Question

Find the area bounded by the parabola y = 2 − x² and the straight line y + x = 0.

Answer 1

The graph of the parabola \[y = 2 - x^2\] and the line \[x + y = 0\] can be given as: 

To find the points of intersection between the parabola and the line let us substitute \[x = - y\]  in \[y = 2 - x^2\]
\[y = 2 - y^2 \]
\[ \Rightarrow y^2 + y - 2 = 0\]
\[ \Rightarrow \left( y - 1 \right)\left( y + 2 \right) = 0\]
\[ \Rightarrow y = 1, - 2\]"
\[\Rightarrow x = - 1, 2\]
Therefore, the points of intersection are \[A( - 1, 1)\] and \[C\left( 2, - 2 \right)\] 
The area of the required region ABCD = \[\int_{- 1}^2 y_1 d x - \int_{- 1}^2 y_2 d x\] where \[y_1 = 2 - x^2\] and \[y_2 = - x\]
Required Area 
\[= \int_{- 1}^2 \left( 2 - x^2 + x \right) d x\]
\[ = \left[ 2x - \frac{x^3}{3} + \frac{x^2}{2} \right]_{- 1}^2 \]
\[ = \left[ \left\{ 2\left( 2 \right) - \frac{\left( 2 \right)^3}{3} + \frac{\left( 2 \right)^2}{2} \right\} - \left\{ 2\left( - 1 \right) - \frac{\left( - 1 \right)^3}{3} + \frac{\left( - 1 \right)^2}{2} \right\} \right]\]

After simplifying we get, \[= \frac{9}{2}\]  square units