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 $\int_a^b [f(x) - g(x)] dx$.
• Power Rule for Integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \ne -1$.
• Fundamental Theorem of Calculus: $\int_a^b F'(x) dx = F(b) - F(a)$.

Step-by-Step Solution

Step 1: Analyze the curves and the region

We are given the curves $y^2 = x$ and $y = |x|$. We need to find the area enclosed between them.

• $y^2 = x$ is a parabola opening to the right, with its vertex at the origin. Since $y^2 \ge 0$, we have $x \ge 0$.
• $y = |x|$ is the absolute value function, which is $y = x$ for $x \ge 0$ and $y = -x$ for $x < 0$.

Since $y^2 = x$ is defined only for $x \ge 0$, we only consider the region where $x \ge 0$. In this region, $|x| = x$, so we are looking for the area between the parabola $y^2 = x$ and the line $y = x$.

Step 2: Find the points of intersection

To find the intersection points, we set $y^2 = x$ and $y = x$ equal to each other. Substituting $y=x$ into $y^2 = x$, we get: $x^2 = x$ $x^2 - x = 0$ $x(x - 1) = 0$ So, $x = 0$ or $x = 1$.

When $x = 0$, $y = 0$. When $x = 1$, $y = 1$. Therefore, the points of intersection are $(0, 0)$ and $(1, 1)$. This means we will integrate from $x = 0$ to $x = 1$.

Step 3: Determine the upper and lower curves

We need to determine which curve is above the other in the interval $[0, 1]$.

• The parabola is given by $y^2 = x$, which means $y = \pm \sqrt{x}$. Since we are in the first quadrant, we only consider $y = \sqrt{x}$.
• The line is given by $y = x$.

For $x \in (0, 1)$, $\sqrt{x} > x$. For example, if $x = 0.25$, then $\sqrt{x} = 0.5$ and $x = 0.25$.

Therefore, $y = \sqrt{x}$ is the upper curve and $y = x$ is the lower curve in the interval $[0, 1]$.

Step 4: Set up the integral

The area enclosed between the curves is given by: $A = \int_0^1 (\sqrt{x} - x) dx$

Step 5: Evaluate the integral

$A = \int_0^1 (x^{1/2} - x) dx$ $A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_0^1$ $A = \left[ \frac{2}{3}x^{3/2} - \frac{1}{2}x^2 \right]_0^1$ $A = \left( \frac{2}{3}(1)^{3/2} - \frac{1}{2}(1)^2 \right) - \left( \frac{2}{3}(0)^{3/2} - \frac{1}{2}(0)^2 \right)$ $A = \left( \frac{2}{3} - \frac{1}{2} \right) - (0 - 0)$ $A = \frac{4}{6} - \frac{3}{6}$ $A = \frac{1}{6}$

Common Mistakes & Tips

• Incorrectly identifying the upper and lower curves: Always test a point in the interval to make sure you have the correct order.
• Forgetting the absolute value when taking the square root: In this case, we only needed the positive root because we're in the first quadrant.
• Not finding the points of intersection correctly: This will lead to incorrect limits of integration.

Summary

We found the area enclosed between the curves $y^2 = x$ and $y = |x|$ by first analyzing the curves, finding their intersection points, determining which curve was above the other, setting up the integral, and then evaluating it. The area is $\frac{1}{6}$.

The final answer is \boxed{1/6}, which corresponds to option (A).