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

The curve = –x2 or x2 = –and the line + 2 = 0

NCERT Exemplar Solutions Class 12 Mathematics Chapter 8 - 28

Solving the two equation, we get

\[x\text{ }-\text{ }{{x}^{2}}~+\text{ }2\text{ }=\text{ }0\]

\[{{x}^{2}}~\text{ }-x\text{ }\text{ }-2\text{ }=\text{ }0\]

\[{{x}^{2}}~\text{ }-2x\text{ }+\text{ }x\text{ }-\text{ }2\text{ }=\text{ }0\]

\[x\left( x\text{ }-\text{ }2 \right)\text{ }+\text{ }1\text{ }\left( x\text{ }-\text{ }2 \right)\text{ }=\text{ }0\]

\[\left( x\text{ }-\text{ }2 \right)\text{ }\left( x\text{ }+\text{ }1 \right)\text{ }=\text{ }0\]

\[x\text{ }=\text{ }-1,\text{ }2\]

The area of the required shaded region

NCERT Exemplar Solutions Class 12 Mathematics Chapter 8 - 29

NCERT Exemplar Solutions Class 12 Mathematics Chapter 8 - 30

Therefore, the required area = 9/2 sq.units