Question

Find the area enclosed by the curves 3x² + 5y = 32 and y = | x − 2 |.

Answer 1

\[3 x^2 + 5y = 32\]  represents a downward opening parabola, symmetrical about negative x-axis
\[\text{ vertex of the parabola is D }\left( 0, \frac{32}{5} \right) . \text{ It cuts the X axis at B }\left( \sqrt{\frac{32}{3}} , 0 \right) \text{ and B' }\left( - \sqrt{\frac{32}{3}} , 0 \right)\]
Also, 
\[y = \left| x - 2 \right| \]
\[ \Rightarrow y = \begin{cases}x - 2 ,& x \geq 2\\2 - x ,& x < 2\end{cases}\]
\[ y = x - 2 , x \geq 2\text{ represents a line , cutting the }x \text{ axis at A } \left( 2, 0 \right)\text{ and the parabola at C }\left( 3, 1 \right)\]
\[\text{ And }y = 2 - x, x < 2\text{ represents a line, cutting the parabola at }\left( - 2, 4 \right)\]
Shaded area AEYCA
\[ = \int_{- 2}^2 \left[ \left( \frac{32 - 3 x^2}{5} \right) - \left( 2 - x \right) \right]dx + \int_2^3 \left[ \left( \frac{32 - 3 x^2}{5} \right) - \left( x - 2 \right) \right]dx \]
\[ = \int_{- 2}^3 \left( \frac{32 - 3 x^2}{5} \right)dx - \left[ \int_{- 2}^2 \left( 2 - x \right)dx + \int_2^3 \left( x - 2 \right)dx \right]\]
\[ = \frac{1}{5} \left[ 32x - x^3 \right]_{- 2}^3 - \left[ 2x - \frac{x^2}{2} \right]_{- 2}^2 - \left[ \frac{x^2}{2} - 2x \right]_2^3 \]
\[ = \frac{1}{5}\left[ 32 \times 3 - 27 + 64 - 8 \right] - \left( 4 - 2 + 4 + 2 \right) - \left( \frac{9}{2} - 6 - 2 + 4 \right)\]
\[ = \frac{1}{5}\left[ 96 - 35 + 64 \right] - \left( 8 \right) - \left( \frac{9}{2} - 4 \right)\]
\[ = \frac{125}{5} - 4 - \frac{9}{2}\]
\[ = 25 - 4 - \frac{9}{2}\]
\[ = 21 - \frac{9}{2}\]
\[ = \frac{42 - 9}{2}\]
\[ = \frac{33}{2}\text{ sq units }\]