Question:

Step 1: Identify the curves and find their intersection points in the first quadrant. The given curves are: 1. Parabola: $y^2 = 8x$. In the first quadrant, this is $y = \sqrt{8x} = 2\sqrt{2x}$. 2. Line: $y = x$. 3. Vertical Line: $x = 2$. First, let's find the intersection points of these curves in the first quadrant: - Intersection of $y^2 = 8x$ and $y=x$: Substitute $y=x$ into $y^2=8x$: $x^2 = 8x$. $x^2 - 8x = 0 \Rightarrow x(x-8) = 0$. So, $x=0$ (giving $y=0$) and $x=8$ (giving $y=8$). The intersection points are $(0,0)$ and $(8,8)$. - Intersection of $y=x$ and $x=2$: Substitute $x=2$ into $y=x$: $y=2$. The intersection point is $(2,2)$. - Intersection of $y^2 = 8x$ and $x=2$: Substitute $x=2$ into $y^2=8x$: $y^2 = 8(2) = 16$. So, $y = \pm 4$. Since we are in the first quadrant, $y=4$. The intersection point is $(2,4)$. Step 2: Visualize the regions bounded by the curves. In the first quadrant, we observe the behavior of $y=\sqrt{8x}$ and $y=x$. At $x=1$, $y=\sqrt{8} \approx 2.8$ for the parabola and $y=1$ for the line. At $x=2$, $y=4$ for the parabola and $y=2$ for the line. At $x=8$, $y=8$ for both. This shows that for $x \in (0,8)$, the curve $y=\sqrt{8x}$ is above the line $y=x$. The total region bounded by $y=\sqrt{8x}$ and $y=x$ is from $x=0$ to $x=8$. The line $x=2$ divides this total region into two smaller regions: - Region 1 ($\alpha_1$): Bounded by $y=\sqrt{8x}$, $y=x$, and $x=2$, from $x=0$ to $x=2$. - Region 2 ($\alpha_2$): Bounded by $y=\sqrt{8x}$, $y=x$, and $x=2$, from $x=2$ to $x=8$. The problem asks for $\alpha$ as the area of the *larger* of these two regions. Step 3: Calculate the area of the first region ($\alpha_1$). The area of Region 1 is given by the integral of the difference between the upper curve ($y=\sqrt{8x}$) and the lower curve ($y=x$) from $x=0$ to $x=2$: $$ \alpha_1 = \int_0^2 (\sqrt{8x} - x) \, dx $$ We can rewrite $\sqrt{8x}$ as $2\sqrt{2}x^{1/2}$. $$ \alpha_1 = \int_0^2 (2\sqrt{2}x^{1/2} - x) \, dx $$ Now, we integrate: $$ \alpha_1 = \left[ 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_0^2 $$ $$ \alpha_1 = \left[ \frac{4\sqrt{2}}{3} x^{3/2} - \frac{x^2}{2} \right]_0^2 $$ Evaluate at the limits: $$ \alpha_1 = \left( \frac{4\sqrt{2}}{3} (2)^{3/2} - \frac{2^2}{2} \right) - \left( \frac{4\sqrt{2}}{3} (0)^{3/2} - \frac{0^2}{2} \right) $$ $$ \alpha_1 = \left( \frac{4\sqrt{2}}{3} (2\sqrt{2}) - \frac{4}{2} \right) - (0) $$ $$ \alpha_1 = \left( \frac{4 \cdot 2 \cdot 2}{3} - 2 \right) = \left( \frac{16}{3} - 2 \right) $$ $$ \alpha_1 = \frac{16 - 6}{3} = \frac{10}{3} $$ Step 4: Calculate the area of the second region ($\alpha_2$). The area of Region 2 is given by the integral of the difference between the upper curve ($y=\sqrt{8x}$) and the lower curve ($y=x$) from $x=2$ to $x=8$: $$ \alpha_2 = \int_2^8 (\sqrt{8x} - x) \, dx $$ Using the same antiderivative as in Step 3: $$ \alpha_2 = \left[ \frac{4\sqrt{2}}{3} x^{3/2} - \frac{x^2}{2} \right]_2^8 $$ Evaluate at the limits: $$ \alpha_2 = \left( \frac{4\sqrt{2}}{3} (8)^{3/2} - \frac{8^2}{2} \right) - \left( \frac{4\sqrt{2}}{3} (2)^{3/2} - \frac{2^2}{2} \right) $$ $$ \alpha_2 = \left( \frac{4\sqrt{2}}{3} (16\sqrt{2}) - \frac{64}{2} \right) - \left( \frac{4\sqrt{2}}{3} (2\sqrt{2}) - \frac{4}{2} \right) $$ $$ \alpha_2 = \left( \frac{128}{3} - 32 \right) - \left( \frac{16}{3} - 2 \right) $$ $$ \alpha_2 = \left( \frac{128 - 96}{3} \right) - \left( \frac{16 - 6}{3} \right) $$ $$ \alpha_2 = \frac{32}{3} - \frac{10}{3} = \frac{22}{3} $$ Step 5: Determine the larger region and calculate $3\alpha$. We have calculated the areas of the two regions: $\alpha_1 = \frac{10}{3}$ $\alpha_2 = \frac{22}{3}$ Comparing these values, $\frac{22}{3} > \frac{10}{3}$. Therefore, the area of the larger region, $\alpha$, is $\frac{22}{3}$. The problem asks for the value of $3\alpha$: $$ 3\alpha = 3 \times \frac{22}{3} = 22 $$ The final answer is $\boxed{22}$.