Question

Find the area enclosed by the curve \[y = - x^2\] and the straight line x + y + 2 = 0. 

Answer 1

The curve \[y = - x^2\] represents a parabola opening towards the negative y-axis.
The straight line x + y + 2 = 0 passes through (−2, 0) and (0, −2).
Solving
\[y = - x^2\] and x + y + 2 = 0, we get
\[x - x^2 + 2 = 0\]
\[ \Rightarrow x^2 - x - 2 = 0\]
\[ \Rightarrow \left( x - 2 \right)\left( x + 1 \right) = 0\]
\[ \Rightarrow x = 2\text{ or }x = - 1\]
Thus, the parabola \[y = - x^2\]  and the straight line x + y + 2 = 0 intersect at A(−1, −1) and B(2, −4).

∴ Required area = Area of the shaded region OABO
\[= \left| \int_{- 1}^2 y_{\text{ line }} dx - \int_{- 1}^2 y_{\text{ parabola }} dx \right|\]
\[ = \left| \int_{- 1}^2 - \left( x + 2 \right)dx - \int_{- 1}^2 - x^2 dx \right|\]
\[ = \left.\left| {- \frac{\left( x + 2 \right)^2}{2}}\right|_{- 1}^2 +\left. \frac{x^3}{3}\right|_{- 1}^2 \right|\]
\[ = \left| - \frac{1}{2}\left( 16 - 1 \right) + \frac{1}{3}\left[ 8 - \left( - 1 \right) \right] \right|\]
\[ = \left| - \frac{15}{2} + 3 \right|\]
\[ = \left| - \frac{9}{2} \right|\]
\[ = \frac{9}{2}\text{ square units }\]