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 $y=2-x^2$.

$$\begin{gathered} y=2-y^2 \\ \Rightarrow y^2+y-2=0 \\ \Rightarrow \left(y-1\right)\left(y+2\right)=0 \\ \Rightarrow y=1,-2 \end{gathered}$$
$\Rightarrow x=-1,2$

Therefore, the points of intersection are $A(-1,1)$ and $C\left(2,-2\right)$.

The area of the required region ABCD =${\int }_{-1}^2{y}_1\mathit{d}x-{\int }_{-1}^2{y}_2\mathit{d}x$ where $y_1=2-x^2$ and $y_2=-x$

Required Area
$$\begin{gathered} ={\int }_{-1}^2\left(2-x^2+x\right)\mathit{d}x \\ ={\left[2x-\frac{x^3}{3}+\frac{x^2}{2}\right]}_{-1}^2 \\ =\left[\left\{2\left(2\right)-\frac{{\left(2\right)}^3}{3}+\frac{{\left(2\right)}^2}{2}\right\}-\left\{2\left(-1\right)-\frac{{\left(-1\right)}^3}{3}+\frac{{\left(-1\right)}^2}{2}\right\}\right] \end{gathered}$$

After simplifying we get,

$=\frac{9}{2}$ square units