The area (in sq. units) bounded by the curves | vedclass

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(C) The given curves are $y = \sqrt{x}$ and $2y - x + 3 = 0$. First,find the intersection point of the curves: $2(\sqrt{x}) - x + 3 = 0$ Let $\sqrt{x}

Vedclass · Accepted Answer

$9$

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(C) The given curves are $y = \sqrt{x}$ and $2y - x + 3 = 0$. First,find the intersection point of the curves: $2(\sqrt{x}) - x + 3 = 0$ Let $\sqrt{x} = t$,then $2t - t^2 + 3 = 0 \implies t^2 - 2t - 3 = 0 (t-3)(t+1) = 0$. Since $t = \sqrt{x} \geq 0$,we have $t = 3$,so $x = 9$ and $y = 3$. The line $2y - x + 3 = 0$ intersects the $X$-axis $(y=0)$ at $x = 3$. The area is given by the integral of the curve $y = \sqrt{x}$ from $x=0$ to $x=9$ minus the area of the triangle formed by the line $2y - x + 3 = 0$ with the $X$-axis from $x=3$ to $x=9$. Area $= \int_0^9 \sqrt{x} \, dx - \int_3^9 \frac{x-3}{2} \, dx$ $= \left[ \frac{2}{3} x^{3/2} \right]_0^9 - \frac{1}{2} \left[ \frac{x^2}{2} - 3x \right]_3^9$ $= \frac{2}{3} (27) - \frac{1}{2} [(\frac{81}{2} - 27) - (\frac{9}{2} - 9)]$ $= 18 - \frac{1}{2} [\frac{27}{2} - (-\frac{9}{2})] = 18 - \frac{1}{2} [\frac{36}{2}] = 18 - 9 = 9$ sq. units.

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

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