Interpret the curves

The given curves are:

$y=\sqrt{x}, \qquad 2y-x+3=0, \qquad y=0$

The line is:

$2y-x+3=0 \implies 2y=x-3 \implies y=\frac{x-3}{2}$

Find the intersection points relevant in the first quadrant

We need the region bounded by:

• the curve $y=\sqrt{x}$,
• the line $y=\dfrac{x-3}{2}$,
• the $x$-axis.

Since the region is in the first quadrant, we find where these meet.

(i) Intersection of $y=\sqrt{x}$ with $x$-axis

$\sqrt{x}=0 \implies x=0$ So point is $(0,0)$.

(ii) Intersection of line with $x$-axis

$\frac{x-3}{2}=0 \implies x=3$ So point is $(3,0)$.

(iii) Intersection of $y=\sqrt{x}$ and $y=\dfrac{x-3}{2}$

Solve: $\sqrt{x}=\frac{x-3}{2}$ Since $\sqrt{x}\ge 0$, we must have $x\ge 3$.

Multiply: $2\sqrt{x}=x-3$ Square both sides: $4x=(x-3)^2=x^2-6x+9$ $x^2-10x+9=0$ $x=1,9$

But $x\ge 3$, so only $x=9$ is valid.

Then: $y=\sqrt{9}=3$ So the point is $(9,3)$.

Understand the bounded region

The closed boundary in the first quadrant is formed by:

• along the curve $y=\sqrt{x}$ from $(0,0)$ to $(9,3)$,
• along the line from $(9,3)$ to $(3,0)$,
• along the $x$-axis from $(3,0)$ back to $(0,0)$.

So the area is split naturally into two parts:

• from $x=0$ to $x=3$, area between $y=\sqrt{x}$ and $y=0$,
• from $x=3$ to $x=9$, area between $y=\sqrt{x}$ and $y=\dfrac{x-3}{2}$.

Set up the area integral

$A=\int_0^3 \sqrt{x}\,dx + \int_3^9 \left(\sqrt{x}-\frac{x-3}{2}\right)dx$

Compute the first integral

Compute the second integral

Add both parts

$A=2\sqrt{3}+(9-2\sqrt{3})=9$

Match with options

$\boxed{9}$

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.