Find the area of the region bounded by the curves $y = x^{2}$ , $y = 2 - x$ and $y = 1$

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Solution

REF.Image.
Area = $\int_{- 2}^{1} (2 - x) d x - \int_{- 2}^{- 1} x^{2} d x - 2$
$I = 2 x - \frac{x^{2}}{2}]_{- 2}^{1} - \frac{1}{3} [x^{3}]_{- 2}^{- 1} - 2$
$= (2 - \frac{1}{2}) - (- 4 - \frac{4}{2}) \frac{- 1}{3} (- 1 + 8) - 2$
$= \frac{3}{2} + 6 - \frac{7}{3} - 2$
$= 4 + \frac{3}{2} + (\frac{- 7}{3})$
$= \frac{24 + 9 - 14}{6}$
$= \frac{19}{6}$ sq unit.

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