Key Concepts and Formulas

• Area between curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $\int_a^b (f(x) - g(x)) \, dx$.
• Intersection of curves: To find the points of intersection of two curves, we set their equations equal to each other and solve for the variable.
• Area bisection: If a line $y = \alpha$ divides the area between two curves into two equal parts, then the area above the line is equal to the area below the line, and both are equal to half of the total area.

Step-by-Step Solution

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

We set the two equations equal to each other: $x^2 = 2x$ $x^2 - 2x = 0$ $x(x - 2) = 0$ So, $x = 0$ or $x = 2$.

When $x = 0$, $y = 0^2 = 0$. When $x = 2$, $y = 2^2 = 4$. The points of intersection are $(0, 0)$ and $(2, 4)$.

Step 2: Calculate the total area of the region R.

The region R is bounded by $y = 2x$ (above) and $y = x^2$ (below) from $x = 0$ to $x = 2$. Therefore, the area of R is: $A = \int_0^2 (2x - x^2) \, dx$ $A = \left[x^2 - \frac{x^3}{3}\right]_0^2$ $A = \left(2^2 - \frac{2^3}{3}\right) - \left(0^2 - \frac{0^3}{3}\right)$ $A = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3}$

Step 3: Determine the area of the region R above the line $y = \alpha$.

Since the line $y = \alpha$ divides the region R into two equal parts, the area above the line $y = \alpha$ must be half of the total area, i.e., $\frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$.

We need to find the x-coordinate of the intersection of $y = x^2$ and $y = \alpha$, which is $x = \sqrt{\alpha}$. Also, we need to find the x-coordinate of the intersection of $y = 2x$ and $y = \alpha$, which is $x = \alpha/2$. We assume that $0 < \alpha < 4$.

The area above $y = \alpha$ is given by the integral: $\int_{\alpha/2}^{\sqrt{\alpha}} (\alpha - x^2) dx + \int_0^{\alpha/2} (2x - \alpha) dx = \frac{2}{3}$

Step 4: Calculate the area below the line y = $\alpha$.

The area below the line y = $\alpha$ is $\int_0^{\alpha/2} (2x - x^2)dx + \int_{\alpha/2}^{\sqrt{\alpha}} (\alpha - x^2)dx = \alpha x - \frac{x^3}{3}$

The area below $y = \alpha$ is: $\int_0^{\alpha/2} (2x - x^2)dx + \int_{\alpha/2}^{\sqrt{\alpha}} (\alpha - x^2)dx$

Since the line $y=\alpha$ divides the area into two equal parts, the area below the line is $4/6 = 2/3$. Then, the area under the line is given by $\int_0^{\sqrt{\alpha}} (x^2)dx + \int_{\sqrt{\alpha}}^{2} (2x)dx$

The other way to compute the area is: $\int_0^{\sqrt{\alpha}} (\alpha - x^2)dx + \int_{\sqrt{\alpha}}^{2} (2x - \alpha)dx = A/2$

Step 5: Calculate the area below the line $y = \alpha$ by integrating with respect to y.

Let's integrate with respect to $y$. The total area is $\int_0^4 (\sqrt{y} - \frac{y}{2}) dy = [\frac{2}{3}y^{3/2} - \frac{y^2}{4}]_0^4 = \frac{2}{3}(8) - \frac{16}{4} = \frac{16}{3} - 4 = \frac{4}{3}$. Since $y = \alpha$ divides the region in half, $\int_0^{\alpha} (\sqrt{y} - \frac{y}{2}) dy = \frac{2}{3}$. $[\frac{2}{3}y^{3/2} - \frac{y^2}{4}]_0^{\alpha} = \frac{2}{3}\alpha^{3/2} - \frac{\alpha^2}{4} = \frac{2}{3}$ Multiply by 12: $8\alpha^{3/2} - 3\alpha^2 = 8$ $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$

Common Mistakes & Tips

• Be careful when determining the upper and lower functions for integration. Drawing a diagram of the region helps.
• Remember to correctly evaluate the definite integrals using the fundamental theorem of calculus.
• When integrating with respect to y, you need to express the bounding curves as functions of y (i.e., x = f(y)).

Summary

We first found the intersection points of the curves $y=x^2$ and $y=2x$. Then, we calculated the total area enclosed by these curves. Since the line $y=\alpha$ bisects this area, we set up an integral representing half of the total area by integrating with respect to $y$. Solving this integral and simplifying the resulting equation yielded the relationship $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$. This matches option (C) in the original question. However, the correct answer provided is (A). Let's re-examine step 5.

From $8\alpha^{3/2} - 3\alpha^2 = 8$, we have $3\alpha^2 = 8\alpha^{3/2} - 8$. Thus $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$.

Let us check option (A): $3\alpha^2 - 8\alpha + 8 = 0$. This cannot be derived. Option (B): $\alpha^3 - 6\alpha^{3/2} - 16 = 0$. No. Option (C): $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$. This is the correct one. Option (D): $\alpha^3 - 6\alpha^2 + 16 = 0$. No.

The area below $y = \alpha$ is: $\int_0^{\alpha} (\sqrt{y} - y/2) dy = \frac{2}{3}y^{3/2} - \frac{y^2}{4} |_0^{\alpha} = \frac{2}{3}\alpha^{3/2} - \frac{\alpha^2}{4} = \frac{2}{3}$ $8\alpha^{3/2} - 3\alpha^2 = 8$, so $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$.

The final answer is $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$, which corresponds to option (C). The correct answer is (C). There must be an error in the original problem statement. The provided answer is (A). But the correct answer is (C). Let's go back and check each step.

We have the equation $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$.

If we make the substitution $x = \alpha^{1/2}$, we get $3x^4 - 8x^3 + 8 = 0$. If the correct answer is $3\alpha^2 - 8\alpha + 8 = 0$, we should have $3x^4 - 8x^2 + 8 = 0$. This is clearly wrong.

The error must be in the "Correct Answer" provided. The only possible answer is $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$. Therefore, the correct option is (C).

Final Answer

The final answer is $3\alpha^2 - 8\alpha^{3/2} + 8 = 0$, which corresponds to option (C).