The area (in square units) bounded by the curves y= x , 2y - x + 3...

First draw the given curves and represent the required area.

Eliminate $x$ to find point of intersection of $y = \sqrt{x}$ and $2y - x + 3 = 0$

$⇒2y-y^{2} + 3 = 0$

$⇒y^{2} - 2y - 3 = 0$

$⇒ \left(y - 3\right)\left(y + 1\right) = 0$

$⇒ y = 3, - 1$
$∴$Required area $A = \mathop{∫}_{0}^{9}\sqrt{x}dx - \dfrac{1}{2} \times 6 \times 3$

$A = {\dfrac{x\dfrac{3}{2}}{32}}_{0}^{9} - 9$

$∴ A = - 9$

Hence, area is~$9sq . units$