The area of the region enclosed by the parabolas y = 4 x - x 2 and 3 y = ( x - 4 ) 2 is equal to [2024]

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Solution

Q.
The area of the region enclosed by the parabolas $y = 4 x - x^{2}$ and $3 y = {(x - 4)}^{2}$ is equal to [2024]

1 6

2 4

3 $\frac{14}{3}$

4 $\frac{32}{9}$

Ans.
(1)

Given, $y = 4 x - x^{2}$ ...(i)

$3 y = {(x - 4)}^{2}$ ...(ii)

From (i),

$3 y = 3 (4 x - x^{2}) = 12 x - 3 x^{2}$

From (ii),

${(x - 4)}^{2} = 12 x - 3 x^{2}$

$\Rightarrow x^{2} - 8 x + 16 = 12 x - 3 x^{2} \Rightarrow x^{2} - 5 x + 4 = 0$

$\Rightarrow (x - 1) (x - 4) = 0 \Rightarrow x = 1, 4$

Points of intersection are $(1, 3)$ and $(4, 0)$

Required area = $\int_{1}^{4} [(4 x - x^{2}) - \frac{{(x - 4)}^{2}}{3}] d x$

$= {[2 x^{2} - \frac{x^{3}}{3} - \frac{1}{3} \frac{{(x - 4)}^{3}}{3}]}_{1}^{4}$

$= [(32 - \frac{64}{3} - \frac{1}{3} (0)) - (2 - \frac{1}{3} - \frac{1}{3} (\frac{- 27}{3}))]$

$= 30 - \frac{64}{3} + \frac{1}{3} - 3 = 27 - 21 = 6 sq. units$

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