)\]
\[\text{ Shaded area = area }\left( OBDCO \right)\text{ and area }\left( OCV'O \right)\]
\[ = \frac{121}{24} + \frac{1}{6}\]
\[ = \frac{125}{24}\text{ sq units }\]