The area (in square units) bounded by the cur | vedclass

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(A) Given curves are $y = \sqrt{x} \implies x = y^2$ $(1)$ and $2y - x + 3 = 0 \implies x = 2y + 3$ $(2)$.To find the intersection points,substitute $

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$9$

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(A) Given curves are $y = \sqrt{x} \implies x = y^2$ $(1)$ and $2y - x + 3 = 0 \implies x = 2y + 3$ $(2)$.To find the intersection points,substitute $x = y^2$ into $(2)$:$2y - y^2 + 3 = 0 \implies y^2 - 2y - 3 = 0 \implies (y - 3)(y + 1) = 0$.Since the region is in the first quadrant,we take $y = 3$. At $y = 3$,$x = 9$.The region is bounded by the $X$-axis $(y=0)$,the curve $x = y^2$ from $y=0$ to $y=3$,and the line $x = 2y+3$ from $y=0$ to $y=3$.The area is given by $\int_{0}^{3} (x_{line} - x_{curve}) \, dy = \int_{0}^{3} ((2y + 3) - y^2) \, dy$.$= [y^2 + 3y - \frac{y^3}{3}]_{0}^{3} = (3^2 + 3(3) - \frac{3^3}{3}) - (0) = 9 + 9 - 9 = 9$ square units.

The area (in square units) bounded by the curves $y = \sqrt{x}$,the line $2y - x + 3 = 0$,and the $X$-axis,lying in the first quadrant,is:

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