Let the area of the bounded region be A. Then 6A is equal to __________. [2025]

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Solution

Q.
Let the area of the bounded region ${(x, y) : 0 \leq 9 x \leq y^{2}, y \geq 3 x - 6}$ be A. Then 6A is equal to __________. [2025]

Ans.
(15)

We have, $0 \leq 9 x \leq y^{2}, y \geq 3 x - 6$

Required area = $A =$ $| \int_{0}^{- 3} \frac{y^{2}}{9} d y + \int_{- 3}^{- 6} (\frac{y + 6}{3}) d y |$

   $=$ $| \frac{1}{9} {[\frac{y^{3}}{3}]}_{0}^{- 3} + \frac{1}{3} {(\frac{y^{2}}{2} + 6 y)}_{- 3}^{- 6} |$

   $=$ $| \frac{1}{9} [- 9 - 0] + \frac{1}{3} [18 - 36 - \frac{9}{2} + 18] |$

   $=$ $| - 1 + \frac{1}{3} [\frac{- 9}{2}] |$

   $=$ $| - 1 - \frac{3}{2} |$ $=$ $| \frac{- 5}{2} |$ $= \frac{5}{2} sq . units$

$∴ 6 A = 6 \times \frac{5}{2} = 15$

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