We are tasked with finding the area enclosed by the curves $y = x^2$, $y = |x|$, and the x-axis. Let's first analyze the region and then sketch the solution.

Step 1: Understanding the curves

1. Curve 1: $y = x^2$ — This is a parabola that opens upwards and has its vertex at the origin (0, 0).
2. Curve 2: $y = |x|$ — This is a V-shaped curve that consists of two linear segments: one for $y = x$ when $x \geq 0$, and the other for $y = -x$ when $x \leq 0$. The vertex is also at the origin.
3. The x-axis acts as a boundary, marking where the curves intersect with $y = 0$.

Step 2: Points of Intersection

We now need to find the points where $y = x^2$ intersects with $y = |x|$. To do this, solve the equations:

• For $x \geq 0$:
  $x^2 = x \quad \Rightarrow \quad x(x - 1) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 1$
• For $x \leq 0$:
  $x^2 = -x \quad \Rightarrow \quad x(x + 1) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = -1$

Thus, the points of intersection are at $x = -1$, $x = 0$, and $x = 1$.

Step 3: Finding the enclosed area

The enclosed region is symmetric about the y-axis, so we can calculate the area for $x \in [0, 1]$, then double it.

The area between $y = |x|$ and $y = x^2$ from $x = 0$ to $x = 1$ is given by the integral:

$\text{Area} = 2 \int_0^1 \left( x - x^2 \right) dx$

Now, let's compute the integral:

$\int_0^1 (x - x^2) dx = \int_0^1 x dx - \int_0^1 x^2 dx$

Evaluating these integrals:

$\int_0^1 x dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$ $\int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$

Thus, the total area for $x \geq 0$ is:

$\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$

Since the region is symmetric, the total enclosed area is:

$\text{Total Area} = 2 \times \frac{1}{6} = \frac{1}{3}$

Step 4: Sketching the Region

The shaded region is the area between the curves $y = x^2$ and $y = |x|$ from $x = -1$ to $x = 1$. You can visualize the V-shape and the parabolic curve forming the boundary of the enclosed area.

I can generate a visual sketch for you, showing the shaded region. Would you like to see it?

Let me know if you want details or have any questions.
Here are five related questions for you:

1. What would the area be if we used the function $y = x^3$ instead of $y = x^2$?
2. How does the symmetry affect the calculation of the area in this problem?
3. Can we use polar coordinates to calculate this area, and would it simplify the process?
4. How would the solution change if the region was bounded by $y = |x|$ and a different quadratic function?
5. What happens to the enclosed area if we extend the range beyond $x = 1$?

Tip: Symmetry often allows you to calculate only a portion of the area and then multiply by a factor to simplify the process.