Key Concepts and Formulas

• Area Under a Curve: The area under a curve $y = f(x)$ from $x = a$ to $x = b$ is given by $\int_a^b |f(x)| dx$. If $f(x) \ge 0$ on $[a, b]$, this simplifies to $\int_a^b f(x) dx$.
• Composition of Functions: $(g \circ f)(x) = g(f(x))$.
• Quadratic Formula/Factoring: Used to find the roots of a quadratic equation $ax^2 + bx + c = 0$.

Step-by-Step Solution

Step 1: Find the roots of the quadratic equation

We are given the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$, and we need to find its roots $\alpha$ and $\beta$, where $\alpha < \beta$. These roots will be the limits of integration.

Why? We need the limits of integration to calculate the definite integral representing the area.

We can factor the quadratic equation: $18x^2 - 9\pi x + \pi^2 = 0$ $18x^2 - 6\pi x - 3\pi x + \pi^2 = 0$ $6x(3x - \pi) - \pi(3x - \pi) = 0$ $(6x - \pi)(3x - \pi) = 0$ This gives us two possible solutions for $x$: $6x - \pi = 0 \implies x = \frac{\pi}{6}$ $3x - \pi = 0 \implies x = \frac{\pi}{3}$ Since $\alpha < \beta$, we have $\alpha = \frac{\pi}{6}$ and $\beta = \frac{\pi}{3}$.

Step 2: Determine the composite function $(g \circ f)(x)$

We are given $g(x) = \cos(x^2)$ and $f(x) = \sqrt{x}$. We need to find $(g \circ f)(x) = g(f(x))$.

Why? The composite function defines the curve whose area we need to calculate.

$(g \circ f)(x) = g(f(x)) = g(\sqrt{x}) = \cos((\sqrt{x})^2) = \cos(x)$ So, $(g \circ f)(x) = \cos(x)$.

Step 3: Set up the definite integral

We want to find the area under the curve $y = \cos(x)$ from $x = \alpha = \frac{\pi}{6}$ to $x = \beta = \frac{\pi}{3}$. Since $\cos(x)$ is positive on the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$, we can directly integrate the function.

Why? We are setting up the integral to calculate the area bounded by the curve and the x-axis.

The area $A$ is given by: $A = \int_{\alpha}^{\beta} \cos(x) dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos(x) dx$

Step 4: Evaluate the definite integral

We need to evaluate the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos(x) dx$.

Why? This will give us the numerical value of the area.

The antiderivative of $\cos(x)$ is $\sin(x)$. So, $A = \left[ \sin(x) \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{6}\right)$ We know that $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Therefore, $A = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2} = \frac{1}{2}(\sqrt{3} - 1)$

Common Mistakes & Tips

• Incorrect Factoring: Double-check your factoring to ensure you obtain the correct roots. A small error can lead to a completely wrong answer.
• Incorrect Composition: Make sure you understand the order of composition $(g \circ f)(x) = g(f(x))$.
• Sign of Cosine: Ensure that $\cos(x)$ is positive on the interval of integration. If it's negative, you need to take the absolute value before integrating. In this case, it was positive, simplifying the process.

Summary

We found the area bounded by the curve $y = (g \circ f)(x)$ and the lines $x = \alpha$, $x = \beta$, and $y = 0$ by first finding the roots $\alpha$ and $\beta$ of the given quadratic equation. Then, we determined the composite function $(g \circ f)(x) = \cos(x)$. Finally, we integrated $\cos(x)$ from $\alpha = \frac{\pi}{6}$ to $\beta = \frac{\pi}{3}$ to find the area, which is $\frac{\sqrt{3} - 1}{2}$.

Final Answer

The final answer is $\boxed{\frac{1}{2}(\sqrt{3} - 1)}$, which corresponds to option (B).