Key Concepts and Formulas

• Area between curves: If $f(x) \ge g(x)$ on the interval $[a, b]$, the area between the curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is given by $\int_a^b [f(x) - g(x)] \, dx$.
• Solving for intersection points: To find where two curves intersect, set their equations equal to each other and solve for $x$ (or $y$).
• Integration of power functions: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

Step-by-Step Solution

Step 1: Find the intersection points of the curves $y^2 = 8x$ and $y = x$.

We need to find the points where the parabola and the line intersect to determine the limits of integration. Substituting $y=x$ into $y^2 = 8x$, we get $x^2 = 8x$, which simplifies to $x^2 - 8x = 0$. Factoring, we have $x(x-8) = 0$, so $x = 0$ or $x = 8$. Since we are only interested in the region in the first quadrant, we consider the intersection points. The intersection points are $(0,0)$ and $(8,8)$.

Step 2: Determine the relevant region and split the area calculation.

The region is bounded by $y^2=8x$, $y=x$, and $x=2$. We need to express $x$ in terms of $y$. From $y^2 = 8x$, we have $x = \frac{y^2}{8}$. Also, we have $x=y$ and $x=2$. Note that the line $x=2$ intersects the parabola $y^2=8x$ when $y^2 = 8(2) = 16$, so $y=4$ (since we are in the first quadrant). The line $x=2$ intersects $y=x$ at $(2,2)$.

We need to find the area of the region bounded by $x = \frac{y^2}{8}$, $x=y$, and $x=2$. The area can be calculated as the area between the line $x=y$ and the parabola $x=\frac{y^2}{8}$ from $y=0$ to $y=2$, plus the area between the line $x=2$ and the parabola $x=\frac{y^2}{8}$ from $y=2$ to $y=4$.

Step 3: Calculate the area $\alpha$ using integration with respect to $y$.

The area $\alpha$ is given by: $\alpha = \int_0^2 \left(y - \frac{y^2}{8}\right) dy + \int_2^4 \left(2 - \frac{y^2}{8}\right) dy$

Let's evaluate the first integral: $\int_0^2 \left(y - \frac{y^2}{8}\right) dy = \left[\frac{y^2}{2} - \frac{y^3}{24}\right]_0^2 = \left(\frac{2^2}{2} - \frac{2^3}{24}\right) - (0) = \frac{4}{2} - \frac{8}{24} = 2 - \frac{1}{3} = \frac{5}{3}$

Now, let's evaluate the second integral: $\int_2^4 \left(2 - \frac{y^2}{8}\right) dy = \left[2y - \frac{y^3}{24}\right]_2^4 = \left(2(4) - \frac{4^3}{24}\right) - \left(2(2) - \frac{2^3}{24}\right) = \left(8 - \frac{64}{24}\right) - \left(4 - \frac{8}{24}\right) = 8 - \frac{8}{3} - 4 + \frac{1}{3} = 4 - \frac{7}{3} = \frac{12-7}{3} = \frac{5}{3}$

Therefore, the total area is: $\alpha = \frac{5}{3} + \frac{5}{3} = \frac{10}{3}$

Step 4: Calculate $3\alpha$.

We are asked to find the value of $3\alpha$. Since $\alpha = \frac{10}{3}$, then $3\alpha = 3 \cdot \frac{10}{3} = 10$.

Common Mistakes & Tips

• Carefully determine the upper and lower limits of integration by finding the intersection points of the curves.
• Make sure you are integrating the difference between the larger function and the smaller function. If you get a negative area, switch the order of subtraction.
• When integrating with respect to $y$, remember to express your functions in the form $x = f(y)$.

Summary

We calculated the area of the region bounded by the parabola $y^2 = 8x$, the line $y=x$, and the line $x=2$ in the first quadrant. We expressed the functions in terms of $y$ and integrated with respect to $y$. The area was found to be $\alpha = \frac{10}{3}$. Finally, we calculated $3\alpha$, which is $3 \cdot \frac{10}{3} = 10$.

Final Answer

The final answer is \boxed{10}.