Question

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

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

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

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

$$\begin{gathered} =\left|{\int }_{-1}^2{y}_{\mathrm{line}}dx-{\int }_{-1}^2{y}_{\mathrm{parabola}}dx\right| \\ =\left|{\int }_{-1}^2-\left(x+2\right)dx-{\int }_{-1}^2-x^{2}dx\right| \\ =\left|{\overline{)-\frac{{\left(x+2\right)}^2}{2}}}_{-1}^2+{\overline{)\frac{x^3}{3}}}_{-1}^2\right| \\ =\left|-\frac{1}{2}\left(16-1\right)+\frac{1}{3}\left[8-\left(-1\right)\right]\right| \\ =\left|-\frac{15}{2}+3\right| \\ =\left|-\frac{9}{2}\right| \\ =\frac{9}{2}\mathrm{square}\mathrm{units} \end{gathered}$$