Key Concepts and Formulas

• Area under a curve: The area of the region bounded by $y = f(x)$, $y = g(x)$, $x = a$, and $x = b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $\int_a^b (f(x) - g(x)) \, dx$.
• Power rule of integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n \ne -1$.
• Definite integral evaluation: If $F(x)$ is the antiderivative of $f(x)$, then $\int_a^b f(x) \, dx = F(b) - F(a)$.

Step-by-Step Solution

Step 1: Find the area A($\alpha$)

The region S($\alpha$) is defined by $y^2 \le x$ and $0 \le x \le \alpha$. This means $-\sqrt{x} \le y \le \sqrt{x}$. Thus, the area A($\alpha$) is the integral of the difference between the upper and lower bounds of $y$ with respect to $x$ from 0 to $\alpha$. $A(\alpha) = \int_0^{\alpha} (\sqrt{x} - (-\sqrt{x})) \, dx = \int_0^{\alpha} 2\sqrt{x} \, dx = 2\int_0^{\alpha} x^{1/2} \, dx$ Applying the power rule of integration: $A(\alpha) = 2 \left[ \frac{x^{3/2}}{3/2} \right]_0^{\alpha} = 2 \cdot \frac{2}{3} \left[ x^{3/2} \right]_0^{\alpha} = \frac{4}{3} (\alpha^{3/2} - 0^{3/2}) = \frac{4}{3} \alpha^{3/2}$

Step 2: Find A($\lambda$) and A(4)

Using the formula derived in Step 1, we can find A($\lambda$) and A(4): $A(\lambda) = \frac{4}{3} \lambda^{3/2}$ $A(4) = \frac{4}{3} (4)^{3/2} = \frac{4}{3} (2^2)^{3/2} = \frac{4}{3} (2^3) = \frac{4}{3} (8) = \frac{32}{3}$

Step 3: Use the given ratio to find $\lambda$

We are given that $A(\lambda) : A(4) = 2 : 5$, which means $\frac{A(\lambda)}{A(4)} = \frac{2}{5}$. Substituting the expressions for $A(\lambda)$ and $A(4)$: $\frac{\frac{4}{3} \lambda^{3/2}}{\frac{32}{3}} = \frac{2}{5}$ $\frac{4 \lambda^{3/2}}{32} = \frac{2}{5}$ $\frac{\lambda^{3/2}}{8} = \frac{2}{5}$ $\lambda^{3/2} = \frac{2}{5} \cdot 8 = \frac{16}{5}$

Step 4: Solve for $\lambda$

To solve for $\lambda$, raise both sides to the power of $\frac{2}{3}$: $\lambda = \left(\frac{16}{5}\right)^{2/3} = \left(\frac{2^4}{5}\right)^{2/3} = \frac{(2^4)^{2/3}}{5^{2/3}} = \frac{2^{8/3}}{5^{2/3}} = \frac{2^{2 + 2/3}}{5^{2/3}} = \frac{2^2 \cdot 2^{2/3}}{5^{2/3}} = 4\left(\frac{2^2}{5^2}\right)^{1/3} = 4\left(\frac{4}{25}\right)^{1/3}$

Common Mistakes & Tips

• Remember to consider both positive and negative roots when dealing with $y^2 \le x$. In this case, it translates to $-\sqrt{x} \le y \le \sqrt{x}$.
• Be careful with fractional exponents. Use exponent rules correctly.
• Double-check the arithmetic when simplifying the expressions.

Summary

We first found the area $A(\alpha)$ under the curve $y^2 \le x$ from $x=0$ to $x=\alpha$. Then, we calculated $A(\lambda)$ and $A(4)$. Finally, we used the given ratio $A(\lambda) : A(4) = 2 : 5$ to solve for $\lambda$, obtaining $\lambda = 4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$. Since the correct answer provided is $2{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$, there is an error in the correct answer. Let's recheck the algebra.

$\lambda^{3/2} = \frac{16}{5}$ $\lambda = \left(\frac{16}{5}\right)^{2/3} = \left(\frac{2^4}{5}\right)^{2/3} = \frac{2^{8/3}}{5^{2/3}} = \frac{2^{2 + 2/3}}{5^{2/3}} = \frac{2^2 \cdot 2^{2/3}}{5^{2/3}} = 4\left(\frac{2^2}{5^2}\right)^{1/3} = 4\left(\frac{4}{25}\right)^{1/3}$

The correct answer should be $4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$. However, the given correct answer is $2{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$. There may be an error in the answer key. Since the problem asks us to provide the answer that we arrived at, we will stick with our calculation.

Final Answer

The final answer is \boxed{4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}}, which corresponds to option (C).