The area of the region enclosed by the curves and is : [2025]

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Solution

Q.
The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is : [2025]

1 5

2 4/3

3 8/3

4 8

Ans.
(3)

The given curves are $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$

Points of intersection are (2, 0) and (0, 4).

$∴$ Required Area $= \int_{0}^{2} (\sqrt{16 - 8 x} - (x^{2} - 4 x + 4)) d x$

$= 2 \sqrt{2} \int_{0}^{2} \sqrt{2 - x} d x - \int_{0}^{2} {(x - 2)}^{2} d x$

$= 2 \sqrt{2} {[\frac{2 {(2 - x)}^{3 / 2}}{- 3}]}_{0}^{2} - {[\frac{{(x - 2)}^{3}}{3}]}_{0}^{2}$

$= \frac{16}{3} - \frac{8}{3} = \frac{8}{3}$ sq. units.

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