Let the area of the region enclosed by the curves y = 3 x , 2 y = 27 - 3 x and y = 3 x - x x be A . Then 10A is equal to [2024]

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Solution

Q.
Let the area of the region enclosed by the curves $y = 3 x, 2 y = 27 - 3 x and y = 3 x - x \sqrt{x}$ be $A .$ Then 10A is equal to [2024]

1 162

2 184

3 154

4 172

Ans.
(1)

Required area,

$A = \int_{0}^{3} (3 x - (3 x - x \sqrt{x})) d x + \int_{3}^{9} \frac{27 - 3 x}{2} - (3 x - x \sqrt{x}) d x$

$= \int_{0}^{3} x^{3 / 2} d x + \int_{3}^{9} \frac{27}{2} - \frac{9 x}{2} + x^{3 / 2} d x$

$= \frac{2}{5} {[x^{5 / 2}]}_{0}^{3} + \frac{27}{2} {[x]}_{3}^{9} - \frac{9}{4} {[x^{2}]}_{3}^{9} + \frac{2}{5} {[x^{5 / 2}]}_{3}^{9}$

$= \frac{2}{5} 3^{5 / 2} + \frac{27}{2} \times 6 - \frac{9}{4} \times 72 + \frac{2}{5} [9^{5 / 2} - 3^{5 / 2}]$

$= 81 - 162 + \frac{2}{5} \cdot 3^{5} = \frac{81}{5}$

So, $10 A = 10 \times \frac{81}{5} = 162$

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