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

We have to find the area enclosed in the first quadrant by the curves $$y=\sqrt{x},\;2y-x+3=0$$ and the X-axis $$y=0$$.

First, we write the straight-line equation in slope-intercept form. Starting from $$2y-x+3=0$$ we add $$x-3$$ on both sides:

$$2y = x-3 \; \Longrightarrow \; y=\dfrac{x-3}{2}.$$

Because we work in the first quadrant we must have $$x\ge 0$$ and $$y\ge 0.$$ For the line $$y=\dfrac{x-3}{2}$$ the ordinate is non-negative only when $$x\ge 3.$$ Hence the segment of this line that interests us starts from its X-intercept. Putting $$y=0$$ gives

$$0 = \dfrac{x-3}{2} \; \Longrightarrow \; x=3.$$

Next we locate the point where the line meets the curve $$y=\sqrt{x}.$$ Equating ordinates,

$$\sqrt{x} \;=\; \dfrac{x-3}{2}.$$

To solve it cleanly we set $$t=\sqrt{x} \;(t\ge 0).$$ Then $$x=t^{2}$$ and the equation becomes

$$2t = t^{2}-3.$$

Moving every term to one side,

$$t^{2}-2t-3 = 0.$$

Using the quadratic-formula $$t=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ with $$a=1,\;b=-2,\;c=-3,$$ we get

$$t=\dfrac{2\pm\sqrt{(-2)^{2}-4(1)(-3)}}{2}=\dfrac{2\pm\sqrt{4+12}}{2}=\dfrac{2\pm4}{2}.$$

This yields $$t=3$$ or $$t=-1.$$ Since $$t=\sqrt{x}\ge 0,$$ we accept only $$t=3.$$ Therefore

$$\sqrt{x}=3 \;\Longrightarrow\; x=9,\qquad y=3.$$

So the three relevant corner points of the bounded region are $$A(0,0),\; B(3,0),\; C(9,3).$$ On $$0\le x\le 3$$ the curve $$y=\sqrt{x}$$ lies above the X-axis, while the line is below the X-axis (so it does not bound the region there). On $$3\le x\le 9$$ both curves are above the axis, and we must take the difference $$\sqrt{x}-\dfrac{x-3}{2}.$$

We therefore split the required area into two parts.

Part 1 : Area under $$y=\sqrt{x}$$ from $$x=0$$ to $$x=3.$$ The integral formula is $$\displaystyle\int_{a}^{b}\sqrt{x}\,dx.$$ We know $$\displaystyle\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2}.$$ Hence

$$A_{1}= \left[\,\frac{2}{3}x^{3/2}\,\right]_{0}^{3} = \frac{2}{3}\bigl(3^{3/2}-0^{3/2}\bigr)=\frac{2}{3}(3\sqrt{3})=2\sqrt{3}.$$

Part 2 : Area between the curves $$y=\sqrt{x}$$ (upper) and $$y=\dfrac{x-3}{2}$$ (lower) from $$x=3$$ to $$x=9.$$ The formula is $$\displaystyle\int_{3}^{9}\Bigl[\sqrt{x}-\frac{x-3}{2}\Bigr]dx.$$ We treat the two integrals separately.

For the first term, using the antiderivative found earlier,

$$\int_{3}^{9}\sqrt{x}\,dx=\Bigl[\frac{2}{3}x^{3/2}\Bigr]_{3}^{9} =\frac{2}{3}\bigl(9^{3/2}-3^{3/2}\bigr) =\frac{2}{3}\bigl(27-3\sqrt{3}\bigr) =18-2\sqrt{3}.$$

For the second term we integrate $$\dfrac{x-3}{2}.$$ We recall the power rule $$\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C$$ for $$n\neq -1.$$ Thus

$$\int\frac{x-3}{2}\,dx =\frac12\int(x-3)\,dx =\frac12\Bigl[\frac{x^{2}}{2}-3x\Bigr] =\frac14x^{2}-\frac32x.$$

Evaluating from $$x=3$$ to $$x=9$$:

$$\Bigl(\frac14x^{2}-\frac32x\Bigr)\Big|_{3}^{9} =\Bigl(\frac14\!\cdot\!81-\frac32\!\cdot\!9\Bigr) -\Bigl(\frac14\!\cdot\!9-\frac32\!\cdot\!3\Bigr) =\Bigl(\frac{81}{4}-\frac{27}{2}\Bigr) -\Bigl(\frac{9}{4}-\frac{9}{2}\Bigr).$$

Changing all fractions to quarters,

$$\frac{81}{4}-\frac{54}{4}=\frac{27}{4},\qquad \frac{9}{4}-\frac{18}{4}=-\frac{9}{4}.$$

Subtracting gives $$\frac{27}{4}-\bigl(-\frac{9}{4}\bigr)=\frac{36}{4}=9.$$ Therefore

$$\int_{3}^{9}\frac{x-3}{2}\,dx = 9.$$

Combining the two contributions for Part 2,

$$A_{2} = \bigl(18-2\sqrt{3}\bigr) - 9 = 9-2\sqrt{3}.$$

Total area :

$$A = A_{1}+A_{2}= 2\sqrt{3} + \bigl(9-2\sqrt{3}\bigr)=9.$$

The enclosed area in the first quadrant is therefore $$9$$ square units.

Hence, the correct answer is Option C.