Question:

Step 1: Determine the intersection point. The given curves are $y = \sqrt{x}$ and $2y - x + 3 = 0$. From the second equation, we can express $y$ in terms of $x$: $y = \frac{x-3}{2}$. To find the intersection points, set the expressions for $y$ equal: $$ \sqrt{x} = \frac{x-3}{2} $$ Square both sides: $$ 4x = (x-3)^2 $$ $$ 4x = x^2 - 6x + 9 $$ $$ x^2 - 10x + 9 = 0 $$ Factor the quadratic equation: $$ (x-1)(x-9) = 0 $$ This gives two possible x-values: $x=1$ and $x=9$. Substitute these values back into $y=\sqrt{x}$ to find the corresponding y-values: For $x=1$, $y=\sqrt{1}=1$. Check with the line equation: $2(1)-1+3 = 4 \ne 0$. So $(1,1)$ is not an intersection point. This is because squaring introduced an extraneous solution. Alternatively, substitute $y=\sqrt{x}$ into the line equation: $2\sqrt{x} - x + 3 = 0$. Let $t = \sqrt{x}$. Since we are in the first quadrant, $t \ge 0$. $$ 2t - t^2 + 3 = 0 $$ $$ t^2 - 2t - 3 = 0 $$ $$ (t-3)(t+1) = 0 $$ Since $t \ge 0$, we must have $t=3$. Therefore, $\sqrt{x}=3$, which implies $x=9$. For $x=9$, $y=\sqrt{9}=3$. Check with the line equation: $2(3)-9+3 = 6-9+3 = 0$. This is consistent. Thus, the curves intersect at the point $(9,3)$. This point defines the upper limit for the region along the x-axis. Step 2: Calculate the area. The area is bounded by $y=\sqrt{x}$, the x-axis ($y=0$), and the vertical line $x=9$ (determined by the intersection point in the first quadrant). The area $A$ is given by the integral of $y=\sqrt{x}$ from $x=0$ to $x=9$: $$ A = \int_0^9 \sqrt{x} \, dx $$ $$ A = \int_0^9 x^{1/2} \, dx $$ $$ A = \left[\frac{x^{3/2}}{3/2}\right]_0^9 $$ $$ A = \left[\frac{2}{3}x^{3/2}\right]_0^9 $$ Evaluate the definite integral: $$ A = \frac{2}{3}(9^{3/2}) - \frac{2}{3}(0^{3/2}) $$ $$ A = \frac{2}{3}(27) - 0 $$ $$ A = 18 $$ The area bounded by the curves is 18 square units.