The area bounded by the curves y = sqrt{x},2y | vedclass

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(A) Given curves are $y = \sqrt{x}$ (or $x = y^2$) and $2y + 3 = x$ (or $y = \frac{x-3}{2}$).To find the intersection points,set $y^2 = 2y + 3$,which

Vedclass · Accepted Answer

$9$

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(A) Given curves are $y = \sqrt{x}$ (or $x = y^2$) and $2y + 3 = x$ (or $y = \frac{x-3}{2}$).To find the intersection points,set $y^2 = 2y + 3$,which gives $y^2 - 2y - 3 = 0$.$(y - 3)(y + 1) = 0$,so $y = 3$ or $y = -1$.Since we are in the $1^{st}$ quadrant,$y = 3$,which implies $x = 9$.The line $2y + 3 = x$ intersects the $x$-axis at $y = 0$,so $x = 3$.The area is bounded by $y = \sqrt{x}$ from $x = 0$ to $x = 3$,and by the difference of the curves from $x = 3$ to $x = 9$.Area $= \int_0^3 \sqrt{x} \, dx + \int_3^9 \left( \sqrt{x} - \frac{x-3}{2} ight) \, dx$$= \left[ \frac{2}{3} x^{3/2} ight]_0^3 + \left[ \frac{2}{3} x^{3/2} - \frac{1}{2} \left( \frac{x^2}{2} - 3x ight) ight]_3^9$$= \frac{2}{3}(3\sqrt{3}) + \left( \left[ \frac{2}{3}(27) - \frac{1}{2} \left( \frac{81}{2} - 27 ight) ight] - \left[ \frac{2}{3}(3\sqrt{3}) - \frac{1}{2} \left( \frac{9}{2} - 9 ight) ight] ight)$$= 2\sqrt{3} + 18 - \frac{1}{2} \left( \frac{27}{2} ight) - 2\sqrt{3} + \frac{1}{2} \left( -\frac{9}{2} ight)$$= 18 - \frac{27}{4} - \frac{9}{4} = 18 - \frac{36}{4} = 18 - 9 = 9 	ext{ sq. units.}$

The area bounded by the curves $y = \sqrt{x}$,$2y + 3 = x$ and the $x$-axis in the $1^{st}$ quadrant is

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