The area enclosed by the curves y 2 + 4 x = 4 and y - 2 x = 2 is [2023]

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Solution

Q.
The area enclosed by the curves $y^{2} + 4 x = 4 and y - 2 x = 2$ is [2023]

1 $\frac{23}{3}$

2 $\frac{22}{3}$

3 $9$

4 $\frac{25}{3}$

Ans.
(3)

Given curves are $y^{2} + 4 x = 4$ and $y - 2 x = 2$

The points of intersection of the given curves are (0, 2) and (- 3, - 4)

$Required area = \int_{- 4}^{2} [(\frac{4 - y^{2}}{4}) - (\frac{y - 2}{2})] d y$

$= \int_{- 4}^{2} \frac{- y^{2} - 2 y + 8}{4} d y = \frac{1}{4} {[- \frac{y^{3}}{3} - y^{2} + 8 y]}_{- 4}^{2}$

$= \frac{1}{4} [(- \frac{8}{3} - 4 + 16) - (\frac{64}{3} - 16 - 32)]$

$= \frac{1}{4} [(\frac{28}{3}) - (- \frac{80}{3})] = \frac{1}{4} (\frac{108}{3}) = \frac{108}{12} = 9 sq. units .$

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