The area of the region is equal to [2025]

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Solution

Q.
The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq | x + 2 |}$ is equal to [2025]

1 5

2 7

3 24/5

4 20/3

Ans.
(4)

We have, $x^{2} + 4 x + 2 \leq y \leq$ $| x + 2 |$

$\Rightarrow {(x + 2)}^{2} - 2 \leq y \leq$ $| x + 2 |$

$∴$ Required area $= \int_{- 4}^{- 2}$ $| (- x - {2) - (x + 2)}^{2} + 2 |$ $d x + \int_{- 2}^{0}$ $| (x + 2) - {(x + 2)}^{2} + 2 |$ $d x$

$= \int_{- 4}^{- 2}$ $| - x - {(x + 2)}^{2} |$ $d x + \int_{- 2}^{0}$ $| x - {(x + 2)}^{2} + 4 |$ $d x$

$= {[| \frac{- x^{2}}{2} - \frac{{(x + 2)}^{3}}{3} |]}_{- 4}^{- 2} + {[| \frac{x^{2}}{2} - \frac{{(x + 2)}^{3}}{3} + 4 x |]}_{- 2}^{0}$

$= \frac{10}{3} + \frac{10}{3} = \frac{20}{3}$ .

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